Chapter H.1, Problem 55E

To determine

To find: the slope of tangent line

Answer to Problem 55E

The slope of the tangent line at $\theta =\pi /3,\pi$ of horizontal is $\left(r,\theta \right)=\left(\frac{3}{2},\frac{\pi }{3}\right);\left(\frac{3}{2},\frac{5\pi }{3}\right);\left(0,\pi \right)$ and vertical is $\left(r,\theta \right)=\left(2,0\right);\left(\frac{1}{2},\frac{2\pi }{3}\right);\left(\frac{1}{2},\frac{4\pi }{3}\right)$ .

Explanation of Solution

Given:

  $r=1+\cos \theta$

Calculation:

Given curve is $r=1+\cos \theta$

  $\begin{array}{l}\frac{dy}{dx}=\frac{\frac{dr}{d\theta }\cdot \sin \theta +r\cos \theta }{\frac{dr}{d\theta }\cos \theta -r\sin \theta }\\ =\frac{\left(-\sin \theta \right)\sin \theta +\left(1+\cos \theta \right)\cos \theta }{\left(-\sin \theta \right)\cos \theta -\left(1+\cos \theta \right)\sin \theta }\\ \left(\begin{array}{l}\because \frac{dr}{d\theta }=\frac{d}{d\theta }\left(1+\cos \theta \right)\\ =0-\sin \theta \\ =-\sin \theta \end{array}\right)\\ =\frac{-{\sin }^2\theta +\cos \theta +{\cos }^2\theta }{-\sin \theta \cdot \cos \theta -\sin \theta -\cos \theta \cdot \sin \theta }\\ \Rightarrow \text{Slope of the tangent }m=\frac{\cos 2\theta +\cos \theta }{-2\sin \theta \cdot \cos \theta -\sin \theta }\mathrm{......}\left(1\right)\\ \left(\because \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta \right)\end{array}$

(a) If the tangent is horizontal, then its

  $\begin{array}{l}\text{Slope = 0}\\ \Rightarrow m=0\\ \Rightarrow \cos 2\theta +\cos \theta =0\left(\text{By}\left(1\right)\right)\\ \Rightarrow 2\cos \frac{3\theta }{2}\cdot \cos \frac{\theta }{2}=0\left(\because \cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cdot \cos \left(\frac{C-D}{2}\right)\right)\\ \Rightarrow \cos \frac{3\theta }{2}=0\left(\text{or}\right)\cos \frac{\theta }{2}=0\\ \Rightarrow \frac{3\theta }{2}=\frac{\pi }{2}\frac{\theta }{2}=\frac{\pi }{2}\\ \Rightarrow \theta =\frac{\pi }{3}\theta =\pi \end{array}$

Here

  $\begin{array}{l}0\le \theta \le 2\pi \Rightarrow \text{other value of }\theta =2\pi -\frac{\pi }{3}\left(<2\pi \right)\\ =\frac{5\pi }{3}\\ \therefore \theta =\frac{\pi }{3},\frac{5\pi }{3},\pi \\ \therefore r=1+\cos \theta \\ =1+\cos \frac{\pi }{3},1+\cos \frac{5\pi }{3},1+\cos \pi \\ =1+\frac{1}{2},1+\cos \left(2\pi -\frac{\pi }{3}\right),1+\left(-1\right)\\ =\frac{3}{2},1+\cos \left(-\frac{\pi }{3}\right),0\\ =\frac{3}{2},1+\frac{1}{2},0\\ =\frac{3}{2},\frac{3}{2},0\\ \therefore \left(r,\theta \right)=\left(\frac{3}{2},\frac{\pi }{3}\right);\left(\frac{3}{2},\frac{5\pi }{3}\right);\left(0,\pi \right)\end{array}$

At these points on the curve, slope of the tangent is zero (i.e., tangent is horizontal).

(b) If the tangent is vertical, then its slope is $\frac{1}{0}$ (undefined)

  $\begin{array}{l}\Rightarrow m=\frac{1}{0}\\ \Rightarrow \text{By}\left(1\right),\frac{\cos 2\theta +\cos \theta }{-2\sin \theta \cos \theta -\sin \theta }=\frac{1}{0}\\ \Rightarrow -2\sin \theta \cos \theta -\sin \theta =0\\ \Rightarrow \sin \theta =-2\sin \theta \cos \theta \\ \Rightarrow \sin \theta \left(1+2\cos \theta \right)=0\\ \Rightarrow \sin \theta =0\left(\text{or}\right)\cos \theta =\frac{-1}{2}\\ \Rightarrow \theta =0\left(\text{or}\right)\theta =\pi -\frac{\pi }{3}=\frac{2\pi }{3}\left(<2\pi \right)\\ \text{And }\theta =2\pi -\frac{2\pi }{3}\left(<2\pi \right)\\ \Rightarrow \theta =\frac{4\pi }{3}\\ \therefore \theta =0,\frac{2\pi }{3},\frac{4\pi }{3}\end{array}$

  $\begin{array}{l}\therefore r=1+\cos \theta =1+\cos 0;1+\cos \frac{2\pi }{3};1+\cos \frac{4\pi }{3}\\ =1+1;1+\left(\frac{-1}{2}\right);1+\left(\frac{-1}{2}\right)\\ =2,\frac{1}{2},\frac{1}{2}\\ \therefore \left(r,\theta \right)=\left(2,0\right);\left(\frac{1}{2},\frac{2\pi }{3}\right);\left(\frac{1}{2},\frac{4\pi }{3}\right)\end{array}$

At these points on the curve, the tangent line is vertical.

Conclusion:

Therefore, the slope of the tangent line at $\theta =\pi /3,\pi$ of horizontal is $\left(r,\theta \right)=\left(\frac{3}{2},\frac{\pi }{3}\right);\left(\frac{3}{2},\frac{5\pi }{3}\right);\left(0,\pi \right)$ and vertical is $\left(r,\theta \right)=\left(2,0\right);\left(\frac{1}{2},\frac{2\pi }{3}\right);\left(\frac{1}{2},\frac{4\pi }{3}\right)$ .