How are the differential elements dx and dy related to the arc length element ds?

Conditions

• $ds$ is the infinitesimal arc length along the curve.
• $\vec{T} = (\cos \alpha, \cos \beta)$ is the unit tangent vector.
• The curve is smooth enough to have a well-defined tangent.

Reasoning, step by step

1. Visualize a small segment of the curve with length $ds$.
2. Project this segment horizontally to obtain $dx$.
3. Project this segment vertically to obtain $dy$.
4. Identify the angles $\alpha$ and $\beta$ that the segment (tangent) makes with the axes.
5. Apply trigonometry to relate the projections to the hypotenuse $ds$: $dx = ds \cos \alpha$ and $dy = ds \cos \beta$.
6. Use these relations to substitute into line integrals.

Example

The script states: 'Zooming in on a small segment of the curve reveals a right triangle formed by the arc length element ds, horizontal projection dx, and vertical projection dy. Based on trigonometry, we establish the bridge between the two types of integrals: dx equals cos α times ds, and dy equals cos β times ds.'

Common misconceptions

• Assuming $dx^2 + dy^2 = ds$ instead of $dx^2 + dy^2 = ds^2$.
• Confusing the direction cosines with the slopes of the curve.
• Thinking that $dx$ and $dy$ are always positive; they can be negative depending on the direction of traversal.

Watch the explanation

Explore next

• What is the geometric significance of the right triangle formed by dx, dy, and ds?
• How does this relationship facilitate the conversion between line integral types?
• Why is the unit tangent vector necessary for defining these projections?