Chapter 11, Problem 49RE

(a)

To determine

To find: The magnitude of the velocity vector at $t=4$ .

Answer to Problem 49RE

The magnitude of the velocity vector at $t=4$ is $\frac{104}{5}$ .

Explanation of Solution

Given:

The position of the particle at time $t\ge 0$ is given by,

  $\begin{array}{c}x\left(t\right)=t^2-2\\ y\left(t\right)=\frac{2}{5}{t}^3\end{array}$

Calculation:

Write the position vector as follows.

  $r\left(t\right)=\langle t^2-2,\frac{2}{5}{t}^3\rangle$

Find the velocity vector of the particle as follows.

  $\begin{array}{c}v\left(t\right)=\frac{d}{dt}\left(r\left(t\right)\right)\\ =\frac{d}{dt}\langle t^2-2,\frac{2}{5}{t}^3\rangle \\ =\langle 2t,\frac{6}{5}{t}^2\rangle \end{array}$

Find the speed of the particle when $t=4$ .

  $\begin{array}{c}v{\left(t\right)}_{t=4}=\langle 2\cdot 4,\frac{6}{5}{\left(4\right)}^2\rangle \\ =\langle 8,\frac{96}{5}\rangle \end{array}$

Find the magnitude of the velocity vector at $t=4$ .

  $\begin{array}{c}\left|v{\left(t\right)}_{t=4}\right|=\sqrt{{\left(8\right)}^2+{\left(\frac{96}{5}\right)}^2}\\ =\frac{104}{5}\end{array}$

Therefore, the magnitude of the velocity vector at $t=4$ is $\frac{104}{5}$ .

(b)

To determine

To find: The total distance traveled by the particle from $t=0$ to $t=4$ .

Answer to Problem 49RE

The total distance traveled by the particle from $t=0$ to $t=4$ is $\frac{4144}{135}$ .

Explanation of Solution

Given:

The position of the particle at time $t\ge 0$ is given by,

  $\begin{array}{c}x\left(t\right)=t^2-2\\ y\left(t\right)=\frac{2}{5}{t}^3\end{array}$

Calculation:

Find the total distance traveled by the particle from $t=0$ to $t=4$ as follows.

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{4}{\int }}\left|v\left(t\right)\right|}={\displaystyle \underset{0}{\overset{4}{\int }}\sqrt{{\left(2t\right)}^2+{\left(\frac{6}{5}{t}^2\right)}^2}dt}\\ d\left(t\right)={\displaystyle \underset{0}{\overset{4}{\int }}\frac{2}{5}t\sqrt{25+9t^2}dt}\\ ={\left[\frac{2}{135}{\left(25+9t^2\right)}^{3/2}\right]}_0^4\\ =\left[\frac{2}{135}{\left(25+96\right)}^{3/2}-{\left(25\right)}^{3/2}\right]\end{array}$

Simplify further.

  $\begin{array}{c}d\left(t\right)=\left[\frac{2}{135}{\left(11\right)}^3-125\right]\\ =\frac{4144}{135}\end{array}$

Therefore, the total distance traveled by the particle from $t=0$ to $t=4$ is $\frac{4144}{135}$ .

(c)

To determine

To find: The $\frac{dy}{dx}$ as a function of $x$ .

Answer to Problem 49RE

The $\frac{dy}{dx}$ as a function of $x$ is $\frac{dy}{dx}=\frac{3}{5}\sqrt{x+2}$ .

Explanation of Solution

Given:

The position of the particle at time $t\ge 0$ is given by,

  $\begin{array}{c}x\left(t\right)=t^2-2\\ y\left(t\right)=\frac{2}{5}{t}^3\end{array}$

Calculation:

Find $\frac{dy}{dx}$ as follows.

  $\begin{array}{c}\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\\ =\frac{6t^2/5}{2t}\\ =\frac{3}{5}t\end{array}$

It is given that $x=t^2-2$ . So,

  $t=\sqrt{x+2}$

Find $\frac{dy}{dx}$ as a function of $x$ .

  $\frac{dy}{dx}=\frac{3}{5}\sqrt{x+2}$

Therefore, the $\frac{dy}{dx}$ as a function of $x$ is $\frac{dy}{dx}=\frac{3}{5}\sqrt{x+2}$ .