Chapter 10, Problem 13RE

a)

To determine

To find: The possible points at which the tangent to the curve $x=-\cos t$ and $y={\cos }^{2}t$ is horizontal.

Answer to Problem 13RE

The point at which the horizontal tangent occurs for the curve $x=-\cos t$ and $y={\cos }^{2}t$ is $\left(0,0\right)$ .

Explanation of Solution

Given information: The equations of the curve are $x=-\cos t$ and $y={\cos }^{2}t$ .

Formula used: The formulas of differentiation for sine and cosine function are:

  $\frac{d}{dt}\left(\sin t\right)=\cos t$ and $\frac{d}{dt}\left(\cos t\right)=-\sin t$

The condition for horizontal tangents to the curve occur, $\frac{dy}{dt}=0$ .

Calculation:

Differentiate the equation $y={\cos }^{2}t$ with respect to $t$ .

  $\frac{dy}{dt}=-2\cos t\sin t$

To find if there are any horizontal tangents, substitute $0$ for $\frac{dy}{dt}$ in the above equation.

  $-2\cos t\sin t=0$

It is evident that at $t=\frac{\pi }{2}$ , the value of $-2\cos t\sin t$ is $0$ .

Substitute $\frac{\pi }{2}$ for $t$ in the equation $x=-\cos t$ .

  $\begin{array}{c}x=-\cos \frac{\pi }{2}\\ =0\end{array}$

Substitute $\frac{\pi }{2}$ for $t$ in the equation $y={\cos }^{2}t$ .

  $\begin{array}{c}y={\cos }^2\frac{\pi }{2}\\ =0\end{array}$

Hence, the points at which the horizontal tangent occurs for the curve are $\left(0,0\right)$ .

b)

To determine

To find: The possible points at which the tangent to the curve $x=-\cos t$ and $y={\cos }^{2}t$ is vertical.

Answer to Problem 13RE

The points at which the vertical tangent occurs for the curve $x=-\cos t$ and $y={\cos }^{2}t$ does not exist.

Explanation of Solution

Given information: The given equations of the curve are $x=-\cos t$ and $y={\cos }^{2}t$ .

Formula used: The formulas of differentiation for sine and cosine function are:$\frac{d}{dt}\left(\sin t\right)=\cos t$ and $\frac{d}{dt}\left(\cos t\right)=-\sin t$

The condition for vertical tangents to occur, $\frac{dx}{dt}=0$ .

Calculation:

Divide equation $y={\cos }^{2}t$ by the equation $x=-\cos t$ .

  $\begin{array}{c}\frac{y}{x}=\frac{{\cos }^{2}t}{-\cos t}\\ =\frac{\cos t}{-1}\end{array}$

As the denominator is $1$ , the vertical tangent does not exist.

Hence, the vertical tangent to the given curves does not exist.