Chapter H.1, Problem 56E

To determine

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

Answer to Problem 56E

Therefore, the horizontal points are

  $\left(r,\theta \right)=\left(0,\frac{\pi }{2}\right),\left(r,\theta \right)=\left(\frac{1}{2},\frac{\pi }{6}\right),\left(r,\theta \right)=\left(\frac{1}{2},\frac{5\pi }{6}\right)\text{ and }\left(r,\theta \right)=\left(2,\frac{3\pi }{2}\right)$ and the vertical points are $\left(r,\theta \right)=\left(0,\frac{\pi }{2}\right),\left(r,\theta \right)=\left(\frac{3}{2},\frac{7\pi }{6}\right)\text{ and }\left(r,\theta \right)=\left(\frac{3}{2},\frac{11\pi }{6}\right)$ .

Explanation of Solution

Given:

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

Calculation:

Consider the curve $r=1-\sin \theta$

The objective is to find the points on the given curve where the tangent line is horizontal or vertical.

Every polar point $\left(r,\theta \right)$ has Cartesian coordinates $\left(x,y\right)$ such that $x=r\cos \theta$ and $y=r\sin \theta$ .

Rewrite the polar coordinate $x=r\cos \theta$ as,

  $\begin{array}{l}x=r\cos \theta \\ =\left(1-\sin \theta \right)\cdot \cos \theta \\ =\cos \theta -\sin \theta \cos \theta \\ =\cos \theta -\frac{1}{2}\sin 2\theta \end{array}$

Differentiate with respect to “ $\theta$ ”,

  $\begin{array}{l}\frac{dx}{d\theta }=\left(-\sin \theta \right)-\frac{1}{2}\cdot 2\cos 2\theta \\ =-\sin \theta -\cos 2\theta \end{array}$

Rewrite the polar coordinate $y=r\sin \theta$ as,

  $\begin{array}{l}y=r\sin \theta \\ =\left(1-\sin \theta \right)\cdot sin\theta \\ =sin\theta -si{n}^2\theta \end{array}$

Differentiate with respect to “ $\theta$ ”,

  $\begin{array}{l}\frac{dy}{d\theta }=\cos \theta -2\sin \theta \cos \theta \\ =\cos \theta -\sin 2\theta \end{array}$

The slope of the tangent line is,

  $\begin{array}{l}\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}\\ =\frac{\cos \theta -\sin 2\theta }{-\sin \theta --\cos 2\theta }\end{array}$

To find the points where the curve at horizontal tangent, set $\frac{dy}{dx}=0$ .

  $\begin{array}{l}\frac{dy}{dx}=0\\ \frac{\cos \theta -\sin 2\theta }{-\sin \theta --\cos 2\theta }=0\\ \cos \theta -2\sin \theta \cos \theta =0\left(\text{Since, }\sin 2\theta =2\sin \theta \cos \theta \right)\\ \cos \theta \left(1-2\sin \theta \right)=0\end{array}$

Either,

  $\begin{array}{l}\cos \theta =0;\text{ or 1}-2\sin \theta =0\\ \theta ={\cos }^{-1}\left(0\right)\text{ or }\sin \theta =\frac{1}{2}\\ \theta =n_1\pi +\frac{\pi }{2}\text{ or }\theta =2n_1\pi +\frac{\pi }{6},2n_2\pi +\frac{5\pi }{6}\end{array}$

As the period of sin and cos function is $2\pi$ .

Hence, the value of $\theta$ are $\theta =\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6}$ and $\frac{3\pi }{2}$ .

Substitute these values of $\theta$ in $r=1-\sin \theta$ .

At $\theta =\frac{\pi }{6}$

The value of the curve is,

  $\begin{array}{l}r=1-\sin \theta \\ r=1-\sin \left(\frac{\pi }{6}\right)\\ =1-\frac{1}{2}\\ =\frac{1}{2}\end{array}$

Hence, the value of the curve is $r=\frac{1}{2}$

At $\theta =\frac{\pi }{2}$

The value of the curve is,

  $\begin{array}{l}r=1-\sin \theta \\ r=1-\sin \left(\frac{\pi }{2}\right)\\ =1-1\\ =0\end{array}$

Hence, the value of the curve is $r=0$ .

At $\theta =\frac{5\pi }{6}$

The value of the curve is,

  $\begin{array}{l}r=1-\sin \theta \\ r=1-\sin \left(\frac{5\pi }{6}\right)\\ =1-\frac{1}{2}\\ =\frac{1}{2}\end{array}$

Hence, the value of the curve is $r=\frac{1}{2}$ .

At $\theta =\frac{3\pi }{2}$

The value of the curve is,

  $\begin{array}{l}r=1-\sin \theta \\ r=1-\sin \left(\frac{3\pi }{2}\right)\\ =1-\left(-1\right)\\ =2\end{array}$

Hence, the value of the curve is $r=2$ .

Hence, the horizontal points are

  $\left(r,\theta \right)=\left(0,\frac{\pi }{2}\right),\left(r,\theta \right)=\left(\frac{1}{2},\frac{\pi }{6}\right),\left(r,\theta \right)=\left(\frac{1}{2},\frac{5\pi }{6}\right)\text{ and }\left(r,\theta \right)=\left(2,\frac{3\pi }{2}\right)$ .

To find the points where the curve has vertical tangent, set $\frac{dx}{dy}=0$ . $\begin{array}{l}\frac{dx}{dy}=0\\ -\sin \theta -\cos 2\theta =0\\ -\sin \theta -\left(1-2{\sin }^2\theta \right)=0\left(\text{Since, cos}2\theta =1-2{\sin }^2\theta \right)\\ 2{\sin }^2\theta -\sin \theta -1=0\\ 2{\sin }^2\theta -2\sin \theta +\sin \theta -1=0\\ \left(2\sin \theta +1\right)\left(\sin \theta -1\right)=0\left(\text{Factor form}\right)\\ \text{Either, }2\sin \theta +1=0\text{ or }\sin \theta -1=0\\ 2\sin \theta =-1\text{ or }\sin \theta =1\\ \sin \theta =-\frac{1}{2}\text{ or }\theta ={\sin }^{-1}\left(1\right)\\ \theta =2n_1\pi -\frac{\pi }{6},2n_2\pi +\frac{7\pi }{6}\text{ or }\theta =2n_1\pi +\frac{\pi }{2}\end{array}$

As the period of sin and cos function is $2\pi$ .

Hence, the values of $\theta$ are $\theta =\frac{\pi }{2},\frac{7\pi }{6}and\frac{11\pi }{6}$ .

At $\theta =\frac{\pi }{2}$

The value of the curve is,

  $\begin{array}{l}r=1-\sin \theta \\ r=1-\sin \left(\frac{\pi }{2}\right)\\ =1-1\\ =0\end{array}$

Here, the value of the curve is 0.

At $\theta =\frac{7\pi }{6}$

The value of the curve is,

  $\begin{array}{l}r=1-\sin \theta \\ r=1-\sin \left(\frac{7\pi }{6}\right)\\ =1-\left(-\frac{1}{2}\right)\\ =\frac{3}{2}\end{array}$

Hence, the value of the curve is $r=\frac{3}{2}$ .

At $\theta =\frac{11\pi }{6}$

The value of the curve is,

  $\begin{array}{l}r=1-\sin \theta \\ r=1-\sin \left(\frac{11\pi }{6}\right)\\ =1-\left(-\frac{1}{2}\right)\\ =\frac{3}{2}\end{array}$

Thus, the value curve is $r=\frac{3}{2}$ .

Hence, the vertical points are $\left(r,\theta \right)=\left(0,\frac{\pi }{2}\right),\left(r,\theta \right)=\left(\frac{3}{2},\frac{7\pi }{6}\right)\text{ and }\left(r,\theta \right)=\left(\frac{3}{2},\frac{11\pi }{6}\right)$ .

The graph of the function is shown below:

  

Conclusion:

Therefore, the horizontal points are

  $\left(r,\theta \right)=\left(0,\frac{\pi }{2}\right),\left(r,\theta \right)=\left(\frac{1}{2},\frac{\pi }{6}\right),\left(r,\theta \right)=\left(\frac{1}{2},\frac{5\pi }{6}\right)\text{ and }\left(r,\theta \right)=\left(2,\frac{3\pi }{2}\right)$ and the vertical points are $\left(r,\theta \right)=\left(0,\frac{\pi }{2}\right),\left(r,\theta \right)=\left(\frac{3}{2},\frac{7\pi }{6}\right)\text{ and }\left(r,\theta \right)=\left(\frac{3}{2},\frac{11\pi }{6}\right)$ .