Calculus / English

How to find the TANGENT PLANE | Linear approximation of multi-variable functions

Dr. Trefor Bazett · YouTube · 9:23

YouTube · Dr. Trefor BazettOriginal video

Selected moment · 0:00 · Chapter markers are time references, not playback progress.

IN THIS VIDEO

01Tangent plane: intuitive meaning
02Point of tangency on $z=f(x,y)$
03Loose goal for finding a tangent plane

FROM THE LEARNING NOTES

z_0=f(x_0,y_0)

Point of tangency on $z=f(x,y)$

SELECTED MOMENT

Title: Tangent Planes

SUGGESTED PREPARATION

Single-variable tangent lines
Graphs of functions of two variables
Vectors in three-dimensional space
Dot product and orthogonality
Equation of a plane in space
Dot product and orthogonality of vectors
Equation of a plane in three-dimensional space
Basic idea of a tangent line in single-variable calculus
Notation for partial derivatives
Single-variable linear approximation
Derivative as slope of tangent line
Basic notation for partial derivatives
Graphs of functions of two variables in 3D
Partial Derivatives
Multivariable Functions
Equation of a Plane

READ & KEEP

The explanation, unpacked.

Reviewed learning material · Video analysis · English

Read the full overview

This 180-second lecture segment introduces tangent planes to graphs of functions of two variables as the multivariable analogue of tangent lines. Using a bell-shaped surface and a marked point $(x_0,y_0,z_0)$ , the lecturer explains that the tangent plane passes through the point, locally "kisses" the graph, and approximates it nearby. He then states the goal informally: find a plane that meets the graph at $z_0=f(x_0,y_0)$ and is "close" to $z=f(x,y)$ nearby, while postponing the precise meaning of "close." The remainder of the clip reviews the general equation of a plane, first as $\vec n\cdot \overrightarrow{P_0P}=0$ and then as $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ , emphasizing that the hard part in the tangent-plane problem is determining the normal vector components $a,b,c$ . This 180-second lecture excerpt derives the tangent-plane formula for a surface $z=f(x,y)$ at a point $(x_0,y_0,z_0)$ . It starts from the plane equation $\vec n\cdot\overline{P_0P}=0$ and its scalar form $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ , then normalizes $c=-1$ to rewrite the plane as the linear function $L(x,y)=a(x-x_0)+b(y-y_0)+z_0$ . By substituting $y=y_0$ , the speaker reduces the problem to a one-variable tangent-line situation and concludes that $a=f_x(x_0,y_0)$ . The clip also flags the exceptional case $c=0$ without fully explaining it. This 180-second clip develops the tangent-plane formula for a two-variable function by reducing the problem to one-variable slices. It first identifies the coefficients in $a(x-x_0)+b(y-y_0)+z_0=z$ as $f_x(x_0,y_0)$ and $f_y(x_0,y_0)$ , then boxes the final formula $f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+z_0=z$ . A 3D diagram interprets the result geometrically: the tangent lines to the slices at fixed $y=y_0$ and fixed $x=x_0$ both lie in the tangent plane. The clip ends with a worked example for $f(x,y)=2-x^2-y^2$ at $(1/2,1/2)$ , computing $z_0=3/2$ , $f_x=f_y=-1$ , and the plane $z=3/2-1(x-1/2)-1(y-1/2)$ . The video concludes a lesson on finding the tangent plane of a multivariable function. It displays the general formula for a tangent plane and provides a step-by-step example calculating the tangent plane for the function $f(x, y) = 2 - x^2 - y^2$ at the point ( $1/2$ , $1/2$ ). The final equation is shown as $z = 3/2 - 1(x - 1/2) - 1(y - 1/2)$ . The remainder of the clip is an outro where the speaker encourages viewers to subscribe and watch more videos in the multivariable calculus playlist.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Title: Tangent Planes 0:03Tangent planes as multivariable tangent lines 0:18Example surface and point of tangency 0:46Different points give different tangent planes 0:54Horizontal tangent plane at the top 1:09Loose goal for a tangent plane 1:42General equation of a plane 2:24Expanding to component form and identifying the hard part 3:00Goal and plane equation 3:15Normalize with

c=-1

3:35Rewrite the plane as

L(x,y)

4:27Plug in

y=y_0

5:32Identify

a=f_x(x_0,y_0)

6:00Deriving the coefficients from one-variable slices 6:50Boxed tangent-plane formula 7:05Geometric interpretation with slice curves 8:02Worked example: tangent plane to

2-x^2-y^2

9:00Tangent Plane Formula and Example 9:03Outro and Call to Action

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The clip opens with the title "TANGENT PLANES," signaling a shift from single-variable tangent lines to surfaces in three dimensions.

The lecturer introduces the main idea: a tangent plane is the multivariable analogue of a tangent line for the graph of a function of two variables.

Using a bell-shaped surface as the example graph, he marks a specific point in red and identifies it as $(x_0,y_0,z_0)$ , where $z_0$ is the height of the function at $(x_0,y_0)$ .

He then gives the geometric intuition: the tangent plane is a plane through that point that "kisses" the graph there and stays close to the surface nearby, so it serves as a local approximation to the graph.

The example emphasizes that the tangent plane depends on the chosen point: move to another location on the surface and the tangent plane changes.

At the top of the displayed surface, the tangent plane becomes horizontal, parallel to the $xy$ -plane, which the lecturer compares to the one-variable situation where a zero derivative gives a horizontal tangent line.

The video next states the goal in deliberately informal terms: find a plane that meets the graph at $z_0=f(x_0,y_0)$ and is "close" to $z=f(x,y)$ nearby, while postponing the exact meaning of "close."

To build toward a formula, the lecturer recalls the general equation of a plane in vector form: if $\vec n$ is a normal vector and $\overrightarrow{P_0P}$ is a vector lying in the plane from a fixed point $P_0$ to a generic point $P$ , then $\vec n\cdot \overrightarrow{P_0P}=0$ .

The reason for the zero dot product is orthogonality: a genuine normal vector must be perpendicular to every vector that lies in the plane.

Writing $\vec n=(a,b,c)$ and $\overrightarrow{P_0P}=(x-x_0,y-y_0,z-z_0)$ , the equation expands to $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .

The clip closes by isolating the remaining difficulty in the tangent-plane problem: the point $(x_0,y_0,z_0)$ is already known, so the essential unknown is the normal vector, i.e. the coefficients $a,b,c$ , that determines the tangent plane.

The clip opens with the stated goal: find a plane that passes through $z_0=f(x_0,y_0)$ and stays close to the surface $z=f(x,y)$ nearby. On the board, the plane is first described geometrically by $\vec n\cdot\overline{P_0P}=0$ , where $\vec n$ is a normal vector, $P_0(x_0,y_0,z_0)$ is a fixed point on the plane, and $P(x,y,z)$ is a variable point on the plane.

The same condition is then written in coordinates as $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ . Here $a,b,c$ are the components of the normal vector, and the equation says that every displacement from $P_0$ to another point of the plane is orthogonal to $(a,b,c)$ .

To prepare for solving the plane for $z$ , the lecturer imposes the simplifying choice $c=-1$ . He explains that if the original $c$ is nonzero, one may divide the whole equation by $c$ so that the coefficient of $z-z_0$ becomes $-1$ . He also notes the exceptional case $c=0$ and asks the viewer to think why that case is irrelevant for a tangent plane to a differentiable function, but he does not finish that explanation in this excerpt.

With $c=-1$ , the equation becomes $a(x-x_0)+b(y-y_0)+z_0=z$ . The bracketed label on the board identifies the right-hand side as a linear function $L(x,y)$ . Conceptually, the tangent plane has now been rewritten as the graph of a function of two variables that is linear in $x$ and $y$ : each variable appears only to the first power and is multiplied by constants.

The lecturer then interprets this formula probabilistically: just as tangent lines approximate nonlinear one-variable functions, tangent planes approximate nonlinear surfaces by simpler linear objects. At this stage the shape of the approximation is known, but the coefficients $a$ and $b$ are still unspecified; they must be chosen so that the plane is genuinely close to $f(x,y)$ near $(x_0,y_0)$ .

To determine $a$ , the board clears down to the linear formula and the instruction Plug in $y=y_0$ appears. Substituting $y=y_0$ into $z=a(x-x_0)+b(y-y_0)+z_0$ kills the $b(y-y_0)$ term, leaving $z=a(x-x_0)+z_0$ . Geometrically, this corresponds to slicing both the surface and the candidate plane by the vertical plane $y=y_0$ .

After the slice, the remaining equation involves only $x$ and $z$ , so it is the equation of a line. The lecturer emphasizes that this is exactly the setting of single-variable tangent-line approximation: once $y$ is frozen at $y_0$ , the problem reduces to approximating a one-variable function of $x$ by a line.

In single-variable calculus, the slope of the tangent line at the point of interest is the derivative there. Translating that fact to the present two-variable setting, the coefficient $a$ must be the derivative of the restricted function obtained from $f(x,y_0)$ at $x=x_0$ . Since the restriction to $y=y_0$ is precisely what a partial derivative with respect to $x$ measures, this derivative is $f_x(x_0,y_0)$ .

The final board statement makes this explicit: Linear approximation with $a=f_x(x_0,y_0)$ . Thus the excerpt completes one half of the tangent-plane construction, showing that the coefficient multiplying $(x-x_0)$ is the $x$ -partial derivative at the base point. The analogous determination of $b$ is not reached within this clip.

The clip opens with the general linear model $a(x-x_0)+b(y-y_0)+z_0=z$ already on the board. The lecturer then fixes $x=x_0$ , which removes the $a(x-x_0)$ term and leaves $b(y-y_0)+z_0=z$ . At this stage $z$ depends only on $y$ , so the problem becomes exactly the one-variable task of finding a linear approximation. The rule inherited from single-variable calculus is that the slope of the approximating line should be the derivative, hence $b=f_y(x_0,y_0)$ .

By symmetry, the earlier board step with $y=y_0$ gives $a(x-x_0)+z_0=z$ , a one-variable relation in $x$ , and therefore $a=f_x(x_0,y_0)$ . The key conceptual move is that each coefficient in the two-variable linear model is determined by freezing the other variable and differentiating the resulting slice.

Once both coefficients are identified, the board clears the intermediate lines and presents the consolidated tangent-plane equation $f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+z_0=z.$ Here $z_0$ is the height of the original surface above the base point $(x_0,y_0)$ , while the two partial derivatives supply the directional slopes in the $x$ - and $y$ -directions.

The lecture then shifts from algebra to geometry. A 3D graph appears with a surface, a translucent plane through a red point, a yellow slice curve obtained by fixing $y=y_0$ , and a blue/cyan slice curve obtained by fixing $x=x_0$ . Each slice has its own tangent line at the red point, and both tangent lines are drawn inside the same plane. This visualizes the claim that the tangent plane is the plane containing the coordinate-direction tangent lines generated by the partial derivatives.

To test the formula, the lecturer introduces the example $f(x,y)=2-x^2-y^2$ at the point $\left(\frac12,\frac12\right)$ . First he computes the base height: $z_0=f\left(\frac12,\frac12\right)=2-\left(\frac12\right)^2-\left(\frac12\right)^2=\frac32.$ Next he computes the two partial derivatives at that point: $f_x\left(\frac12,\frac12\right)=-2\left(\frac12\right)=-1, \qquad f_y\left(\frac12,\frac12\right)=-2\left(\frac12\right)=-1.$ Substituting these three numbers into the general formula yields the explicit tangent plane $z=\frac32-1\left(x-\frac12\right)-1\left(y-\frac12\right).$

A small notation caveat remains visible on the board: the example header is written as ` $f(x)=2-x^2-y^2$ `, but the spoken explanation and all computations treat it as a two-variable function $f(x,y)$ . Mathematically, the worked example is consistent with the tangent-plane formula for a surface in three variables.

The speaker finalizes the lesson by pointing to the screen, which displays the general formula for a tangent plane: $f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) + z_0 = z$ . Below it, a complete example is worked out for the function $f(x, y) = 2 - x^2 - y^2$ at the point ( $1/2$ , $1/2$ ). The calculations show finding the z-coordinate ( $z_0 = 3/2$ ), the partial derivatives with respect to x and y (both equal to -1 at the given point), and substituting these into the formula to get the final equation: $z = 3/2 - 1(x - 1/2) - 1(y - 1/2)$ .

The scene transitions to an office setting. The speaker addresses the audience directly, asking them to leave questions in the comments. He humorously refers to the 'YouTube algorithm' and asks viewers to like the video. He then points out that this video is part of a larger playlist on multivariable calculus and encourages viewers to check out the other videos before signing off.

Knowledge cards

Tangent plane: intuitive meaning

For the graph of a function of two variables, the tangent plane at a point is the plane analogue of a tangent line. It passes through the point on the graph and locally matches, or "kisses," the surface there, giving a nearby linear approximation to the graph.

Point of tangency on $z=f(x,y)$

The relevant point is a point on the graph, written $(x_0,y_0,z_0)$ , with the vertical coordinate determined by the function value: $z_0=f(x_0,y_0)$ .

z_0=f(x_0,y_0)

Loose goal for finding a tangent plane

The video states two requirements: the desired plane must meet the graph at $(x_0,y_0,z_0)$ , and it must be "close" to $z=f(x,y)$ nearby. The exact meaning of "close" is intentionally deferred in this clip.

\text{meet at } z_0=f(x_0,y_0),\ \text{and be close to } z=f(x,y) \text{ nearby}

Vector equation of a plane

A plane can be described using a normal vector $\vec n$ and a fixed point $P_0$ in the plane. A point $P$ lies in the plane exactly when the displacement vector $\overrightarrow{P_0P}$ is perpendicular to $\vec n$ , which is expressed by a zero dot product.

\vec{n}\cdot \overrightarrow{P_0P}=0

Component equation of a plane

If the normal vector has components $(a,b,c)$ and the plane passes through $(x_0,y_0,z_0)$ , then the equation of the plane is obtained by expanding the dot product into coordinates.

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

What remains hard for a tangent plane

Once the tangency point is known, substituting $(x_0,y_0,z_0)$ into the plane formula is straightforward. The central remaining task is finding the normal vector components $a,b,c$ appropriate to the graph of the function at that point.

Special case: horizontal tangent plane

At the top of the example surface, the tangent plane is parallel to the $xy$ -plane. The lecturer compares this to the single-variable case where a zero derivative produces a horizontal tangent line.

Plane from a point and a normal vector

A plane can be characterized by a fixed point $P_0$ on it and a normal vector $\vec n$ . A point $P$ lies in the plane exactly when the displacement vector $\overline{P_0P}$ is perpendicular to $\vec n$ , which is expressed by the dot-product equation $\vec n\cdot\overline{P_0P}=0$ .

\vec n \cdot \overline{P_0P}=0

Scalar equation of a plane

Writing the normal vector as $(a,b,c)$ and the points as $P_0(x_0,y_0,z_0)$ and $P(x,y,z)$ turns the geometric condition into the coordinate equation $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ . This is the standard linear equation of a plane in three variables.

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

Why choose $c=-1$

If the coefficient $c$ of $(z-z_0)$ is nonzero, the whole plane equation can be rescaled so that $c=-1$ . This normalization makes it possible to solve the equation for $z$ and view the plane as a function of $x$ and $y$ . The speaker explicitly notes the exceptional case $c=0$ and treats it as irrelevant for the tangent-plane setting, without completing that justification in this excerpt.

c=-1

Tangent plane as a linear function $L(x,y)$

After setting $c=-1$ , the plane equation becomes $z=a(x-x_0)+b(y-y_0)+z_0$ . The right-hand side is a linear function of $x$ and $y$ , so the tangent plane can be understood as the graph of a linear approximation to the surface near $(x_0,y_0)$ .

L(x,y)=a(x-x_0)+b(y-y_0)+z_0

Restricting to the slice $y=y_0$

Substituting $y=y_0$ into the linear approximation removes the $b(y-y_0)$ term and yields $z=a(x-x_0)+z_0$ . Geometrically this is the intersection of the plane with the vertical slice $y=y_0$ , reducing the problem to a one-variable line.

a(x-x_0)+z_0=z

From tangent line to partial derivative

Once $y$ is fixed at $y_0$ , the remaining expression is the equation of a line in $x$ and $z$ . In single-variable calculus the slope of the tangent line is the derivative at the point, so here the coefficient $a$ must be the derivative of the restricted function $f(x,y_0)$ at $x=x_0$ . That derivative is exactly the partial derivative $f_x(x_0,y_0)$ .

a=f_x(x_0,y_0)

General linear model for a tangent plane

Before identifying the slopes, the video writes the plane in the form $a(x-x_0)+b(y-y_0)+z_0=z$ . Here $(x_0,y_0,z_0)$ is the point of tangency, and $a,b$ are the unknown directional coefficients to be determined.

a(x-x_0)+b(y-y_0)+z_0=z

Partial derivatives come from one-variable slices

Fixing $y=y_0$ leaves a one-variable relation in $x$ , so the coefficient of $(x-x_0)$ must be $f_x(x_0,y_0)$ . Fixing $x=x_0$ leaves a one-variable relation in $y$ , so the coefficient of $(y-y_0)$ must be $f_y(x_0,y_0)$ . This is the clip’s bridge from single-variable linear approximation to the multivariable formula.

a=f_x(x_0,y_0),\quad b=f_y(x_0,y_0)

Final tangent-plane formula

After substituting the identified coefficients, the lecture states the general equation of the tangent plane at $(x_0,y_0,z_0)$ as a linear expression in the two coordinate increments.

f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+z_0=z

Geometric meaning of the tangent plane

The 3D diagram shows that the tangent plane contains the tangent line to the slice with $y=y_0$ and also the tangent line to the slice with $x=x_0$ . Thus the plane is built from the two coordinate-direction tangent lines at the point of tangency.

Worked example setup

The example applies the formula to the surface $z=2-x^2-y^2$ at the point $\left(\frac12,\frac12\right)$ . The board header writes ` $f(x)=2-x^2-y^2$ `, but the actual computation treats the function as two-variable.

f(x,y)=2-x^2-y^2,\quad (x_0,y_0)=\left(\frac12,\frac12\right)

Example: compute the base height

The first numerical step is to evaluate the function at the base point, giving the $z$ -coordinate of the point of tangency.

z_0=f\left(\frac12,\frac12\right)=2-\left(\frac12\right)^2-\left(\frac12\right)^2=\frac32

Example: compute the partial derivatives

The next step differentiates with respect to each variable and evaluates at the base point. Both partial derivatives turn out to be $-1$ .

f_x\left(\frac12,\frac12\right)=-1,\quad f_y\left(\frac12,\frac12\right)=-1

Example: explicit tangent plane

Substituting $z_0=\frac32$ and both partial derivatives equal to $-1$ into the general formula gives the final plane equation for the example surface.

z=\frac32-1\left(x-\frac12\right)-1\left(y-\frac12\right)

General Formula for Tangent Plane

The equation of the tangent plane to a surface $z = f(x, y)$ at a specific point ( $x_0$ , $y_0$ , $z_0$ ) is given by the formula involving the partial derivatives of the function at that point.

f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) + z_0 = z

Example: Tangent Plane Calculation

To find the tangent plane for $f(x, y) = 2 - x^2 - y^2$ at ( $1/2$ , $1/2$ ), first calculate $z_0 = f(1/2, 1/2) = 3/2$ . Then find the partial derivatives $f_x = -2x$ and $f_y = -2y$ , and evaluate them at the point to get $f_x(1/2, 1/2) = -1$ and $f_y(1/2, 1/2) = -1$ . Substitute these values into the general formula to get the final equation.

z = \frac{3}{2} - 1\left(x - \frac{1}{2}\right) - 1\left(y - \frac{1}{2}\right)

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 33

f

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer refers to "the graph of a function" and later writes the surface as $z=f(x,y)$ .
Formula
Observation
On-screen text at about 01:12 includes $z_0 = f(x_0, y_0)$ and $z = f(x,y)$ .

Symbol

Meaning

A scalar-valued function of two variables whose graph is the surface $z=f(x,y)$ in three-dimensional space.

Domain

Two-variable real function; the video does not state explicit differentiability or continuity assumptions.

$x_0$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the red point is on the graph of the function at " $x_0, y_0$ " with height " $z_0$ ".
Formula
Observation
The goal text uses $(x_0,y_0)$ and $z_0=f(x_0,y_0)$ ; the plane formula uses $(x-x_0)$ . The diagram labels $P_0(x_0,y_0,z_0)$ .

Symbol

$x_0$

Meaning

The $x$ -coordinate of the specified point of tangency on the graph of $f$ .

Domain

Real coordinate value.

$y_0$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer names the point of tangency using " $x_0, y_0$ " and height " $z_0$ ".
Formula
Observation
The goal text uses $(x_0,y_0)$ and $z_0=f(x_0,y_0)$ ; the plane formula uses $(y-y_0)$ . The diagram labels $P_0(x_0,y_0,z_0)$ .

Symbol

$y_0$

Meaning

The $y$ -coordinate of the specified point of tangency on the graph of $f$ .

Domain

Real coordinate value.

$z_0$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the point has height " $z_0$ ".
Formula
Observation
The goal text states $z_0 = f(x_0,y_0)$ ; the plane formula uses $(z-z_0)$ . The diagram labels $P_0(x_0,y_0,z_0)$ .

Symbol

$z_0$

Meaning

The height, or $z$ -coordinate, of the point on the graph of $f$ where the tangent plane is taken.

Domain

Real value determined by $z_0=f(x_0,y_0)$ .

x

Clear evidence

Shown in the video

Evidence

Formula
Observation
The expanded plane equation contains $(x-x_0)$ .
Diagram
Observation
The generic point in the plane diagram is labeled $P(x,y,z)$ .

Symbol

Meaning

The $x$ -coordinate of a generic point $P$ lying in the plane being described.

Domain

Real coordinate variable.

y

Clear evidence

Shown in the video

Evidence

Formula
Observation
The expanded plane equation contains $(y-y_0)$ .
Diagram
Observation
The generic point in the plane diagram is labeled $P(x,y,z)$ .

Symbol

Meaning

The $y$ -coordinate of a generic point $P$ lying in the plane being described.

Domain

Real coordinate variable.

z

Clear evidence

Shown in the video

Evidence

Formula
Observation
The expanded plane equation contains $(z-z_0)$ .
Diagram
Observation
The generic point in the plane diagram is labeled $P(x,y,z)$ .

Symbol

Meaning

The $z$ -coordinate of a generic point $P$ lying in the plane being described.

Domain

Real coordinate variable.

$\vec{n}$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer describes the equation of a plane using "the dot product between the normal vector" and a vector in the plane.
Formula
Observation
The displayed equation is $\vec{n}\cdot \overrightarrow{P_0P}=0$ .
Diagram
Observation
A pink/magenta arrow perpendicular to the blue plane is labeled $\vec n$ .

Symbol

$\vec{n}$

Meaning

A normal vector to the plane, i.e. a vector perpendicular to every vector lying in the plane.

Domain

A three-dimensional vector; its components are later denoted $a,b,c$ .

$P_0$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the vector emanates from a point " $P_0$ " out to " $P$ ".
Formula
Observation
The displayed equation uses $\overrightarrow{P_0P}$ .
Diagram
Observation
The fixed point on the blue plane is labeled $P_0(x_0,y_0,z_0)$ .

Symbol

$P_0$

Meaning

A fixed reference point on the plane, here the point of tangency $(x_0,y_0,z_0)$ .

Domain

Point in three-dimensional space.

P

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer refers to a vector from $P_0$ out to $P$ .
Formula
Observation
The displayed equation uses $\overrightarrow{P_0P}$ .
Diagram
Observation
A green arrow from $P_0$ to a generic point is labeled $P(x,y,z)$ .

Symbol

Meaning

A generic point in the plane, with coordinates $(x,y,z)$ .

Domain

Point in three-dimensional space.

$\overrightarrow{P_0P}$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer calls it "the vector that emanates from a point P0 out to P".
Formula
Observation
The displayed equation is $\vec{n}\cdot \overrightarrow{P_0P}=0$ .
Diagram
Observation
A green arrow lies in the blue plane from $P_0(x_0,y_0,z_0)$ to $P(x,y,z)$ .

Symbol

$\overrightarrow{P_0P}$

Meaning

The displacement vector from the fixed point $P_0$ to a generic point $P$ in the plane.

Domain

Three-dimensional vector equal componentwise to $(x-x_0,y-y_0,z-z_0)$ in the subsequent expansion.

a

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, "If we give the normal vector the components A, B, and C".
Formula
Observation
The expanded equation begins with $a(x-x_0)$ .

Symbol

Meaning

The first component of the normal vector $\vec n$ in the plane equation.

Domain

Real scalar component.

Knowledge points · 17

Intuitive definition of a tangent plane to a surface $z=f(x,y)$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer introduces tangent planes as "the multivariable generalization of a tangent line that we saw in single variable calculus" and asks how to define and compute one.
Audio
Observation
He describes the tangent plane as a plane that goes through the point and "sort of kisses the graph of this function at a particular point," and says that nearby it is a close value and a good approximation for the graph.
Diagram
Observation
A semi-transparent gray plane touches a yellow-green bell-shaped surface at a red point in a 3D coordinate system.

Definition

Explanation

For the graph of a two-variable function, the video presents a tangent plane at a point as the plane analogue of a tangent line in single-variable calculus. It passes through the point on the graph and locally "kisses" the surface, giving a close linear approximation to the graph near that point.

Formula

Conditions

The object is the graph of a function of two variables, written in the video as $z=f(x,y)$ .
A specific point $(x_0,y_0,z_0)$ on the graph is chosen.
The approximation statement is local: the plane is close to the graph "nearby" the point.

Prerequisites

Vector equation of a plane using a normal vector

Point of tangency on the graph

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says he has specified a particular point in red, which is a spot on the graph of the function at $x_0,y_0$ with height $z_0$ .
Diagram
Observation
A red dot marks the contact point on the surface.
Formula
Observation
Later text gives the same point condition as $z_0=f(x_0,y_0)$ .

Definition

Explanation

The point where the tangent plane is taken is a point on the graph of the function. In the video it is denoted by coordinates $(x_0,y_0,z_0)$ , with the vertical coordinate determined by the function value at the input point.

Formula

(x_0,y_0,z_0),\quad z_0=f(x_0,y_0)

Conditions

The point lies on the graph of $f$ .
The height is the function value at $(x_0,y_0)$ .
The video does not separately state differentiability assumptions.

Prerequisites

Intuitive definition of a tangent plane to a surface $z=f(x,y)$

Loose goal for constructing a tangent plane

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, "let's begin with the goal," then states two requirements: the tangent plane and graph meet at $x_0,y_0,z_0$ , and nearby the tangent plane is close to the graph in some sense.
Formula
Observation
On-screen text reads: Goal: A plane that meets at $z_0 = f(x_0, y_0)$ and is "close" to $z = f(x, y)$ "near by".

Uncertainties

The phrase "close" is explicitly left informal in the video; the lecturer says he will put a pin in the exact meaning and return later.

Definition

Explanation

The video frames the problem as finding a plane satisfying two conditions: it must pass through the point of tangency on the graph, and it must approximate the graph closely in a neighborhood of that point. The precise meaning of "close" is deferred rather than defined within this clip.

Formula

\text{Goal: a plane through } z_0=f(x_0,y_0) \text{ that is close to } z=f(x,y) \text{ nearby}

Conditions

The plane must meet the graph at $(x_0,y_0,z_0)$ .
The closeness requirement is local around that point.
The exact quantitative meaning of closeness is not given in this segment.

Prerequisites

Point of tangency on the graph

Vector equation of a plane using a normal vector

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says one way to give the equation of a plane is as the dot product between the normal vector and a generic vector lying in the plane, namely the vector from $P_0$ to $P$ , and that this dot product is zero.
Formula
Observation
Displayed text: Equation of Plane: $\vec{n}\cdot \overrightarrow{P_0P}=0$ .
Diagram
Observation
A blue plane is shown with a pink/magenta normal vector $\vec n$ perpendicular to it and a green in-plane vector from $P_0(x_0,y_0,z_0)$ to $P(x,y,z)$ .

Formula

Explanation

A plane can be described as the set of points $P$ such that the displacement vector from a fixed point $P_0$ in the plane to $P$ is orthogonal to a normal vector $\vec n$ . Orthogonality is expressed by a zero dot product.

Formula

\vec{n}\cdot \overrightarrow{P_0P}=0

Conditions

$P_0$ is a known point on the plane.
$\vec n$ is a normal vector to the plane.
$P$ is a generic point being tested for membership in the plane.
The equation encodes perpendicularity between $\vec n$ and every in-plane displacement vector from $P_0$ .

Prerequisites

Component form of the plane equation

Component form of the plane equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says that if the normal vector has components $A,B,C$ , then expanding gives $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$ .
Formula
Observation
Displayed text: $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ .

Formula

Explanation

Writing the normal vector as $(a,b,c)$ and the displacement vector as $(x-x_0,y-y_0,z-z_0)$ turns the dot-product equation of a plane into a scalar linear equation in the coordinates of a generic point.

Formula

a(x - x_0) + b(y - y_0) + c(z - z_0) = 0

Conditions

$(x_0,y_0,z_0)$ is a point on the plane.
$(a,b,c)$ are the components of a normal vector to the plane.
$(x,y,z)$ denotes a generic point in the plane.

Prerequisites

Vector equation of a plane using a normal vector

Vector form of a plane equation

Clear evidence

Shown in the video

Evidence

Formula
Observation
Equation of Plane: $\vec n \cdot \overline{P_0P}=0$ .
Diagram
Observation
Blue plane with perpendicular pink normal vector $\vec n$ , point $P_0(x_0,y_0,z_0)$ , and point $P(x,y,z)$ .

Definition

Explanation

The video presents a plane as the set of points $P$ such that the displacement vector from a fixed point $P_0$ on the plane to $P$ is orthogonal to the plane’s normal vector $\vec n$ . Orthogonality is expressed by the dot product being zero.

Formula

\vec n \cdot \overline{P_0P}=0

Conditions

A fixed point $P_0$ lies on the plane.
$\vec n$ is a normal vector to the plane.
$P$ is any point on the plane.

Prerequisites

$\vec n$
$P_0(x_0,y_0,z_0)$
$P(x,y,z)$

Scalar component form of a plane

Clear evidence

Shown in the video

Evidence

Formula
Observation
Displayed scalar equation: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .
Audio
Observation
Speaker discusses the coefficient of $z-z_0$ as $c$ and divides through to normalize it.

Formula

Explanation

Expanding the normal-vector description into coordinates gives a linear equation in $x,y,z$ with coefficients $a,b,c$ coming from the normal vector and constants determined by the point $P_0(x_0,y_0,z_0)$ .

Formula

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

Conditions

$(a,b,c)$ are the components of a normal vector.
$(x_0,y_0,z_0)$ is a point on the plane.
$(x,y,z)$ ranges over points of the plane.

Prerequisites

Vector form of a plane equation
$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$

Choosing $c=-1$ to solve the plane for $z$

Clear evidence

Shown in the video

Evidence

Caption evidence
Observation
Choose $c=-1$ .
Audio
Observation
Speaker says if $c$ is nonzero, one can divide through so the coefficient of $z-z_0$ equals $-1$ .

Method

Explanation

When the normal vector has nonzero third component, the plane equation can be rescaled so that $c=-1$ . This makes it possible to isolate $z$ and rewrite the plane as a function of $x$ and $y$ .

Formula

c=-1

Conditions

The original coefficient $c$ is nonzero.
The plane is not vertical in the sense relevant to solving for $z$ .

Prerequisites

Scalar component form of a plane
$c=-1$

Tangent plane as a linear function $L(x,y)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
Displayed equation: $a(x-x_0)+b(y-y_0)+z_0=z$ .
Caption evidence
Observation
Label under the expression: A linear function $L(x,y)$ .
Audio
Observation
Speaker says the height $z$ depends on $x$ and $y$ linearly, with variables raised only to the first power and multiplied by constants.

Definition

Explanation

After setting $c=-1$ , the plane equation becomes $z=a(x-x_0)+b(y-y_0)+z_0$ . The right-hand side is a linear function of $x$ and $y$ , so the plane can be interpreted as the graph of a linear approximation to the surface.

Formula

L(x,y)=a(x-x_0)+b(y-y_0)+z_0

Conditions

The plane has been normalized so that the coefficient of $z-z_0$ is $-1$ .
$a$ and $b$ are still undetermined constants at this stage.

Prerequisites

Choosing $c=-1$ to solve the plane for $z$
$L(x,y)=a(x-x_0)+b(y-y_0)+z_0$

Restricting the linear approximation to $y=y_0$

Clear evidence

Shown in the video

Evidence

Caption evidence
Observation
Plug in $y=y_0$ .
Formula
Observation
Resulting displayed equation: $a(x-x_0)+z_0=z$ .
Audio
Observation
Speaker says plugging in $y=y_0$ makes the second term go away and restricts to the plane where $y=y_0$ .

Method

Explanation

Substituting $y=y_0$ into $L(x,y)$ removes the $b(y-y_0)$ term and leaves a one-variable linear expression in $x$ . Geometrically, this corresponds to cutting the surface and plane by the vertical plane $y=y_0$ .

Formula

a(x-x_0)+z_0=z

Conditions

Start from $z=a(x-x_0)+b(y-y_0)+z_0$ .
Set $y=y_0$ .

Prerequisites

Tangent plane as a linear function $L(x,y)$
$L(x,y)=a(x-x_0)+b(y-y_0)+z_0$

Coefficient $a$ equals the $x$ -partial derivative

Clear evidence

Shown in the video

Evidence

Caption evidence
Observation
Linear approximation with $a=f_x(x_0,y_0)$ .
Audio
Observation
Speaker says the answer from single-variable calculus is that $a$ is the derivative, here the partial of $f$ with respect to $x$ evaluated at the point.

Formula

Explanation

Once $y$ is fixed at $y_0$ , the problem reduces to approximating a single-variable function of $x$ . In one-variable calculus the slope of the tangent line is the derivative, so the coefficient $a$ must be $f_x(x_0,y_0)$ .

Formula

a=f_x(x_0,y_0)

Conditions

The function is being approximated near $(x_0,y_0)$ .
$y$ is held fixed at $y_0$ while differentiating with respect to $x$ .
The relevant partial derivative exists.

Prerequisites

Restricting the linear approximation to $y=y_0$
$f_x(x_0,y_0)$

General form of the tangent-plane equation

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board displays $a(x-x_0)+b(y-y_0)+z_0=z$ and then substitutes $a=f_x(x_0,y_0)$ , $b=f_y(x_0,y_0)$ .
Audio
Observation
The lecturer explains that fixing one variable leaves a single-variable problem whose slope is given by the corresponding derivative.

Formula

Explanation

The video presents the tangent plane as a linear equation in $x$ and $y$ built from the base point $(x_0,y_0,z_0)$ and two directional coefficients. Those coefficients are then identified as the partial derivatives at the base point.

Formula

a(x-x_0)+b(y-y_0)+z_0=z

Conditions

A base point $(x_0,y_0)$ is fixed.
$z_0$ is the function value at that point.
The coefficients are chosen to match the one-variable slice approximations.

Prerequisites

Partial derivatives as slopes of coordinate slices

Claims and conditions · 9

Tangent planes generalize tangent lines to several variables

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer states that tangent planes are "the multivariable generalization of a tangent line that we saw in single variable calculus."

Proposition

Statement

In the context of graphs of functions of two variables, a tangent plane plays the role analogous to a tangent line for graphs of functions of one variable.

Hypotheses

The surface is the graph of a function of two variables.
The comparison is to single-variable calculus tangent lines.

Quantifiers

Stated informally for the topic of multivariable functions; no formal quantifier is given in the clip.

Tangent plane approximates the graph locally

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the tangent plane goes through the point, kisses the graph there, and nearby is a close value and a good approximation for the graph of the function.

Uncertainties

The word "good" is qualitative; the clip does not provide the formal epsilon-type or differentiability condition that would make this precise.

Proposition

Statement

Near the chosen point, the tangent plane is close to the graph of the function and therefore serves as a local approximation to that graph.

Hypotheses

A point $(x_0,y_0,z_0)$ on the graph is selected.
The plane under discussion is the tangent plane at that point.

Quantifiers

Local statement around the point of tangency; the exact neighborhood notion is not formalized in this clip.

Different points generally give different tangent planes

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says he could have a tangent plane at one spot, move it around the front, and at a different location get a different tangent plane.

Proposition

Statement

Moving the point of tangency on the surface changes the tangent plane associated with that point.

Hypotheses

The surface is fixed.
The point of tangency is varied along the surface.

Quantifiers

Informal universal-style claim over locations on the displayed surface; no exceptional cases are discussed.

At the top of the example surface, the tangent plane is horizontal

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says that at the top of the surface, the tangent plane is parallel to the $xy$ -plane, analogously to taking a derivative equal to zero and getting a horizontal tangent line in single-variable calculus.
Diagram
Observation
The tangent plane at the peak is drawn horizontally.

Uncertainties

The video does not explicitly name the top point as a critical point or state partial-derivative conditions in this clip.

Proposition

Statement

For the displayed bell-shaped surface, the tangent plane at the top point is parallel to the $xy$ -plane, matching the single-variable idea of a horizontal tangent line when the derivative is zero.

Hypotheses

The point considered is the top of the displayed surface.
The tangent plane exists at that point.

Quantifiers

Stated for the specific example point at the top of the shown surface.

A normal vector has zero dot product with every vector in the plane

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says that if the normal is truly going to be normal, it had to have a zero dot product with any vector that lay in the plane.

Proposition

Statement

A vector is normal to a plane precisely when its dot product with any vector lying in that plane is zero.

Hypotheses

A plane is given.
A vector is claimed to be normal to that plane.
Another vector lies in the plane.

Quantifiers

Universal over vectors lying in the plane.

Exceptional case $c=0$ for normalizing the plane equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says the normalization always works unless $c=0$ , then asks the viewer to pause and think why that is not relevant when asking for a tangent plane to a differentiable function.

Uncertainties

The video does not state the full explanation inside this clip; it only poses the question.

Proposition

Statement

The step of dividing through to make the coefficient of $z-z_0$ equal to $-1$ is valid when $c\neq 0$ ; the speaker indicates that the case $c=0$ is not relevant for the tangent plane to a differentiable function, but leaves the reason as a pause-and-think prompt.

Hypotheses

The plane equation is written as $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .
One wants to solve the plane for $z$ by normalizing $c$ to $-1$ .

Quantifiers

For a plane with coefficient $c$ of $(z-z_0)$ , the normalization works if $c\neq 0$ ; the excluded case is $c=0$ .

After fixing $y=y_0$ , the coefficient $a$ is determined by one-variable tangent-line logic

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says that in single-variable calculus the answer was that $a$ was just the derivative at the point being discussed.
Caption evidence
Observation
Later text identifies $a=f_x(x_0,y_0)$ .

Proposition

Statement

When $y$ is fixed at $y_0$ , the remaining expression is the equation of a line in $x$ and $z$ , so the coefficient $a$ must be the derivative of the resulting one-variable function, namely the partial derivative of $f$ with respect to $x$ at $(x_0,y_0)$ .

Hypotheses

The plane has been written as $z=a(x-x_0)+b(y-y_0)+z_0$ .
Substitute $y=y_0$ to obtain $z=a(x-x_0)+z_0$ .
The target surface is $z=f(x,y)$ and the approximation is taken at $(x_0,y_0)$ .

Quantifiers

For the restricted one-variable problem obtained by holding $y=y_0$ , the linear coefficient $a$ equals $f_x(x_0,y_0)$ .

Multivariable linear approximation inherits the single-variable derivative rule

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer states: “We have inherited the notion of a good approximation from single variable calculus to determine the a and the b that we need in multivariable calculus.”

Proposition

Statement

The coefficients in the multivariable linear approximation are obtained by applying the single-variable rule “slope equals derivative” to each coordinate slice.

Hypotheses

The function can be restricted to a one-variable slice by fixing the other input.
The relevant one-variable derivative exists at the base point.

Quantifiers

For each coordinate direction at the chosen base point.

Coordinate-slice tangent lines lie in the tangent plane

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the tangent line to the yellow curve lives inside the tangent plane, and likewise the tangent line to the blue curve lives within the tangent plane at the particular point.
Diagram
Observation
Both colored tangent lines are drawn lying in the translucent plane at the red point.

Proposition

Statement

At the point of tangency, the tangent line to the slice with $y=y_0$ and the tangent line to the slice with $x=x_0$ both lie in the tangent plane.

Hypotheses

The tangent plane is the linear approximation plane at that point.
The slices are taken through the same base point.

Quantifiers

At the specific point shown in the diagram.

Derivations and proofs · 10

Expanding the dot-product plane equation into component form

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, "From this particular equation that we have, we can expand it. If we give the normal vector the components A, B, and C, then we're going to get the formula A times X minus X0 plus B times Y minus Y0 and C times Z minus Z0, and that the sum of all these three things is zero."
Formula
Observation
The screen shows the transition from $\vec{n}\cdot \overrightarrow{P_0P}=0$ to $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .

Proof

Steps

Expression
$\vec{n}\cdot \overrightarrow{P_0P}=0$
Explanation
Start from the vector equation of a plane using a normal vector and an in-plane displacement vector.
Justification
This is the displayed plane equation introduced immediately before the expansion.
Shown in the video
Expression
$\vec n=(a,b,c),\qquad \overrightarrow{P_0P}=(x-x_0,\,y-y_0,\,z-z_0)$
Explanation
Write the normal vector and the displacement vector in components relative to the fixed point $P_0(x_0,y_0,z_0)$ and generic point $P(x,y,z)$ .
Justification
The lecturer says to give the normal vector components $A,B,C$ ; the component form of the displacement follows from the labeled points in the diagram.
Shown in the video
Expression
$(a,b,c)\cdot (x-x_0,\,y-y_0,\,z-z_0)=0$
Explanation
Substitute the component expressions into the dot product.
Justification
Direct substitution into the previous equation.
Derived from the video
Expression
$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
Explanation
Evaluate the dot product to obtain the scalar linear equation of the plane.
Justification
Definition of the Euclidean dot product in three dimensions.
Shown in the video

Conclusion

The vector equation $\vec n\cdot \overrightarrow{P_0P}=0$ is equivalent to the component equation $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .

From general plane equation to a linear function of $x$ and $y$

Clear evidence

Shown in the video

Evidence

Formula
Observation
Sequence on screen: $\vec n\cdot\overline{P_0P}=0$ , then $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ , then Choose $c=-1$ , then $a(x-x_0)+b(y-y_0)+z_0=z$ .
Audio
Observation
Speaker explains dividing through so the coefficient of $z-z_0$ becomes $-1$ , then says the formula cleans up to $a(x-x_0)+b(y-y_0)+z_0=z$ .

Proof

Steps

Expression
$\vec n \cdot \overline{P_0P}=0$
Explanation
Start with the geometric condition that a point $P$ lies in the plane exactly when the displacement from $P_0$ to $P$ is perpendicular to the normal vector.
Justification
Definition of a plane by a point and a normal vector.
Shown in the video
Expression
$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
Explanation
Write the same condition in coordinates using components of $\vec n=(a,b,c)$ and coordinates of $P_0$ and $P$ .
Justification
Coordinate expansion of the dot product.
Shown in the video
Expression
$c=-1$
Explanation
Rescale the equation so the coefficient of $z-z_0$ is $-1$ .
Justification
Allowed when $c\neq 0$ ; the speaker explicitly notes the exception $c=0$ .
Shown in the video
Expression
$a(x-x_0)+b(y-y_0)+z_0=z$
Explanation
Substitute $c=-1$ into the scalar plane equation and rearrange to isolate $z$ .
Justification
Algebraic substitution and rearrangement.
Shown in the video

Conclusion

With $c=-1$ , the plane can be written as $z=L(x,y)=a(x-x_0)+b(y-y_0)+z_0$ , a linear function of $x$ and $y$ .

Determining $a$ by restricting to $y=y_0$

Clear evidence

Shown in the video

Evidence

Caption evidence
Observation
Plug in $y=y_0$ .
Formula
Observation
Displayed reduced equation: $a(x-x_0)+z_0=z$ .
Caption evidence
Observation
Linear approximation with $a=f_x(x_0,y_0)$ .
Audio
Observation
Speaker says the second term goes away, the result is the equation of a line, and in single-variable calculus $a$ is the derivative, here the partial with respect to $x$ evaluated at the point.

Proof

Steps

Expression
$z=a(x-x_0)+b(y-y_0)+z_0$
Explanation
Begin with the linear function obtained from the normalized plane equation.
Justification
Previous derivation from the plane equation with $c=-1$ .
Shown in the video
Expression
$y=y_0$
Explanation
Hold $y$ fixed at the base-point value.
Justification
This reduces the two-variable approximation problem to a one-variable slice through the surface.
Shown in the video
Expression
$z=a(x-x_0)+z_0$
Explanation
The term $b(y-y_0)$ vanishes because $y-y_0=0$ .
Justification
Direct substitution.
Shown in the video
Expression
$a=f_x(x_0,y_0)$
Explanation
Identify $a$ as the slope of the tangent line to the one-variable function obtained from $f(x,y_0)$ at $x=x_0$ .
Justification
Single-variable tangent-line rule plus the definition of the partial derivative with respect to $x$ .
Shown in the video

Conclusion

The coefficient of $(x-x_0)$ in the tangent-plane linearization is $a=f_x(x_0,y_0)$ .

Deriving the $x$ -coefficient from the $y=y_0$ slice

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows `Plug in $y = y_0$ `, then $a(x-x_0)+z_0=z$ , then `Linear approximation with $a = f_x(x_0,y_0)$ `.
Audio
Observation
The lecturer says that after plugging in $y=y_0$ , $z$ depends only on $x$ , and the single-variable answer is to make the slope the derivative.

Intuitive argument

Steps

Expression
$a(x-x_0)+b(y-y_0)+z_0=z$
Explanation
Start from the general linear equation in two variables.
Justification
Displayed on the board at the beginning of the clip.
Shown in the video
Expression
$a(x-x_0)+z_0=z$
Explanation
Substitute $y=y_0$ , so the $b(y-y_0)$ term vanishes and the relation becomes one-variable in $x$ .
Justification
Explicit board step labeled `Plug in $y = y_0$ `.
Shown in the video
Expression
$a=f_x(x_0,y_0)$
Explanation
Choose the remaining coefficient to be the derivative of the slice with respect to $x$ at the base point.
Justification
The lecturer invokes the single-variable rule that the approximating line’s slope is the derivative.
Shown in the video

Conclusion

The coefficient multiplying $(x-x_0)$ in the tangent-plane formula is $f_x(x_0,y_0)$ .

Deriving the $y$ -coefficient from the $x=x_0$ slice

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows `Plug in $x = x_0$ `, then $b(y-y_0)+z_0=z$ , then `Linear approximation with $b = f_y(x_0,y_0)$ `.
Audio
Observation
The lecturer says that plugging in $x=x_0$ gives a different linear equation where $z$ depends only on $y$ , and the derivative is now with respect to $y$ .

Intuitive argument

Steps

Expression
$a(x-x_0)+b(y-y_0)+z_0=z$
Explanation
Begin again from the general two-variable linear form.
Justification
Same board setup used for both slice arguments.
Shown in the video
Expression
$b(y-y_0)+z_0=z$
Explanation
Substitute $x=x_0$ , eliminating the $a(x-x_0)$ term and leaving a one-variable relation in $y$ .
Justification
Explicit board step labeled `Plug in $x = x_0$ `.
Shown in the video
Expression
$b=f_y(x_0,y_0)$
Explanation
Set the coefficient equal to the partial derivative with respect to $y$ at the base point.
Justification
The lecturer applies the single-variable derivative-as-slope rule to the $y$ -slice.
Shown in the video

Conclusion

The coefficient multiplying $(y-y_0)$ in the tangent-plane formula is $f_y(x_0,y_0)$ .

Combining the two slice results into the tangent-plane equation

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board clears the intermediate lines and displays the boxed equation $f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+z_0=z$ .
Audio
Observation
The lecturer says, “Putting it all together, I can get the equation …” and reads out the two partial-derivative terms plus $z_0$ .

Proof

Steps

Expression
$a=f_x(x_0,y_0)$
Explanation
Use the result from the $y=y_0$ slice.
Justification
Derived earlier in the clip from the one-variable approximation rule.
Shown in the video
Expression
$b=f_y(x_0,y_0)$
Explanation
Use the result from the $x=x_0$ slice.
Justification
Derived earlier in the clip from the one-variable approximation rule.
Shown in the video
Expression
$f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+z_0=z$
Explanation
Substitute the identified coefficients back into the general linear form.
Justification
Direct substitution into $a(x-x_0)+b(y-y_0)+z_0=z$ .
Shown in the video

Conclusion

The tangent plane at $(x_0,y_0,z_0)$ is given by $f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+z_0=z$ .

Computing the base height $z_0$ in the example

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $z_0=f\left(\frac12,\frac12\right)=2-\left(\frac12\right)^2-\left(\frac12\right)^2=\frac32$ .
Audio
Observation
The lecturer asks for the height of the function at the point and says plugging in gives $\frac32$ .

Numerical verification

Steps

Expression
$f(x,y)=2-x^2-y^2$
Explanation
Start from the example function.
Justification
Given in the example statement on the board and in speech.
Shown in the video
Expression
$z_0=f\left(\frac12,\frac12\right)$
Explanation
Evaluate the function at the specified base point.
Justification
The lecturer says the first thing to figure out is $z_0$ , the height at the point.
Shown in the video
Expression
$2-\left(\frac12\right)^2-\left(\frac12\right)^2=\frac32$
Explanation
Substitute and simplify to obtain the numerical height.
Justification
Explicit arithmetic shown on the board.
Shown in the video

Conclusion

For the example, $z_0=\frac32$ .

Computing the partial derivatives in the example

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $f_x\left(\frac12,\frac12\right)=-2\left(\frac12\right)=-1$ and $f_y\left(\frac12,\frac12\right)=-2\left(\frac12\right)=-1$ .
Audio
Observation
The lecturer says the partial with respect to $x$ is $-2x$ evaluated at $x=\frac12$ , and similarly the partial with respect to $y$ is $-2y$ evaluated at $y=\frac12$ .

Uncertainties

The differentiation step from $2-x^2-y^2$ to $-2x$ and $-2y$ is stated orally and shown only in evaluated form on the board.

Numerical verification

Steps

Expression
$f_x(x,y)=-2x$
Explanation
Differentiate $2-x^2-y^2$ with respect to $x$ while treating $y$ as constant.
Justification
Stated by the lecturer as “the partial with respect to $x$ … is going to be the same thing as $-2x$ .
Shown in the video
Expression
$f_x\left(\frac12,\frac12\right)=-2\left(\frac12\right)=-1$
Explanation
Evaluate the $x$ -partial at the base point.
Justification
Explicit board computation.
Shown in the video
Expression
$f_y(x,y)=-2y$
Explanation
Differentiate $2-x^2-y^2$ with respect to $y$ while treating $x$ as constant.
Justification
Stated by the lecturer as “if I take the partial with respect to $y$ here, this is $-2y$ .
Shown in the video
Expression
$f_y\left(\frac12,\frac12\right)=-2\left(\frac12\right)=-1$
Explanation
Evaluate the $y$ -partial at the base point.
Justification
Explicit board computation.
Shown in the video

Conclusion

At $\left(\frac12,\frac12\right)$ , both partial derivatives equal $-1$ .

Assembling the example tangent plane

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $z=\frac32-1\left(x-\frac12\right)-1\left(y-\frac12\right)$ .
Audio
Observation
The lecturer says, “Take those three points, plug it in, and finally I get the formula …” and reads the same expression.

Numerical verification

Steps

Expression
$f_x\left(\frac12,\frac12\right)(x-\tfrac12)+f_y\left(\frac12,\frac12\right)(y-\tfrac12)+z_0=z$
Explanation
Start from the general tangent-plane formula specialized to the example point.
Justification
This is the boxed formula established earlier in the lecture.
Shown in the video
Expression
$(-1)\left(x-\frac12\right)+(-1)\left(y-\frac12\right)+\frac32=z$
Explanation
Substitute the computed values $f_x=-1$ , $f_y=-1$ , and $z_0=\frac32$ .
Justification
Direct substitution of the three previously computed quantities.
Shown in the video
Expression
$z=\frac32-1\left(x-\frac12\right)-1\left(y-\frac12\right)$
Explanation
Rewrite the equation with $z$ isolated on the left.
Justification
Algebraic rearrangement shown on the board.
Shown in the video

Conclusion

The tangent plane for the example is $z=\frac32-1\left(x-\frac12\right)-1\left(y-\frac12\right)$ .

Derivation of the Example's Tangent Plane Equation

Clear evidence

Shown in the video

Evidence

Formula
Observation
The sequence of equations on the screen shows the step-by-step derivation.

Proof

Steps

Expression
$z_0 = f\left(\frac{1}{2}, \frac{1}{2}\right) = 2 - \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{3}{2}$
Explanation
Evaluate the function $f(x, y) = 2 - x^2 - y^2$ at the point ( $1/2$ , $1/2$ ) to find the z-coordinate of the point of tangency.
Justification
Definition of the function and point evaluation.
Shown in the video
Expression
$f_x\left(\frac{1}{2}, \frac{1}{2}\right) = -2\left(\frac{1}{2}\right) = -1$
Explanation
Calculate the partial derivative with respect to x, which is -2x, and evaluate it at $x = 1/2$ .
Justification
Rules of partial differentiation.
Shown in the video
Expression
$f_y\left(\frac{1}{2}, \frac{1}{2}\right) = -2\left(\frac{1}{2}\right) = -1$
Explanation
Calculate the partial derivative with respect to y, which is -2y, and evaluate it at $y = 1/2$ .
Justification
Rules of partial differentiation.
Shown in the video
Expression
$z = \frac{3}{2} - 1\left(x - \frac{1}{2}\right) - 1\left(y - \frac{1}{2}\right)$
Explanation
Substitute the calculated values of $z_0$ , $f_x$ , $f_y$ , $x_0$ , and $y_0$ into the general tangent plane formula.
Justification
Substitution into the general formula.
Shown in the video

Conclusion

The equation of the tangent plane for the given function at the specified point is $z = 3/2 - 1(x - 1/2) - 1(y - 1/2)$ .

Worked examples · 3

Visual example: tangent plane to a bell-shaped surface

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, "in this example, I have the graph of a function," identifies a red point on the graph at $x_0,y_0$ with height $z_0$ , and describes the tangent plane as kissing the graph and approximating it nearby.
Diagram
Observation
A yellow-green bell-shaped surface in 3D coordinates is shown with a red point and a semi-transparent tangent plane.
Audio
Observation
He then says the tangent plane can be moved to another location to get a different tangent plane, and at the top the tangent plane is parallel to the $xy$ -plane.

Uncertainties

The explicit formula for the bell-shaped surface is not shown in the sampled visual evidence; only the qualitative shape and the general description as a graph of a function are clear.

Problem

Use a displayed surface $z=f(x,y)$ to understand what a tangent plane at a point means and how it changes with the point.

Given

A bell-shaped surface representing the graph of a function of two variables.
A red point on the surface with coordinates $(x_0,y_0,z_0)$ .
A semi-transparent plane passing through that point.

Goal

Identify the geometric meaning of the tangent plane and compare tangent planes at different points on the surface.

Steps

Expression
Explanation
Choose a specific point on the graph, marked in red, with coordinates $(x_0,y_0,z_0)$ .
Justification
The lecturer explicitly introduces the red point as the spot on the graph where the tangent plane will be taken.
Shown in the video
Expression
Explanation
Draw a plane through that point so that it locally matches the surface, described as "kissing" the graph there.
Justification
This is the lecturer's intuitive definition of the tangent plane in the example.
Shown in the video
Expression
Explanation
Interpret the plane as a nearby approximation to the graph around the chosen point.
Justification
The lecturer states that nearby the tangent plane is a close value and a good approximation for the graph.
Shown in the video
Expression
Explanation
Move the point of tangency to another location on the surface and note that the tangent plane changes.
Justification
The lecturer says that at a different location one gets a different tangent plane.
Shown in the video
Expression
Explanation
At the top of the surface, observe that the tangent plane is horizontal, i.e. parallel to the $xy$ -plane.
Justification
The lecturer explicitly compares this to a horizontal tangent line in single-variable calculus when the derivative is zero.
Shown in the video

Answer

The example shows a tangent plane as the local linear surface through a point of the graph; different points yield different tangent planes, and at the top of the displayed surface the tangent plane is horizontal.

Verification

Verification is visual and conceptual within the clip: the plane passes through the marked point, appears tangent to the surface, and becomes horizontal at the top as stated by the lecturer.

Tangent plane to $f(x,y)=2-x^2-y^2$ at $\left(\frac12,\frac12\right)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board introduces `Ex: $f(x)=2-x^2-y^2$ at ( $1/2$ , $1/2$ )` and then computes $z_0$ , $f_x$ , $f_y$ , and the final plane equation.
Audio
Observation
The lecturer says the example function is $2-x^2-y^2$ at the point $x_0=\frac12$ , $y_0=\frac12$ , then works through the substitutions.

Uncertainties

The displayed notation ` $f(x)$ ` conflicts with the two-variable expression and speech; the worked mathematics clearly treats it as $f(x,y)$ .

Problem

Find the tangent plane / linear approximation of the surface $z=2-x^2-y^2$ at the point $\left(\frac12,\frac12\right)$ .

Given

$f(x,y)=2-x^2-y^2$
Base point $(x_0,y_0)=\left(\frac12,\frac12\right)$
General formula $f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+z_0=z$

Goal

Compute $z_0$ , $f_x(x_0,y_0)$ , $f_y(x_0,y_0)$ , and write the explicit tangent-plane equation.

Steps

Expression
$z_0=f\left(\frac12,\frac12\right)=2-\left(\frac12\right)^2-\left(\frac12\right)^2=\frac32$
Explanation
Evaluate the function at the base point to get the height of the surface.
Justification
Direct substitution into the given formula for $f$ .
Shown in the video
Expression
$f_x\left(\frac12,\frac12\right)=-2\left(\frac12\right)=-1$
Explanation
Compute the partial derivative with respect to $x$ and evaluate it at the base point.
Justification
The lecturer states the $x$ -partial is $-2x$ , then substitutes $x=\frac12$ .
Shown in the video
Expression
$f_y\left(\frac12,\frac12\right)=-2\left(\frac12\right)=-1$
Explanation
Compute the partial derivative with respect to $y$ and evaluate it at the base point.
Justification
The lecturer states the $y$ -partial is $-2y$ , then substitutes $y=\frac12$ .
Shown in the video
Expression
$z=\frac32-1\left(x-\frac12\right)-1\left(y-\frac12\right)$
Explanation
Substitute the three computed quantities into the tangent-plane formula and solve for $z$ .
Justification
Uses the general formula derived earlier in the lecture.
Shown in the video

Answer

$z=\frac32-1\left(x-\frac12\right)-1\left(y-\frac12\right)$

Verification

The result matches the final boxed board expression at the end of the clip.

Finding the Tangent Plane of a Paraboloid

Clear evidence

Shown in the video

Evidence

Formula
Observation
The entire example problem and its solution are written on the screen.

Problem

Find the equation of the tangent plane to the surface defined by $f(x, y) = 2 - x^2 - y^2$ at the point ( $1/2$ , $1/2$ ).

Given

Function: $f(x, y) = 2 - x^2 - y^2$
Point of tangency: ( $x_0$ , $y_0$ ) = ( $1/2$ , $1/2$ )

Goal

Determine the equation of the tangent plane at the given point.

Steps

Expression
$z_0 = \frac{3}{2}$
Explanation
Calculate the z-coordinate of the point of tangency.
Justification
Evaluating the function at the given x and y coordinates.
Shown in the video
Expression
$f_x = -1, f_y = -1$
Explanation
Calculate the partial derivatives with respect to x and y at the given point.
Justification
Differentiating the function and evaluating at the given point.
Shown in the video
Expression
$z = \frac{3}{2} - 1\left(x - \frac{1}{2}\right) - 1\left(y - \frac{1}{2}\right)$
Explanation
Construct the final equation using the general formula.
Justification
Substituting the known values into the tangent plane equation.
Shown in the video

Answer

$z = \frac{3}{2} - 1\left(x - \frac{1}{2}\right) - 1\left(y - \frac{1}{2}\right)$

Verification

The steps follow the standard procedure for finding a tangent plane using partial derivatives.

Visual events · 12

Opening title card

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A chalkboard-style title card displays the words "TANGENT PLANES".

Objects

Black chalkboard background
Title text "TANGENT PLANES"

Changes

The title text appears on screen during the opening seconds.

Invariants

No mathematical formula is shown yet.
The segment functions as a topic introduction.

Interpretation

The clip announces the subject as tangent planes before moving to the lecturer and example.

Animated example of a tangent plane to a surface

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A 3D coordinate system with axes $x,y,z$ shows a yellow-green bell-shaped surface and a semi-transparent gray plane touching it at a red point.
Audio
Observation
The lecturer explains that the red point is on the graph at $(x_0,y_0)$ with height $z_0$ , and the tangent plane kisses the graph there and approximates it nearby.

Uncertainties

The exact algebraic formula for the surface is not visible in the sampled evidence.

Objects

Coordinate axes $x,y,z$
Bell-shaped surface $z=f(x,y)$
Red point of tangency
Semi-transparent tangent plane

Changes

The tangent plane is shown intersecting the surface at the highlighted point.
The lecturer gestures toward the plane while describing local approximation.

Invariants

The red point remains the chosen point of tangency during this portion.
The surface remains the graph of a two-variable function.

Interpretation

The picture illustrates the geometric meaning of a tangent plane: a flat surface passing through a point of the graph and matching the graph locally.

Changing the point of tangency changes the tangent plane

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The tangent plane shifts upward along the surface and ends as a horizontal plane at the top point.
Audio
Observation
The lecturer says he could move the tangent plane around and get a different tangent plane, and at the top it is parallel to the $xy$ -plane.

Objects

Same bell-shaped surface
Moving tangent plane
Top point of the surface
Horizontal plane at the peak

Changes

The plane is repositioned from a side point to the top of the surface.
At the top, the plane becomes horizontal.

Invariants

The underlying surface stays the same.
The plane always represents a tangent plane at whichever point is selected.

Interpretation

The animation demonstrates point-dependence of tangent planes and the special case of a horizontal tangent plane at the maximum-like top point.

On-screen statement of the goal

Clear evidence

Shown in the video

Evidence

Formula
Observation
Text appears reading: Goal: A plane that meets at $z_0 = f(x_0, y_0)$ and is "close" to $z = f(x, y)$ "near by".
Audio
Observation
The lecturer verbally restates the two-part goal: meeting at the point and being close nearby.

Uncertainties

The quotation marks around "close" and "near by" signal that the condition is intentionally informal at this stage.

Objects

Text block with the goal statement
Lecturer speaking beside the text

Changes

The visual focus shifts from the 3D example to a textual formulation of the problem.

Invariants

The same point $(x_0,y_0,z_0)$ and surface $z=f(x,y)$ remain the objects of study.

Interpretation

The clip converts the geometric intuition into a two-condition target definition for the tangent plane.

Diagram of a general plane with normal vector and in-plane displacement

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A blue plane appears with a pink/magenta vector labeled $\vec n$ perpendicular to it and a green vector from $P_0(x_0,y_0,z_0)$ to $P(x,y,z)$ lying in the plane.
Formula
Observation
The equation $\vec{n}\cdot \overrightarrow{P_0P}=0$ is displayed above the diagram.
Formula
Observation
Later the expanded equation $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ is added.

Objects

Blue plane
Fixed point $P_0(x_0,y_0,z_0)$
Generic point $P(x,y,z)$
Normal vector $\vec n$
Displacement vector $\overrightarrow{P_0P}$
Equation text

Changes

The diagram introduces the general plane equation in vector form.
The formula is then expanded into component form.

Invariants

The normal vector remains perpendicular to the plane.
The displacement vector remains in the plane from $P_0$ to $P$ .

Interpretation

The picture supplies the geometric basis for the algebraic plane formulas used to seek a tangent plane.

Geometric picture of the plane equation

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A blue plane is shown with axes $x,y,z$ , a pink normal vector $\vec n$ , a red point $P_0(x_0,y_0,z_0)$ , and a green point $P(x,y,z)$ connected by a vector.

Objects

blue plane
pink vector $\vec n$
point $P_0(x_0,y_0,z_0)$
point $P(x,y,z)$
coordinate axes $x,y,z$

Changes

The diagram is present only at the beginning and then disappears while the algebra remains on screen.

Invariants

The normal vector is depicted as perpendicular to the plane.
$P_0$ is shown on the plane and $P$ is another point on the plane.

Interpretation

The picture encodes the condition $\vec n\cdot\overline{P_0P}=0$ : every displacement vector between two points of the plane is orthogonal to the normal.

Board-text progression from plane equation to partial derivative

Clear evidence

Shown in the video

Evidence

Caption evidence
Observation
Text appears in sequence: Choose $c=-1$ ; then $a(x-x_0)+b(y-y_0)+z_0=z$ with label A linear function $L(x,y)$ ; then Plug in $y=y_0$ ; then $a(x-x_0)+z_0=z$ ; then Linear approximation with $a=f_x(x_0,y_0)$ .

Objects

equation lines
labels under equations

Changes

The board first shows the general plane equation.
It then adds the normalization $c=-1$ .
Next it rewrites the plane as a linear function $L(x,y)$ .
Then it substitutes $y=y_0$ and simplifies.
Finally it identifies $a$ as $f_x(x_0,y_0)$ .

Invariants

The base point $(x_0,y_0,z_0)$ remains fixed throughout.
The structure stays linear in the displayed variables.

Interpretation

The visual sequence tracks the logical reduction from a three-dimensional plane equation to a two-variable linear approximation and then to a one-variable slice that reveals the meaning of the coefficient $a$ .

Board build-up of the coefficient identification

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A chalkboard-style layout accumulates the general equation, the two substitution steps, and the identifications $a=f_x(x_0,y_0)$ and $b=f_y(x_0,y_0)$ .

Objects

General equation $a(x-x_0)+b(y-y_0)+z_0=z$
Line `Plug in $y = y_0$ `
Reduced equation $a(x-x_0)+z_0=z$
Line `Linear approximation with $a = f_x(x_0,y_0)$ `
Line `Plug in $x = x_0$ `
Reduced equation $b(y-y_0)+z_0=z$
Line `Linear approximation with $b = f_y(x_0,y_0)$ `

Changes

The board first shows the full two-variable linear form.
It then adds the $y=y_0$ reduction and identifies $a$ .
Next it adds the $x=x_0$ reduction and identifies $b$ .
Finally the intermediate lines are cleared and replaced by the consolidated formula.

Invariants

The base point $(x_0,y_0)$ remains fixed throughout.
The constant term $z_0$ remains in the equation.

Interpretation

The visual sequence encodes the logic that each coefficient is found by freezing the other variable and applying the one-variable derivative rule.

Consolidated tangent-plane formula on screen

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The board clears previous text and displays a single boxed equation at the top left.

Objects

Boxed equation $f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+z_0=z$

Changes

Earlier intermediate lines disappear.
The final formula is isolated in a box for emphasis.

Invariants

The structure remains linear in $(x-x_0)$ and $(y-y_0)$ with constant term $z_0$ .

Interpretation

The boxing marks the transition from derivation to the usable general result.

Geometric visualization of the tangent plane

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A 3D plot appears with axes $x,y,z$ , a downward-opening surface, a translucent plane, a red point, a yellow curve with yellow tangent line, and a blue/cyan curve with blue/cyan tangent line.
Audio
Observation
The lecturer explicitly ties the yellow slice to fixed $y_0$ and the blue slice to fixed $x_0$ , saying both tangent lines lie in the tangent plane.

Uncertainties

The second slice color is spoken as blue but appears cyan/blue-green on screen.

Objects

Surface graph of a two-variable function
Translucent tangent plane
Red point of tangency
Yellow slice curve at fixed $y=y_0$
Yellow tangent line in the $x$ -direction
Blue/cyan slice curve at fixed $x=x_0$
Blue/cyan tangent line in the $y$ -direction
Coordinate axes $x,y,z$

Changes

The diagram is introduced after the algebraic formula is established.
The speaker points out first the yellow slice and then the blue/cyan slice.
The visual emphasis is on both tangent lines lying in the same plane.

Invariants

All highlighted curves and lines pass through the same red base point.
The plane remains the common object containing both directional tangent lines.

Interpretation

The picture shows that the tangent plane is determined by the two coordinate-direction tangent lines obtained from partial derivatives.

Worked example written step by step on the board

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The 3D graphic disappears and the board returns with the example statement and sequential computations.

Uncertainties

The example header visibly reads ` $f(x)=2-x^2-y^2$ `, although the rest of the work treats it as a function of two variables.

Objects

Example header `Ex: $f(x)=2-x^2-y^2$ at ( $1/2$ , $1/2$ )`
Computation of $z_0$
Computation of $f_x(1/2,1/2)$
Computation of $f_y(1/2,1/2)$
Final plane equation

Changes

The example statement appears first.
Then $z_0=\frac32$ is added.
Then the two partial-derivative evaluations are added.
Finally the explicit tangent-plane equation is written.

Invariants

The base point remains $\left(\frac12,\frac12\right)$ throughout the example.
The general formula from earlier is being instantiated rather than changed.

Interpretation

The board turns the abstract formula into a concrete numerical tangent-plane computation.

Scene Transition

Clear evidence

Shown in the video

Evidence

Animation
Observation
The video cuts from a blackboard background to an office setting.

Objects

Blackboard with math formulas
Office desk
Monitor displaying 'SUBSCRIBE'

Changes

Background changes from blackboard to office
Speaker's clothing changes

Invariants

The speaker remains the same person

Interpretation

This visual transition marks the end of the mathematical instruction and the beginning of the video's outro and call to action.

Misconceptions · 6

Mistaking the informal "close" condition for a finished definition

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the tangent plane should be close to the graph nearby, but adds, "what exactly I mean by that, let's put a pin in it and we're going to come back to it later."
Formula
Observation
The on-screen goal puts "close" and "near by" in quotation marks.

Misconception

One might think the displayed goal already gives the full rigorous definition of a tangent plane because it says the plane is "close" to the graph nearby.

Clarification

In this clip the lecturer explicitly postpones the exact meaning of "close"; the statement is only a loose goal, not yet a formal criterion.

Thinking the tangency point alone determines the tangent plane

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the point is easy to figure out by plugging $x_0,y_0,z_0$ into the formula, but the harder part is the $a,b,c$ , i.e. how to come up with the normal vector to the graph of the function.

Misconception

One might assume that once the point $(x_0,y_0,z_0)$ is known, the tangent plane is automatically determined.

Clarification

The video emphasizes that the point supplies only the location term in the plane equation; the essential remaining task is finding the normal vector components $a,b,c$ .

Assuming one can always normalize $c$ to $-1$

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says the method always works unless $c=0$ and asks viewers to think why that case is not relevant for a tangent plane to a differentiable function.

Uncertainties

The clip raises the issue but does not fully explain it before moving on.

Misconception

One might think the coefficient of $z-z_0$ can always be made equal to $-1$ in a plane equation.

Clarification

That step requires $c\neq 0$ . The speaker explicitly flags $c=0$ as the exceptional case and treats it as irrelevant in the tangent-plane setting for a differentiable function, without giving the full justification in this excerpt.

Confusing the tangent plane with the original nonlinear surface

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says what this is is a linear function, with $z$ depending on $x$ and $y$ in a linear way, variables raised only to the power of one with multiplicative constants out front.

Misconception

One might think the expression for the plane is itself the original complicated function.

Clarification

The video distinguishes the nonlinear function $z=f(x,y)$ from its linear approximation $L(x,y)=a(x-x_0)+b(y-y_0)+z_0$ , which is built to approximate the surface near the point.

Notation mismatch in the example header

Clear evidence

Shown in the video

Evidence

Formula
Observation
The example header on the board reads `Ex: $f(x)=2-x^2-y^2$ at ( $1/2$ , $1/2$ )`.
Audio
Observation
The lecturer nevertheless speaks and computes as if the function has two inputs, evaluating $f(\frac12,\frac12)$ and taking both $f_x$ and $f_y$ .

Misconception

One might read ` $f(x)=2-x^2-y^2$ ` literally and think the example concerns a one-variable function.

Clarification

Within the clip, the spoken explanation and all computations treat the example as the two-variable function $f(x,y)=2-x^2-y^2$ ; the displayed ` $f(x)$ ` is inconsistent notation.

Color naming versus on-screen hue

Approximate timing

Shown in the video

Evidence

Audio
Observation
The lecturer calls the second slice the “blue curve”.
Diagram
Observation
On screen that curve and its tangent line appear cyan/blue-green rather than pure blue.

Uncertainties

This is a visual description mismatch, not a mathematical error.

Misconception

A viewer may expect the second slice to be exactly blue from the narration.

Clarification

Mathematically the important point is that it is the slice with $x=x_0$ ; the on-screen color is better described as cyan/blue-green even though the lecturer says blue.

Concept relations · 16

Intuitive definition of a tangent plane to a surface $z=f(x,y)$ → Tangent planes generalize tangent lines to several variables

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer directly calls tangent planes the multivariable generalization of a tangent line from single-variable calculus.

Generalizes

Explanation

The tangent-plane concept in the clip is presented as extending the one-variable tangent-line idea to graphs of functions of two variables.

Tangent plane approximates the graph locally → Intuitive definition of a tangent plane to a surface $z=f(x,y)$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer's definition of the tangent plane includes the claim that nearby it is a good approximation to the graph.

Proof dependency

Explanation

The intuitive definition of the tangent plane relies on the local-approximation property stated in the clip.

Loose goal for constructing a tangent plane → Point of tangency on the graph

Clear evidence

Shown in the video

Evidence

Formula
Observation
The goal text explicitly uses $z_0=f(x_0,y_0)$ .
Audio
Observation
The lecturer says the plane must meet the graph at the specific point $(x_0,y_0,z_0)$ .

Application

Explanation

The loose goal statement applies the point-of-tangency condition as the first requirement for the desired plane.

Vector equation of a plane using a normal vector → Intuitive definition of a tangent plane to a surface $z=f(x,y)$

Clear evidence

Shown in the video

Evidence

Audio
Observation
After defining the goal for a tangent plane, the lecturer says tangent planes are planes and reviews the general formula for a plane.
Formula
Observation
The displayed plane equations are introduced in the same sequence as the method for finding a tangent plane.

Application

Explanation

The general plane equation is the algebraic tool the video brings in to construct the tangent plane once the correct normal vector is found.

Vector equation of a plane using a normal vector → Component form of the plane equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says, "From this particular equation that we have, we can expand it" and then gives the component formula.
Formula
Observation
The screen shows both $\vec{n}\cdot \overrightarrow{P_0P}=0$ and $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ in sequence.

Equivalent

Explanation

The component equation is obtained by expanding the dot-product equation, so the two forms express the same plane condition.

At the top of the example surface, the tangent plane is horizontal → Tangent planes generalize tangent lines to several variables

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer compares the horizontal tangent plane at the top to taking a derivative equal to zero and getting a horizontal tangent line in single-variable calculus.

Special case

Explanation

The horizontal tangent plane at the top is presented as the multivariable analogue of the one-variable case where the derivative vanishes.

Vector form of a plane equation → Scalar component form of a plane

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board moves from $\vec n\cdot\overline{P_0P}=0$ to $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .

Equivalent

Explanation

The scalar equation is the coordinate form of the vector dot-product condition for a plane.

Choosing $c=-1$ to solve the plane for $z$ → Tangent plane as a linear function $L(x,y)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
After Choose $c=-1$ , the equation becomes $a(x-x_0)+b(y-y_0)+z_0=z$ , labeled A linear function $L(x,y)$ .

Application

Explanation

Normalizing $c$ to $-1$ is the algebraic step that lets the plane be solved for $z$ and viewed as a linear function of $x$ and $y$ .

Tangent plane as a linear function $L(x,y)$ → Restricting the linear approximation to $y=y_0$

Clear evidence

Shown in the video

Evidence

Formula
Observation
Substituting $y=y_0$ changes $a(x-x_0)+b(y-y_0)+z_0=z$ into $a(x-x_0)+z_0=z$ .
Audio
Observation
Speaker says this is like restricting to the plane where $y=y_0$ and that the result is the equation of a line.

Special case

Explanation

Fixing $y=y_0$ turns the two-variable linear approximation into a one-variable linear function, i.e. a line in the $xz$ -slice.

Restricting the linear approximation to $y=y_0$ → Coefficient $a$ equals the $x$ -partial derivative

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says in single-variable calculus $a$ was the derivative at the point, and here it is the partial of $f$ with respect to $x$ evaluated at $(x_0,y_0)$ .
Caption evidence
Observation
Linear approximation with $a=f_x(x_0,y_0)$ .

Application

Explanation

The one-variable tangent-line rule is applied to the slice $y=y_0$ , yielding the partial derivative formula for $a$ .

Tangent plane as a linear function $L(x,y)$ → $z_0 = f(x_0, y_0)$ , $\quad z = f(x,y)$

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says tangent planes are an approximation for a nonlinear function, replacing it with a linear thing that is simpler to understand.

Application

Explanation

The linear function $L(x,y)$ is precisely the device used to meet the stated goal of finding a plane close to $z=f(x,y)$ near $(x_0,y_0)$ .

Partial derivatives as slopes of coordinate slices → General form of the tangent-plane equation

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer derives $a$ and $b$ by fixing one variable and using the single-variable derivative rule.
Formula
Observation
The board links `Plug in $y = y_0$ ` to $a=f_x(x_0,y_0)$ and `Plug in $x = x_0$ ` to $b=f_y(x_0,y_0)$ .

Proof dependency

Explanation

The general plane formula depends on identifying its coefficients through the slice interpretation of partial derivatives.

Find an answer · 20

What is a tangent plane to the graph of a function of two variables?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer introduces tangent planes and gives the intuitive definition using a point on the graph and local approximation.

Knowledge points

Intuitive definition of a tangent plane to a surface $z=f(x,y)$
Point of tangency on the graph
Tangent plane approximates the graph locally

How does a tangent plane generalize the tangent line from single-variable calculus?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer explicitly compares tangent planes to tangent lines from single-variable calculus.

Knowledge points

Tangent planes generalize tangent lines to several variables
At the top of the example surface, the tangent plane is horizontal

What two conditions define the goal when looking for a tangent plane?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The on-screen goal text states the two conditions for the desired plane.
Audio
Observation
The lecturer verbally explains the same two-part goal.

Knowledge points

Loose goal for constructing a tangent plane
Point of tangency on the graph

What is the equation of a plane using a normal vector and a point on the plane?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The clip displays both the vector and component forms of the plane equation.
Audio
Observation
The lecturer explains the dot-product construction and then expands it.

Knowledge points

Vector equation of a plane using a normal vector
Component form of the plane equation
Expanding the dot-product plane equation into component form

Why does the plane equation use $\vec n\cdot \overrightarrow{P_0P}=0$ ?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says a true normal must have zero dot product with any vector lying in the plane.
Diagram
Observation
The diagram shows $\vec n$ perpendicular to the plane and $\overrightarrow{P_0P}$ inside it.

Knowledge points

Vector equation of a plane using a normal vector
A normal vector has zero dot product with every vector in the plane

Once the point of tangency is known, what remains to determine the tangent plane?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the point is easy to plug in, but the harder part is finding $a,b,c$ , the normal vector to the graph.

Knowledge points

Component form of the plane equation
Thinking the tangency point alone determines the tangent plane

Why is the tangent plane at the top of the example surface horizontal?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says the tangent plane at the top is parallel to the $xy$ -plane and compares it to a horizontal tangent line when the derivative is zero.
Diagram
Observation
The plane at the peak is drawn horizontally.

Knowledge points

At the top of the example surface, the tangent plane is horizontal
Changing the point of tangency changes the tangent plane

Does the tangent plane change if you choose a different point on the surface?

Clear evidence

Shown in the video

Evidence

Audio
Observation
The lecturer says moving to a different location gives a different tangent plane.

Knowledge points

Different points generally give different tangent planes
Changing the point of tangency changes the tangent plane

Why does the video choose $c=-1$ in the plane equation?

Clear evidence

Shown in the video

Evidence

Caption evidence
Observation
Choose $c=-1$ .
Audio
Observation
Speaker explains dividing through so the coefficient of $z-z_0$ becomes $-1$ .

Knowledge points

Choosing $c=-1$ to solve the plane for $z$
Scalar component form of a plane
Exceptional case $c=0$ for normalizing the plane equation

How is a tangent plane rewritten as a linear function $L(x,y)$ ?

Clear evidence

Shown in the video

Evidence

Formula
Observation
$a(x-x_0)+b(y-y_0)+z_0=z$ .
Caption evidence
Observation
A linear function $L(x,y)$ .

Knowledge points

Tangent plane as a linear function $L(x,y)$
From general plane equation to a linear function of $x$ and $y$

What happens when you plug in $y=y_0$ in the tangent-plane formula?

Clear evidence

Shown in the video

Evidence

Caption evidence
Observation
Plug in $y=y_0$ .
Formula
Observation
$a(x-x_0)+z_0=z$ .

Knowledge points

Restricting the linear approximation to $y=y_0$
Determining $a$ by restricting to $y=y_0$

Why is the coefficient $a$ equal to $f_x(x_0,y_0)$ ?

Clear evidence

Shown in the video

Evidence

Audio
Observation
Speaker says the answer in single-variable calculus was that $a$ was the derivative at the point.
Caption evidence
Observation
Linear approximation with $a=f_x(x_0,y_0)$ .

Knowledge points

Coefficient $a$ equals the $x$ -partial derivative
After fixing $y=y_0$ , the coefficient $a$ is determined by one-variable tangent-line logic
Determining $a$ by restricting to $y=y_0$

Coverage and review notes

Covered · Opening title card announces the topic "TANGENT PLANES"; no formula is shown yet.

Covered · The lecturer introduces tangent planes through a 3D example, defines the point of tangency, explains local approximation, and shows that moving the point changes the plane, including the horizontal case at the top.

Covered · The clip states the two-part goal for a tangent plane and explicitly leaves the meaning of "close" informal for later.

Covered · The lecturer reviews the general equation of a plane in vector and component form, explains why the dot product is zero, expands the formula, and identifies the normal vector as the difficult remaining ingredient for a tangent plane.

Covered · Opening board shows the goal, the vector plane equation, the scalar plane equation, and the geometric diagram.

Covered · Diagram disappears; speaker introduces the simplifying assumption about $c$ and the nonzero condition.

Covered · Text Choose $c=-1$ appears and the speaker explains rescaling the plane equation.

Covered · The plane is rewritten as $a(x-x_0)+b(y-y_0)+z_0=z$ and identified as a linear function $L(x,y)$ .

Covered · Earlier text clears; the board shows Plug in $y=y_0$ and the reduced equation $a(x-x_0)+z_0=z$ while the speaker interprets it as a line.

Covered · The board adds Linear approximation with $a=f_x(x_0,y_0)$ and the speaker connects this to single-variable derivatives.

Covered · The clip derives the two coefficients by fixing one variable at a time and applying the single-variable derivative rule.

Covered · The intermediate board content is cleared and the consolidated tangent-plane formula is displayed.

Covered · A 3D diagram interprets the formula geometrically using coordinate-direction slice curves and their tangent lines.

Covered · Transition interval: the geometric diagram disappears and the lecture moves to a worked example.

Covered · The example computes $z_0$ , both partial derivatives, and the explicit tangent plane for $2-x^2-y^2$ at $(1/2,1/2)$ .

Covered · Mathematical content showing the general formula and a worked example for finding a tangent plane.

Covered · Outro segment with no new mathematical content; speaker asks for likes, comments, and subscriptions.

Explore the knowledge in this video

Reviewed subject paths

Tangent plane

How to find the TANGENT PLANE | Linear approximation of multi-variable functions

The explanation, unpacked.

Chapters

Learning script

Knowledge cards

Tangent plane: intuitive meaning

Point of tangency on z=f(x,y)z=f(x,y)z=f(x,y)

Loose goal for finding a tangent plane

Vector equation of a plane

Component equation of a plane

What remains hard for a tangent plane

Special case: horizontal tangent plane

Plane from a point and a normal vector

Scalar equation of a plane

Why choose c=−1c=-1c=−1

Tangent plane as a linear function L(x,y)L(x,y)L(x,y)

Restricting to the slice y=y0y=y_0y=y0​

From tangent line to partial derivative

General linear model for a tangent plane

Partial derivatives come from one-variable slices

Final tangent-plane formula

Geometric meaning of the tangent plane

Worked example setup

Example: compute the base height

Example: compute the partial derivatives

Example: explicit tangent plane

General Formula for Tangent Plane

Example: Tangent Plane Calculation

Detailed learning notes

f

x0x_0x0​

y0y_0y0​

z0z_0z0​

x

y

z

n⃗\vec{n}n

P0P_0P0​

P

P0P→\overrightarrow{P_0P}P0​P​

a

Intuitive definition of a tangent plane to a surface z=f(x,y)z=f(x,y)z=f(x,y)

Point of tangency on the graph

Loose goal for constructing a tangent plane

Vector equation of a plane using a normal vector

Component form of the plane equation

Vector form of a plane equation

Scalar component form of a plane

Choosing c=−1c=-1c=−1 to solve the plane for zzz

Tangent plane as a linear function L(x,y)L(x,y)L(x,y)

Restricting the linear approximation to y=y0y=y_0y=y0​

Coefficient aaa equals the xxx-partial derivative

General form of the tangent-plane equation

Tangent planes generalize tangent lines to several variables

Tangent plane approximates the graph locally

Different points generally give different tangent planes

At the top of the example surface, the tangent plane is horizontal

A normal vector has zero dot product with every vector in the plane

Exceptional case c=0c=0c=0 for normalizing the plane equation

After fixing y=y0y=y_0y=y0​, the coefficient aaa is determined by one-variable tangent-line logic

Multivariable linear approximation inherits the single-variable derivative rule

Coordinate-slice tangent lines lie in the tangent plane

Expanding the dot-product plane equation into component form

From general plane equation to a linear function of xxx and yyy

Determining aaa by restricting to y=y0y=y_0y=y0​

Deriving the xxx-coefficient from the y=y0y=y_0y=y0​ slice

Deriving the yyy-coefficient from the x=x0x=x_0x=x0​ slice

Combining the two slice results into the tangent-plane equation

Computing the base height z0z_0z0​ in the example

Computing the partial derivatives in the example

Assembling the example tangent plane

Derivation of the Example's Tangent Plane Equation

Visual example: tangent plane to a bell-shaped surface

Tangent plane to f(x,y)=2−x2−y2f(x,y)=2-x^2-y^2f(x,y)=2−x2−y2 at (12,12)\left(\frac12,\frac12\right)(21​,21​)

Finding the Tangent Plane of a Paraboloid

Opening title card

Animated example of a tangent plane to a surface

Changing the point of tangency changes the tangent plane

On-screen statement of the goal

Diagram of a general plane with normal vector and in-plane displacement

Point of tangency on $z=f(x,y)$

Why choose $c=-1$

Tangent plane as a linear function $L(x,y)$

Restricting to the slice $y=y_0$

$x_0$

$y_0$

$z_0$

$\vec{n}$

$P_0$

$\overrightarrow{P_0P}$

Intuitive definition of a tangent plane to a surface $z=f(x,y)$

Choosing $c=-1$ to solve the plane for $z$

Tangent plane as a linear function $L(x,y)$

Restricting the linear approximation to $y=y_0$

Coefficient $a$ equals the $x$ -partial derivative

Exceptional case $c=0$ for normalizing the plane equation

After fixing $y=y_0$ , the coefficient $a$ is determined by one-variable tangent-line logic

From general plane equation to a linear function of $x$ and $y$

Determining $a$ by restricting to $y=y_0$

Deriving the $x$ -coefficient from the $y=y_0$ slice

Deriving the $y$ -coefficient from the $x=x_0$ slice

Computing the base height $z_0$ in the example

Tangent plane to $f(x,y)=2-x^2-y^2$ at $\left(\frac12,\frac12\right)$

Assuming one can always normalize $c$ to $-1$

Intuitive definition of a tangent plane to a surface $z=f(x,y)$ → Tangent planes generalize tangent lines to several variables

Tangent plane approximates the graph locally → Intuitive definition of a tangent plane to a surface $z=f(x,y)$

Vector equation of a plane using a normal vector → Intuitive definition of a tangent plane to a surface $z=f(x,y)$

Choosing $c=-1$ to solve the plane for $z$ → Tangent plane as a linear function $L(x,y)$

Tangent plane as a linear function $L(x,y)$ → Restricting the linear approximation to $y=y_0$

Restricting the linear approximation to $y=y_0$ → Coefficient $a$ equals the $x$ -partial derivative

Tangent plane as a linear function $L(x,y)$ → $z_0 = f(x_0, y_0)$ , $\quad z = f(x,y)$

Why does the plane equation use $\vec n\cdot \overrightarrow{P_0P}=0$ ?

Why does the video choose $c=-1$ in the plane equation?

How is a tangent plane rewritten as a linear function $L(x,y)$ ?

What happens when you plug in $y=y_0$ in the tangent-plane formula?

Why is the coefficient $a$ equal to $f_x(x_0,y_0)$ ?