Chapter 10.1, Problem 26E

$\text{(a)}$

To determine

To find: The points of horizontal.

Answer to Problem 26E

The answer: $(-2,4)$ and $(-2,-2)$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=-2+3\cos t,\\ y=1+3\sin t\end{array}$

Calculation:

Since need to find the points at which the tangent line to the curve is horizontal, need to find the relative extrema of the function $y=1+3\sin t$

Know that function $y$ has a relative extremum at point $t=t_0$ .

So,

  $\frac{dy}{dt}\left(t_0\right)=0$

To use the previous result,

  $\begin{array}{c}\frac{dy}{dt}=3\cos t\\ \Rightarrow \frac{dy}{dt}\left(\frac{\pi }{2}+k\pi \right)=0\end{array}$

For

   , know that, $\cos t=\cos (t+2k\pi )$ for then the function

  $y=1+3\sin t$ has the relative extrema at points

  $\begin{array}{c}t=\pi /2\\ t=3\pi /2\end{array}$

In this exercise,

  $x=-2+3\cos t$

Then,

  $\begin{array}{c}x(\pi /2)=-2+3\cos (\pi /2)\\ =-2+3\cdot 0\\ =-2,\\ y(\pi /2)=1+3\sin (\pi /2)\\ =1+3\cdot 1\\ =4\text{, }\\ x(3\pi /2)=-2+3\cos (3\pi /2)\\ =-2+3\cdot 0\\ =-2\\ y(3\pi /2)=1+3\sin (3\pi /2)\\ =1+3\cdot (-1)\\ =-2\end{array}$

which means that the points at which the tangent line to the curve is horizontal are equal to $(-2,4)$ and $(-2,-2)$ .

Therefore, the required horizontal points are $(-2,4)$ and $(-2,-2)$ .

$\text{(b)}$

To determine

To find: The points of vertical.

Answer to Problem 26E

The answer: $(1,1)$ and $(-5,1)$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=-2+3\cos t,\\ y=1+3\sin t\end{array}$

Calculation:

Since need to find the points at which the tangent line to the curve is vertical, need to find the relative extrema of the function $x=-2+3\cos t$

Know that function $y$ has a relative extremum at point $t=t_0$ .

So,

  $\frac{dy}{dt}\left(t_0\right)=0$

To use the previous result,

  $\begin{array}{c}\frac{dx}{dt}=3\cdot (-\sin t)\\ =-3\sin t\\ \Rightarrow \frac{dx}{dt}(k\pi )=0\end{array}$

For

   , know that, $\sin t=\sin (t+2k\pi )$ for then the function

  $x=-2+3\cos t$

has the relative extrema at points

  $\begin{array}{c}t=0\\ t=\pi \end{array}$

In this exercise,

  $y=1+3\sin t$

Then,

  $\begin{array}{c}x(0)=-2+3\cdot \cos (0)\\ =-2+3\cdot 1\\ =1\text{, }\\ y(0)=1+3\cdot \sin (0)\\ =1+3\cdot 0\\ =1\\ x(\pi )=-2+3\cdot \cos (\pi )\\ =-2+3\cdot (-1)\\ =-5\\ y(\pi )=1+3\cdot \sin (\pi )\\ =1+3\cdot 0\\ =1\end{array}$

which means that the points at which the tangent line to the curve is vertical are equal to $(1,1)$ and $(-5,1)$ .

Therefore, the required vertical points are $(1,1)$ and $(-5,1)$ .