Chapter 10, Problem 14RE

a)

To determine

To find: The possible points at which the tangent to the curve $x=4\cos t$ and $y=9\sin t$ is horizontal.

Answer to Problem 14RE

The points at which the horizontal tangent occurs for the curve $x=4\cos t$ and $y=9\sin t$ are $\left(0,-9\right)$ and $\left(0,9\right)$ .

Explanation of Solution

Given information: The given equations of the curves are $x=4\cos t$ and $y=9\sin t$ .

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 $y=9\sin t$ with respect to $t$ .

  $\frac{dy}{dt}=9\cos t$

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

  $\begin{array}{c}9\cos t=0\\ \cos t=0\\ t={\cos }^{-1}\left(0\right)\\ t=\frac{\pi }{2}\end{array}$

Substitute $\frac{\pi }{2}$ for $t$ in the equation $y=9\sin t$ .

  $\begin{array}{c}y=9\sin \frac{\pi }{2}\\ =9\end{array}$

Substitute $\frac{3\pi }{2}$ for $t$ in the equation $y=9\sin t$ .

  $\begin{array}{c}y=9\sin \frac{3\pi }{2}\\ =-9\end{array}$

Therefore, the points at which the horizontal tangent occurs for the curve are $\left(0,-9\right)$ and $\left(0,9\right)$ .

b)

To determine

To find: The possible points at which the tangent to the curve $x=4\cos t$ and $y=9\sin t$ is vertical.

Answer to Problem 14RE

The points at which the vertical tangent occurs for the curve $x=4\cos t$ and $y=9\sin t$ are $\left(-4,0\right)$ and $\left(4,0\right)$ .

Explanation of Solution

Given information: The given equations of the curve are $x=4\cos t$ and $y=9\sin t$ .

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:

Differentiate the equation $x=4\cos t$ with respect to $t$ .

  $\frac{dx}{dt}=-4\sin t$

At $t=0$ , and $t=\pi$

  $-4\sin t=0$ .

Substitute $0$ for $t$ in the equation $x=4\cos t$ .

  $\begin{array}{c}x=4\cos 0\\ =4\end{array}$

Substitute $\pi$ for $t$ in the equation $x=4\cos t$ .

  $\begin{array}{c}x=4\cos \pi \\ =-4\end{array}$

Therefore, the points at which the vertical tangent occur are $\left(-4,0\right)$ and $\left(4,0\right)$ .