Chapter 11.1, Problem 7E

(a)

To determine

To calculate: The value of $\frac{dy}{dx}$ .

Answer to Problem 7E

The required value of $\frac{dy}{dx}$ is $\sqrt{3}$ .

Explanation of Solution

Given information:

The parametric equations: $x=4\sin t$ , $y=2\cos t$

Calculation:

The parametric equations are $x=4\sin t$ , $y=2\cos t$ .

Differentiate the above equations with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left[4\sin t\right]\\ =4\cos t\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left[2\cos t\right]\\ =-2\sin t\end{array}$

Now,

  $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{-2\sin t}{4\cos t}\\ =-\frac{1}{2}\tan t\end{array}$

Hence, the required value of $\frac{dy}{dx}$ is $-\frac{1}{2}\tan t$ .

(b)

To determine

To calculate: The value of $\frac{d^{2}y}{d{x}^2}$ .

Answer to Problem 7E

The required value of $\frac{d^{2}y}{d{x}^2}$ is $-\frac{1}{8}{\sec }^{3}t$ .

Explanation of Solution

Given information:

The parametric equations: $x=4\sin t$ , $y=2\cos t$

Calculation:

The parametric equations are $x=4\sin t$ , $y=2\cos t$ .

Differentiate the above equations with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left[4\sin t\right]\\ =4\cos t\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left[2\cos t\right]\\ =-2\sin t\end{array}$

Now,

  $\begin{array}{c}\frac{dy}{dx}=y^{\prime }\\ =\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{-2\sin t}{4\cos t}\\ =-\frac{1}{2}\tan t\end{array}$

Again,

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\left(\frac{\frac{d{y}^{\prime }}{dt}}{\frac{dx}{dt}}\right)\\ =\frac{-\frac{1}{2}{\sec }^{2}t}{4\cos t}\\ =-\frac{1}{2}{\sec }^{2}t\times \frac{1}{4}\sec t\\ =-\frac{1}{8}{\sec }^{3}t\end{array}$

Hence, the required value of $\frac{d^{2}y}{d{x}^2}$ is $-\frac{1}{8}{\sec }^{3}t$ .