What does the gradient ∇u represent for a function $u(x,y)$?

To expand (5,2)·∇u, substitute ∇$u = (u_x, u_y)$ and apply the definition of the Euclidean dot product: multiply corresponding components and add them. This yields $5u_x + 2u_y$.

Conditions: ∇$u = (u_x, u_y)$.; The dot product is defined as the sum of the products of corresponding components.

The directional derivative formula a*(∂f/∂x) + b*(∂f/∂y) structurally matches the definition of a dot product between the vector [a,b] and the vector [∂f/∂x, ∂f/∂y]. Since the second vector is defined as the gradient ∇f, the expression simplifies to w·∇f.

Conditions: The function $f(x,y)$ is differentiable.; The direction vector is $w=[a,b]$.; The gradient ∇f = [∂$f/x$, ∂$f/y$] exists.

Rewriting the PDE as a dot product with the gradient allows the equation to be interpreted as a statement about directional derivatives. Specifically, it shows that the directional derivative of u in the direction (5,2) is zero, which implies that u is constant along curves in that direction.

Conditions: The PDE is $5u_x+2u_y=0$.; ∇u is the gradient vector ($u_x$,$u_y$).

The directional derivative in the direction of a vector w is equal to the dot product of w and the gradient vector ∇f. The gradient vector contains the partial derivatives of the function as its components.

Conditions: The function $f(x,y)$ is differentiable.; The direction vector is $w=[a,b]$.; The gradient ∇f = [∂f/∂x, ∂f/∂y] exists.

Yes, the dot-product formula w·∇f generalizes naturally to higher dimensions. If the input has five variables, both the direction vector and the gradient vector simply expand to have five components, maintaining the same structural relationship.

Conditions: The function is differentiable in higher dimensions.; The direction vector and gradient are defined in the same dimensional space.