Chapter 11.1, Problem 26E

(a)

To determine

To calculate: The points at which the tangent line to the curve is horizontal.

Answer to Problem 26E

The required points are $\left(-2,4\right)$ and $\left(-2,-2\right)$ .

Explanation of Solution

Given information:

The parametric equations: $x=-2+3\cos t$ , $y=1+3\sin t$

Calculation:

The parametric equations are $x=-2+3\cos t$ , $y=1+3\sin t$ .

Differentiate the above equations with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left[-2+3\cos t\right]\\ =-3\sin t\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left[1+3\sin t\right]\\ =3\cos t\end{array}$

Now, $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{3\cos t}{-3\sin t}\\ =-\cot t\end{array}$

Also, $\begin{array}{c}\frac{dx}{dy}=\frac{\frac{dx}{dt}}{\frac{dy}{dt}}\\ =\frac{-3\sin t}{3\cos t}\\ =-\tan t\end{array}$

To find the point at which the tangent line to the curve is horizontal, use the expression $\frac{dy}{dx}$ .

  $\begin{array}{c}\frac{dy}{dx}=-\cot t\\ =0\end{array}$

So, $t=\frac{\pi }{2},\frac{3\pi }{2}$

At $t=\frac{\pi }{2}$ ; $\begin{array}{c}\left(x,y\right)=\left[\left(-2+0\right),\left(1+3\right)\right]\\ =\left(-2,4\right)\end{array}$

And, at $t=\frac{3\pi }{2}$ ; $\begin{array}{c}\left(x,y\right)=\left[\left(-2+0\right),\left(1-3\right)\right]\\ =\left(-2,-2\right)\end{array}$

Hence, the required points are $\left(-2,4\right)$ and $\left(-2,-2\right)$ .

(b)

To determine

To calculate: The points at which the tangent line to the curve is vertical.

Answer to Problem 26E

The required points are $\left(1,1\right)$ and $\left(-5,1\right)$ .

Explanation of Solution

Given information:

The parametric equations: $x=-2+3\cos t$ , $y=1+3\sin t$

Calculation:

The parametric equations are $x=-2+3\cos t$ , $y=1+3\sin t$ .

Differentiate the above equations with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left[-2+3\cos t\right]\\ =-3\sin t\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left[1+3\sin t\right]\\ =3\cos t\end{array}$

Now, $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{3\cos t}{-3\sin t}\\ =-\cot t\end{array}$

Also, $\begin{array}{c}\frac{dx}{dy}=\frac{\frac{dx}{dt}}{\frac{dy}{dt}}\\ =\frac{-3\sin t}{3\cos t}\\ =-\tan t\end{array}$

To find the point at which the tangent line to the curve is horizontal, use the expression $\frac{dy}{dx}$ .

  $\begin{array}{c}\frac{dx}{dy}=-\tan t\\ =0\end{array}$

So, $t=0,\pi$

At $t=0$ ; $\begin{array}{c}\left(x,y\right)=\left[\left(-2+-3\right),\left(1+0\right)\right]\\ =\left(1,1\right)\end{array}$

And, at $t=\pi$ ; $\begin{array}{c}\left(x,y\right)=\left[\left(-2-3\right),\left(1+0\right)\right]\\ =\left(-5,1\right)\end{array}$

Hence, the required points are $\left(1,1\right)$ and $\left(-5,1\right)$ .