Chapter 10.1, Problem 13E

$\text{(a)}$

To determine

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

Answer to Problem 13E

The answer:

  $\frac{dy}{dx}=\sin t$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=\tan t,\\ y=\sec t\end{array}$

Calculation:

The given equations are

  $\begin{array}{c}x=\tan t,\\ y=\sec t\end{array}$

Hence, know that,

  $\begin{array}{c}\tan t=\sin t/\cos t\\ \sec t=1/\cos t\end{array}$

Then,

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

Because,

  ${\sin }^{2}t+{\cos }^{2}t=1$

Consider that to use the formulas for parametric differentiation:

  $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

To use the previous result,

  $\begin{array}{c}\frac{dy}{dx}=\frac{\sec t\tan t}{{\sec }^{2}t}\\ =\frac{\tan t}{\sec t}\\ =\frac{\sin t/\cos t}{1/\cos t}\\ =\frac{\sin t\cos t}{\cos t}\\ =\sin t\end{array}$

Therefore, the required value of $\frac{dy}{dx}=\sin t$ .

$\text{(b)}$

To determine

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

Answer to Problem 13E

The answer:

  $\frac{d^{2}y}{d{x}^2}={\cos }^{3}t$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=\tan t,\\ y=\sec t\end{array}$

Calculation:

The given equations are

  $\begin{array}{c}x=\tan t,\\ y=\sec t\end{array}$

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,

  $y^{\prime }=\sin t$

Then,

  $d{y}^{\prime }/dt=\cos t$

Since from part $\text{(a)}$

  $dx/dt={\sec }^{2}t$

Know that,

  $\sec t=1/\cos t$

To use the previous result,

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

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