Chapter 11, Problem 51RE

(a)

To determine

To calculate: The position vector of the particle.

Answer to Problem 51RE

The required position vector is $\overrightarrow{r}\left(t\right)=\langle t^2+\pi ,\cos \left(t^2+\pi \right)\rangle$ .

Explanation of Solution

Given information:

The curve: $y=\cos x$

The x -component of acceleration = 2

At time t = 0 the position of the particle = $\left(\pi ,-1\right)$

At time t = 0 the velocity of the particle = $\left(0,0\right)$

Calculation:

The given curve along which the particle is moving is $y=\cos x$ .

The x -component of acceleration is 2.

So, for $t=0$ ; the acceleration vector of the particle is $\overrightarrow{a}\left(t\right)=\langle 2,{\overrightarrow{a}}_2\rangle$ .

The velocity of the particle is $\overrightarrow{v}\left(0,0\right)=\langle 0,0\rangle$ .

And, the position of the particle is $\overrightarrow{r}\left(0\right)=\langle \pi ,-1\rangle$ .

It is known that the first derivative of the velocity vector $\overrightarrow{v}\left(t\right)=\langle {\overrightarrow{v}}_1\left(t\right),{\overrightarrow{v}}_2\left(t\right)\rangle$ is the acceleration vector $\overrightarrow{a}\left(t=\langle {\overrightarrow{a}}_1\left(t\right),{\overrightarrow{a}}_2\left(t\right)\rangle \right)$ that is,

  $\frac{d{\overrightarrow{v}}_1\left(t\right)}{dt}={\overrightarrow{a}}_1\left(t\right)$ and $\frac{d\left({\overrightarrow{v}}_2\left(t\right)\right)}{dt}={\overrightarrow{a}}_2\left(t\right)$

Now, $\begin{array}{c}\frac{d{\overrightarrow{v}}_1\left(t\right)}{dt}=2\\ {\overrightarrow{v}}_1\left(t\right)={\displaystyle \int \left(2\right)dt}\\ {\overrightarrow{v}}_1\left(t\right)=2t+C\end{array}$

The velocity component of the particle at $t=0$ is ${\overrightarrow{v}}_1=0$ .

Substitute 0 for ${\overrightarrow{v}}_1\left(t\right)$ and 0 for t in the expression ${\overrightarrow{v}}_1\left(t\right)=2t+C$ .

  $\begin{array}{c}0=2\left(0\right)+C\\ C+0=0\\ C=0\end{array}$

Substitute 0 for C in the expression ${\overrightarrow{v}}_1\left(t\right)=2t+C$ .

  $\begin{array}{c}{\overrightarrow{v}}_1\left(t\right)=2t+0\\ {\overrightarrow{v}}_1\left(t\right)=2t\end{array}$

Thus, the velocity component of the particle is ${\overrightarrow{v}}_1\left(t\right)=2t$ .

It is known that the first derivative of the position vector $\overrightarrow{r}\left(t\right)=\langle \overrightarrow{x}\left(t\right),\overrightarrow{y}\left(t\right)\rangle$ is the velocity vector $\overrightarrow{v}\left(t\right)=\langle {\overrightarrow{v}}_1\left(t\right),{\overrightarrow{v}}_2\left(y\right)\rangle$ that is,

  $\frac{d\overrightarrow{x}\left(t\right)}{dt}={\overrightarrow{v}}_1\left(t\right)$ and $\frac{d\overrightarrow{y}\left(t\right)}{dt}={\overrightarrow{v}}_2\left(t\right)$

  $\begin{array}{c}\frac{d\overrightarrow{x}\left(t\right)}{dt}=2t\\ \overrightarrow{x}\left(t\right)={\displaystyle \int \left(2t\right)dt}\\ \overrightarrow{x}\left(t\right)=t^2+C\end{array}$

The position of the particle at t = 0 is $x=\pi$ .

Substitute 0 for t and $\pi$ for $\overrightarrow{x}$ in the expression $\overrightarrow{x}\left(t\right)=t^2+C$ .

  $\begin{array}{c}\pi ={\left(0\right)}^2+C\\ C=\pi \end{array}$

Substitute $\pi$ for C in the expression $x\left(t\right)=t^2+C$ .

  $\overrightarrow{x}\left(t\right)=t^2+\pi$

Also, $\overrightarrow{y}\left(t\right)=\cos \left(t^2+\pi \right)$

Hence, the required position vector is $\overrightarrow{r}\left(t\right)=\langle t^2+\pi ,\cos \left(t^2+\pi \right)\rangle$ .

(b)

To determine

To calculate: The speed of the particle when it is at the point $\left(4,\cos 4\right)$ .

Answer to Problem 51RE

The required of the particle is 2.324.

Explanation of Solution

Given information:

The curve: $y=\cos x$

The x -component of acceleration = 2

At time t = 0 the position of the particle = $\left(\pi ,-1\right)$

At time t = 0 the velocity of the particle = $\left(0,0\right)$

Calculation:

The given curve along which the particle is moving is $y=\cos x$ .

The x -component of acceleration is 2.

So, for $t=0$ ; the acceleration vector of the particle is $\overrightarrow{a}\left(t\right)=\langle 2,{\overrightarrow{a}}_2\rangle$ .

The velocity of the particle is $\overrightarrow{v}\left(0,0\right)=\langle 0,0\rangle$ .

And, the position of the particle is $\overrightarrow{r}\left(0\right)=\langle \pi ,-1\rangle$ .

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

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

So, $\begin{array}{c}\overrightarrow{v}\left(t\right)=\frac{d}{dt}\langle t^2+\pi ,\cos \left(t^2+\pi \right)\rangle \\ =\langle \frac{d}{dt}\left(t^2+\pi \right),\frac{d}{dt}\left(\cos \left(t^2+\pi \right)\right)\rangle \\ =\langle 2t,-2t\sin \left(t^2+\pi \right)\rangle \end{array}$

At the point $\left(4,\cos 4\right)$ ;

  $\begin{array}{c}{t}^2+\pi =4\\ t=\sqrt{4-\pi }\end{array}$

Also, it is known that the magnitude of the velocity vector is the speed of the particle and the magnitude of a vector $\langle \overrightarrow{a},\overrightarrow{b}\rangle$ is:

  $\left|\langle \overrightarrow{a},\overrightarrow{b}\rangle \right|=\sqrt{{\overrightarrow{a}}^2+{\overrightarrow{b}}^2}$

Thus, the magnitude of the velocity vector at $t=\sqrt{4-\pi }$ is:

  ${\sqrt{{\left(2t\right)}^2+{\left(-2t\sin \left(t^2+\pi \right)\right)}^2}|}_{t=\sqrt{4-\pi }}=2.324$

Hence, the required of the particle is 2.324.