Chapter 10.3, Problem 74E

To determine

To prove: The expression $L={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}}d\theta$ by substitution of $x=r\cos \theta$ and $y=r\sin \theta$ in the arc length formula.

Answer to Problem 74E

It is proved that $L={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}}d\theta$ by substitutions $x=r\cos \theta$ and $y=r\sin \theta$ .

Explanation of Solution

Given information: The parametric form of the arc length formula of polar curve is $L={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2}}d\theta$ .

Calculation:

Substitute $r\cos \theta$ for $x$ and $r\sin \theta$ for $y$ in ${\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2$ .

  $\begin{array}{c}{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2={\left(\frac{d\left(r\cos \theta \right)}{d\theta }\right)}^2+{\left(\frac{d\left(r\sin \theta \right)}{d\theta }\right)}^2\\ ={\left(\frac{dr}{d\theta }\cdot \cos \theta +r\left(-\sin \theta \right)\right)}^2+{\left(\frac{dr}{d\theta }\cdot \sin \theta +r\cdot \cos \theta \right)}^2\\ ={\left(\frac{dr}{d\theta }\cdot \cos \theta \right)}^2+{\left(-r\sin \theta \right)}^2-2\frac{dr}{d\theta }\cdot \cos \theta \left(r\sin \theta \right)\\ +{\left(\frac{dr}{d\theta }\cdot \sin \theta \right)}^2+{\left(r\cos \theta \right)}^2+2\frac{dr}{d\theta }\cdot \sin \theta \left(r\cdot \cos \theta \right)\\ ={\left(\frac{dr}{d\theta }\right)}^2+r^2\end{array}$

Substitute ${\left(\frac{dr}{d\theta }\right)}^2+r^2$ for ${\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2$ in the formula of arc length of the curve.

  $L={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{{\left(\frac{dr}{d\theta }\right)}^2+r^2}}d\theta$

Hence, it is proved that $L={\displaystyle {\int }_{\alpha }^{\beta }\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}}d\theta$ by substitutions $x=r\cos \theta$ and $y=r\sin \theta$ .