Chapter 11.2, Problem 50E

(a)

To determine

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

Answer to Problem 50E

The required magnitude of the velocity vector is 32.985.

Explanation of Solution

Given information:

The position of the particle: $\overrightarrow{x}=t^2-3$ , $\overrightarrow{y}=\frac{2}{3}{t}^3$

Calculation:

The given position of the particle is $\overrightarrow{r}\left(t\right)=\langle t^2-3,\frac{2}{3}{t}^3\rangle$ .

It is known that for a position vector $\overrightarrow{r}\left(t\right)$ of a particle, the first derivative of $\overrightarrow{r}\left(t\right)$ is the velocity vector of the particle.

So, $\overrightarrow{v}\left(t\right)=\frac{d\left(\overrightarrow{r}\left(t\right)\right)}{dt}$

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

Also, it is known that the magnitude of the velocity vector of a particle is known as the speed of the particle.

The magnitude of a vector $\langle \overrightarrow{a},\overrightarrow{b}\rangle$ is $\left|\langle \overrightarrow{a},\overrightarrow{b}\rangle \right|=\sqrt{{\overrightarrow{a}}^2+{\overrightarrow{b}}^2}$ .

Substitute $2t$ for $\overrightarrow{a}$ and $2t^2$ for $\overrightarrow{b}$ in the above formula.

  $\sqrt{{\overrightarrow{a}}^2+{\overrightarrow{b}}^2}=\sqrt{{\left(2t\right)}^2+{\left(\left(2t^2\right)\right)}^2}$

Substitute 4 for t in the above expression.

  $\begin{array}{c}\sqrt{{\left(2t\right)}^2+{\left(\left(2t^2\right)\right)}^2}=\sqrt{{\left(2\left(4\right)\right)}^2+{\left(\left(2{\left(4\right)}^2\right)\right)}^2}\\ =\sqrt{64+1024}\\ =\sqrt{1088}\\ \approx 32.985\end{array}$

Hence, the required magnitude of the velocity vector is 32.985.

(b)

To determine

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

Answer to Problem 50E

The required total distance is 46.062.

Explanation of Solution

Given information:

The position of the particle: $\overrightarrow{x}=t^2-3$ , $\overrightarrow{y}=\frac{2}{3}{t}^3$

Calculation:

The given position of the particle is $\overrightarrow{r}\left(t\right)=\langle t^2-3,\frac{2}{3}{t}^3\rangle$ .

It is known that for a position vector $\overrightarrow{r}\left(t\right)$ of a particle, the first derivative of $\overrightarrow{r}\left(t\right)$ is the velocity vector of the particle.

So, $\overrightarrow{v}\left(t\right)=\frac{d\left(\overrightarrow{r}\left(t\right)\right)}{dt}$

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

The velocity of the particle which moves along a path is $\overrightarrow{v}\left(t\right)=\langle {\overrightarrow{v}}_1\left(t\right),{\overrightarrow{v}}_2\left(t\right)\rangle$ .

To find the total distance traveled by the particle, use the formula ${\displaystyle \underset{a}{\overset{b}{\int }}\left|\overrightarrow{v}\left(t\right)\right|dt}={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{{\left({\overrightarrow{v}}_1\left(t\right)\right)}^2+{\left({\overrightarrow{v}}_2\left(t\right)\right)}^2}}dt$

So,

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{4}{\int }}\left|\overrightarrow{v}\left(t\right)\right|dt}={\displaystyle \underset{0}{\overset{4}{\int }}\sqrt{{\left(2t\right)}^2+{\left(2t^2\right)}^2}}dt\\ \approx 46.062\end{array}$

Hence, the required total distance is 46.062.

(c)

To determine

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

Answer to Problem 50E

The required value of $\frac{dy}{dx}$ is ${\left(3+x\right)}^{\frac{1}{2}}$ .

Explanation of Solution

Given information:

The position of the particle: $\overrightarrow{x}=t^2-3$ , $\overrightarrow{y}=\frac{2}{3}{t}^3$

Calculation:

The given position of the particle is $\overrightarrow{r}\left(t\right)=\langle t^2-3,\frac{2}{3}{t}^3\rangle$ .

Rewrite the x -component as:

  $\begin{array}{c}x=t^2-3\\ \sqrt{t^2}=\sqrt{x+3}\\ t=\sqrt{3+x}\end{array}$

Substitute $\sqrt{3+x}$ for t in the y -component that is, $y=\frac{2}{3}{t}^3$ .

  $y=\frac{2}{3}{\left(3+x\right)}^{\frac{3}{2}}$

Differentiate the above expression with respect to x .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(\frac{2}{3}{\left(3+x\right)}^{\frac{3}{2}}\right)\\ =\frac{\cancel{2}}{\cancel{3}}\cdot \frac{\cancel{3}}{\cancel{2}}{\left(3+x\right)}^{\frac{3}{2}-1}\\ ={\left(3+x\right)}^{\frac{1}{2}}\end{array}$

Hence, the required value of $\frac{dy}{dx}$ is ${\left(3+x\right)}^{\frac{1}{2}}$ .