Chapter 11.1, Problem 13E

(a)

To determine

To find: $\frac{dy}{dx}$

Answer to Problem 13E

Therequired value of $\frac{dy}{dx}$ is $\sin t$

Explanation of Solution

Given Information:

  $x=\tan t$  ........ (1)

  $y=\sec t$  ........ (2)

Calculation:

Differentiating equation (1) and (2) with respect to $t$

  $\begin{array}{l}x=\tan t\\ \frac{dx}{dt}=\sec t\end{array}$  ........ (3)

  $\begin{array}{l}y=\sec t\\ \frac{dy}{dt}=\sec t\tan t\end{array}$  ........ (4)

Now,

  $\begin{array}{c}\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}\\ =\frac{\sec t\tan t}{{\sec }^{2}t}\\ =\frac{\tan t}{\sec t}\\ =\sin t\end{array}$

Conclusion:

Hence, the required value of $\frac{dy}{dx}$ is $\sin t$

(b)

To determine

To find: $\frac{d^{2}y}{d{x}^2}$

Answer to Problem 13E

The required value of $\frac{d^{2}y}{d{x}^2}$ is ${\cos }^{3}t$

Explanation of Solution

Given Information:

  $x=\tan t$  ........ (1)

  $y=\sec t$  ........ (2)

Formula Used:

  ${\sec }^{2}t=\frac{1}{{\cos }^{2}t}$  ........ (3)

Calculation:

Differentiating equation (1) and (2) with respect to $t$

  $\begin{array}{l}x=\tan t\\ \frac{dx}{dt}=\sec t\end{array}$  ........ (3)

  $\begin{array}{l}y=\sec t\\ \frac{dy}{dt}=\sec t\tan t\end{array}$  ........ (4)

Now,

  $\begin{array}{c}\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}\\ =\frac{\sec t\tan t}{{\sec }^{2}t}\\ =\frac{\tan t}{\sec t}\\ =\sin t\end{array}$

Again differentiating the equation

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\frac{\left(\frac{d}{dx}\right)\left(\frac{dy}{dx}\right)}{\left(\frac{dx}{dt}\right)}\\ =\frac{\cos t}{{\sec }^{2}t}\\ ={\cos }^{3}t\end{array}$

Conclusion:

Hence, the required value of $\frac{d^{2}y}{d{x}^2}$ is ${\cos }^{3}t$