How is the line integral of the second kind converted into an integral with respect to arc length using direction cosines?

Conditions

• The curve $L$ is smooth or piecewise smooth.
• $\vec{T} = (\cos \alpha, \cos \beta)$ is the unit tangent vector at a point on the curve.
• $ds$ is the differential of arc length.

Reasoning, step by step

1. Identify the unit tangent vector $\vec{T}$ at a point on the curve $L$.
2. Define the direction cosines $\cos \alpha$ and $\cos \beta$ as the components of $\vec{T}$ along the x and y axes.
3. Establish the geometric relationship between the differential elements: $dx = \cos \alpha \, ds$ and $dy = \cos \beta \, ds$.
4. Substitute these expressions into the original line integral $\int_L P dx + Q dy$.
5. Factor out $ds$ to obtain the final form: $\int_L (P \cos \alpha + Q \cos \beta) ds$.
6. Recognize that the integrand $(P \cos \alpha + Q \cos \beta)$ is the dot product $\vec{F} \cdot \vec{T}$.
7. Conclude that the integral represents the accumulation of the tangential component of the field over the arc length.

Example

The script states: 'Substituting these relationships back into the original expression allows us to transform the integral... Factoring out ds yields the general conversion formula: $\int _L P dx + Q dy$ becomes $\int _L (P \cos α + Q \cos β)$ ds.'

Common misconceptions

• Confusing the direction cosines with the coordinates of the point on the curve.
• Assuming $dx$ and $dy$ are independent of $ds$; they are projections of $ds$.
• Believing the conversion formula applies to scalar line integrals (first kind) without modification; it converts second kind to first kind.

Watch the explanation

Explore next

• What is the geometric interpretation of the term $(P \cos \alpha + Q \cos \beta)$?
• How does reversing the direction of the curve affect the value of the integral?
• Why is the unit tangent vector essential for this conversion?