Chapter 10.1, Problem 7E

$\text{(a)}$

To determine

To find: The value of $dy/dx$ .

Answer to Problem 7E

The answer:

  $\frac{dy}{dx}=-\frac{1}{2}\tan t$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=4\sin t,\\ y=2\cos t\end{array}$

Calculation:

The given equations are

  $\begin{array}{c}x=4\sin t,\\ y=2\cos t\end{array}$

Consider that to use the formulas for parametric differentiation:

  $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

To use the previous result,

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

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

$\text{(b)}$

To determine

To find: The value of $d^{2}y/d{x}^2$ in terms of $t$ .

Answer to Problem 7E

The answer:

  $\frac{d^{2}y}{d{x}^2}=-\frac{1}{8}{\sec }^{3}t$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=4\sin t,\\ y=2\cos t\end{array}$

Calculation:

The given equations are

  $\begin{array}{c}x=4\sin t,\\ y=2\cos t\end{array}$

To use the formulas for parametric differentiation, see that

  $\frac{d^{2}y}{d{x}^2}=\frac{d{y}^{\prime }/dt}{dx/dt}$

Where,

  $y^{\prime }=dy/dx$

Hence from part $\text{(a)}$ know that,

  $\tan t=\sin t/\cos t\text{,}$

Then,

  $\begin{array}{c}\frac{d{y}^{\prime }}{dt}=-\frac{1}{2}\cdot \frac{\cos t\cdot \cos t-\sin t\cdot (-\sin t)}{{\cos }^{2}t}\\ =-\frac{1}{2}\cdot \frac{{\cos }^{2}t+{\sin }^{2}t}{{\cos }^{2}t}\\ =-\frac{1}{2}\cdot \frac{1}{{\cos }^{2}t}\\ =-\frac{1}{2}{\sec }^{2}t\end{array}$

Because,

  $\begin{array}{c}{\sin }^{2}t+{\cos }^{2}t=1\\ 1/\cos t=\sec t\end{array}$

To use the previous result,

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\frac{-{\sec }^{2}t/2}{4\cos t}\\ =\frac{-1}{2}\cdot \frac{1}{4}\cdot {\sec }^{2}t\cdot \sec t\\ =-\frac{1}{8}{\sec }^{3}t\end{array}$

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