Chapter 11.1, Problem 9E

(a)

To determine

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

Answer to Problem 9E

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

Explanation of Solution

Given information:

The parametric equations: $x=-\sqrt{t+1}$ , $y=\sqrt{3}t$

Calculation:

The parametric equations are $x=-\sqrt{t+1}$ , $y=\sqrt{3}t$ .

Differentiate the above equations with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left[-\sqrt{t+1}\right]\\ =-\frac{1}{2\sqrt{t+1}}\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left[\sqrt{3}t\right]\\ =\frac{3}{2\sqrt{3}t}\end{array}$

Now,

  $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{\frac{3}{2\sqrt{3}t}}{\frac{-1}{2\sqrt{t+1}}}\\ =-\frac{3\sqrt{t+1}}{\sqrt{3}t}\\ =-\sqrt{3+\frac{3}{t}}\end{array}$

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

(b)

To determine

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

Answer to Problem 9E

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

Explanation of Solution

Given information:

The parametric equations: $x=-\sqrt{t+1}$ , $y=\sqrt{3}t$

Calculation:

The parametric equations are $x=-\sqrt{t+1}$ , $y=\sqrt{3}t$ .

Differentiate the above equations with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left[-\sqrt{t+1}\right]\\ =-\frac{1}{2\sqrt{t+1}}\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left[\sqrt{3}t\right]\\ =\frac{3}{2\sqrt{3}t}\end{array}$

Now,

  $\begin{array}{c}\frac{dy}{dx}=y^{\prime }\\ =\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{\frac{3}{2\sqrt{3}t}}{\frac{-1}{2\sqrt{t+1}}}\\ =-\frac{3\sqrt{t+1}}{\sqrt{3}t}\\ =-\sqrt{3+\frac{3}{t}}\end{array}$

Again,

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\left(\frac{\frac{d{y}^{\prime }}{dt}}{\frac{dx}{dt}}\right)\\ =\left(\frac{\frac{3}{\left(2t^2\sqrt{3+\frac{3}{t}}\right)}}{\frac{-1}{\left(2\sqrt{t+1}\right)}}\right)\\ =-\frac{3\sqrt{t+1}}{t^2\sqrt{3+\frac{3}{t}}}\\ =-\frac{\sqrt{3}}{t^{\frac{3}{2}}}\end{array}$

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