Chapter 11.1, Problem 8E

(a)

To determine

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

Answer to Problem 8E

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

Explanation of Solution

Given information:

The parametric equations: $x=\cos t$ , $y=\sqrt{3}\cos t$

Calculation:

The parametric equations are $x=\cos t$ , $y=\sqrt{3}\cos t$ .

Differentiate the above equations with respect to t .

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

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left[\sqrt{3}\cos t\right]\\ =-\sqrt{3}\sin t\end{array}$

Now,

  $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{-\sqrt{3}\sin t}{-\sin t}\\ =\sqrt{3}\end{array}$

Hence, the required value of $\frac{dy}{dx}$ is $\sqrt{3}$ .

(b)

To determine

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

Answer to Problem 8E

The required value of $\frac{d^{2}y}{d{x}^2}$ is $0$ .

Explanation of Solution

Given information:

The parametric equations: $x=\cos t$ , $y=\sqrt{3}\cos t$

Calculation:

The parametric equations are $x=\cos t$ , $y=\sqrt{3}\cos t$ .

Differentiate the above equations with respect to t .

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

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left[\sqrt{3}\cos t\right]\\ =-\sqrt{3}\sin t\end{array}$

Now,

  $\begin{array}{c}\frac{dy}{dx}=y^{\prime }\\ =\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{-\sqrt{3}\sin t}{-\sin t}\\ =\sqrt{3}\end{array}$

Again, $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\left(\frac{\frac{d{y}^{\prime }}{dt}}{\frac{dx}{dt}}\right)\\ =\frac{0}{-\sin t}\\ =0\end{array}$

Hence, the required value of $\frac{d^{2}y}{d{x}^2}$ is $0$ .