Chapter H.1, Problem 58E

To determine

To find: the curves $r=a\sin \theta$ and $r=a\cos \theta$ intersect at right angles.

Answer to Problem 58E

The two curves intersect right angles.

Explanation of Solution

Given:

The curves are $r=a\sin \theta$ and $r=a\cos \theta$

Calculation:

Given curve are $r=a\sin \theta$ and $r=a\cos \theta$ both the curves intersect when $\theta =\frac{\pi }{4}$

For the first curve $r=a\sin \theta$

Now we use the product rule of the parametric curves.

  $\begin{array}{l}m=\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }\\ =\frac{\frac{dr}{d\theta }\sin \theta +rcos\theta }{\frac{dr}{d\theta }\cos \theta -r\sin \theta }\left[\because \frac{dy}{d\theta }=\frac{dr}{d\theta }\sin \theta +rcos\theta ,\frac{dx}{d\theta }=\frac{dr}{d\theta }\cos \theta -r\sin \theta \right]\end{array}$

And given that

  $\begin{array}{l}r=a\sin \theta \\ \Rightarrow r\cos \theta =a\sin \theta \cos \theta ,r\sin \theta =a{\sin }^2\theta \\ =\frac{\frac{d\left(a\sin \theta \right)}{d\theta }\sin \theta +a\sin \theta \cos \theta }{\frac{d\left(a\sin \theta \right)}{d\theta }\cos \theta -a{\sin }^2\theta }\\ =\frac{a\cos \theta \sin \theta +a\sin \theta \cos \theta }{a{\cos }^2\theta -a{\sin }^2\theta }\\ =\frac{2a\cos \theta \sin \theta }{a\left({\cos }^2\theta -{\sin }^2\theta \right)}\\ =\frac{\sin 2\theta }{\cos 2\theta }\left[\begin{array}{l}\because 2\cos \theta \sin \theta =\sin 2\theta \\ {\cos }^2\theta -{\sin }^2\theta =\cos 2\theta \end{array}\right]\\ =\tan 2\theta \end{array}$

For the second curve $r=a\cos \theta$

Now we use the product rule of the parametric curves.

  $\begin{array}{l}{m}_2=\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }\\ =\frac{\frac{dr}{d\theta }\sin \theta +rcos\theta }{\frac{dr}{d\theta }\cos \theta -r\sin \theta }\end{array}$

We have

  $\begin{array}{l}r=a\cos \theta \\ \Rightarrow r\cos \theta =a{\cos }^2\theta ,r\sin \theta =a\sin \theta \cos \theta \\ =\frac{\frac{d\left(a\cos \theta \right)}{d\theta }\sin \theta +a\cos \theta \cos \theta }{\frac{d\left(a\cos \theta \right)}{d\theta }\cos \theta -a\cos \theta \sin \theta }\\ =\frac{-a\sin \theta \sin \theta +a{\cos }^2\theta }{-a\sin \theta \cos \theta -a\cos \theta \sin \theta }\\ =\frac{-a{\sin }^2\theta +a{\cos }^2\theta }{-2a\sin \theta \cos \theta }\\ =\frac{a\left({\cos }^2\theta -{\sin }^2\theta \right)}{-a\sin 2\theta }\left[\because 2\sin \theta \cos \theta =\sin 2\theta \right]\\ =\frac{a\cos 2\theta }{-a\sin 2\theta }\left[\because {\cos }^2\theta -{\sin }^2\theta =\cos 2\theta \right]\\ =\tan 2\theta \cdot \frac{1}{\tan 2\theta }\\ =-1\end{array}$

So, the two curves intersect right angles.

Conclusion:

Therefore, the the two curves intersect right angles.