Chapter 10.1, Problem 23E

$\text{(a)}$

To determine

To find: The points of horizontal.

Answer to Problem 23E

The answer: $(2,0)$ and $(2,-2)$

Explanation of Solution

Given information:

The equations are:

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

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 $y=-1+\sin t$ .

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,

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

For

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

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

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

In this exercise,

  $x=2+\cos t$

Then,

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

which means that the points at which the tangent line to the curve is horizontal are equal to $(2,0)$ and $(2,-2)$ .

Therefore, the required horizontal points are $(2,0)$ and $(2,-2)$ .

$\text{(b)}$

To determine

To find: The points of vertical.

Answer to Problem 23E

The answer:

  $(3,-1)$ and $(1,-1)$

Explanation of Solution

Given information:

The equations are:

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

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 $y=-1+\sin t$ .

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,

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

For

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

  $x=2+\cos t$

has the relative extrema at points

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

In this exercise,

  $y=-1+\sin t$

Then,

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

which means that the points at which the tangent line to the curve is vertical are equal to $(3,-1)$ and $(1,-1)$ .

Therefore, the required vertical points are $(3,-1)$ and $(1,-1)$ .