Chapter 11.2, Problem 45E

(a)

To determine

To calculate: The speed of the point at $t=2$ .

Answer to Problem 45E

The required speed of the point is 2.399.

Explanation of Solution

Given information:

The position of the point: $\overrightarrow{x}=5\cos \left(\frac{\pi t}{6}\right)$ , $\overrightarrow{y}=3\sin \left(\frac{\pi t}{6}\right)$

Calculation:

The given position of the point is $\overrightarrow{x}=5\cos \left(\frac{\pi t}{6}\right)$ , $\overrightarrow{y}=3\sin \left(\frac{\pi t}{6}\right)$ .

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, $\begin{array}{c}\overrightarrow{v}\left(t\right)=\frac{d}{dt}\langle 5\cos \left(\frac{\pi t}{6}\right),3\sin \left(\frac{\pi t}{6}\right)\rangle \\ =\langle \frac{d}{dt}\left(5\cos \left(\frac{\pi t}{6}\right)\right),\frac{d}{dt}\left(3\sin \left(\frac{\pi t}{6}\right)\right)\rangle \\ =\langle -\frac{5}{6}\pi \sin \left(\frac{\pi }{6}t\right),\frac{1}{2}\pi \cos \left(\frac{\pi }{6}t\right)\rangle \end{array}$

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

At $t=2$ ; substitute $-\frac{5}{6}\pi \sin \left(\frac{\pi }{6}t\right)$ for $\overrightarrow{a}$ and $\frac{1}{2}\pi \cos \left(\frac{\pi }{6}t\right)$ for $\overrightarrow{b}$ in the above formula.

  $\begin{array}{c}\sqrt{{\overrightarrow{a}}^2+{\overrightarrow{b}}^2}=\sqrt{{\left(-\frac{5}{6}\pi \sin \left(\frac{\pi }{6}t\right)\right)}^2+{\left(\frac{1}{2}\pi \cos \left(\frac{\pi }{6}t\right)\right)}^2}\\ =\pi \sqrt{\frac{7}{12}}\\ \approx 2.399\end{array}$

Hence, the required speed of the point is 2.399.

(b)

To determine

To calculate: The acceleration of the point at $t=2$ .

Answer to Problem 45E

The required acceleration of the point is $\langle -\frac{5{\pi }^2}{72},-\frac{{\pi }^2\sqrt{3}}{24}\rangle$ .

Explanation of Solution

Given information:

The position of the point: $\overrightarrow{x}=5\cos \left(\frac{\pi t}{6}\right)$ , $\overrightarrow{y}=3\sin \left(\frac{\pi t}{6}\right)$

Calculation:

The given position of the point is $\overrightarrow{x}=5\cos \left(\frac{\pi t}{6}\right)$ , $\overrightarrow{y}=3\sin \left(\frac{\pi t}{6}\right)$ .

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

So, $\begin{array}{c}\overrightarrow{a}\left(t\right)=\frac{d^2\left(\overrightarrow{r}\left(t\right)\right)}{d{t}^2}\\ \overrightarrow{a}\left(t\right)=\frac{d\left(\overrightarrow{v}\left(t\right)\right)}{dt}\end{array}$

Now,

  $\begin{array}{c}\overrightarrow{a}\left(t\right)=\frac{d}{dt}\langle -\frac{5}{6}\pi \sin \left(\frac{\pi }{6}t\right),\frac{1}{2}\pi \cos \left(\frac{\pi }{6}t\right)\rangle \\ =\langle \frac{d}{dt}\left(-\frac{5}{6}\pi \sin \left(\frac{\pi }{6}t\right)\right),\frac{d}{dt}\left(\frac{1}{2}\pi \cos \left(\frac{\pi }{6}t\right)\right)\rangle \\ =\langle -\frac{5}{36}{\pi }^2\cos \left(\frac{\pi }{6}t\right),-\frac{1}{12}{\pi }^2\sin \left(\frac{\pi }{6}t\right)\rangle \\ =\langle -\frac{5{\pi }^2}{72},-\frac{{\pi }^2\sqrt{3}}{24}\rangle \end{array}$

Hence, the required acceleration of the point is $\langle -\frac{5{\pi }^2}{72},-\frac{{\pi }^2\sqrt{3}}{24}\rangle$ .

(c)

To determine

To determine: The equation in x and y that defines curve on which the point moves.

Answer to Problem 45E

The required equation is $\frac{x^2}{25}+\frac{y^2}{9}=1$ .

Explanation of Solution

Given information:

The position of the point: $\overrightarrow{x}=5\cos \left(\frac{\pi t}{6}\right)$ , $\overrightarrow{y}=3\sin \left(\frac{\pi t}{6}\right)$

Calculation:

The given position of the point is $\overrightarrow{x}=5\cos \left(\frac{\pi t}{6}\right)$ , $\overrightarrow{y}=3\sin \left(\frac{\pi t}{6}\right)$ .

Use the trigonometric identity ${\sin }^{2}t+{\cos }^{2}t=1$ .

Write the expressions $\overrightarrow{x}=5\cos \left(\frac{\pi t}{6}\right)$ and $\overrightarrow{y}=3\sin \left(\frac{\pi t}{6}\right)$ as:

  $\begin{array}{l}{x}^2={\left(5\cos \left(\frac{\pi t}{6}\right)\right)}^2\\ x^2=25{\cos }^2\left(\frac{\pi t}{6}\right)\\ \frac{x^2}{25}={\cos }^2\left(\frac{\pi t}{6}\right)\end{array}$

Also,

  $\begin{array}{l}{y}^2={\left(3\sin \left(\frac{\pi t}{6}\right)\right)}^2\\ y^2=9{\sin }^2\left(\frac{\pi t}{6}\right)\\ \frac{y^2}{9}={\sin }^2\left(\frac{\pi t}{6}\right)\end{array}$

Add the above two equations:

  $\begin{array}{c}\frac{x^2}{25}+\frac{y^2}{9}={\cos }^2\left(\frac{\pi t}{6}\right)+{\sin }^2\left(\frac{\pi t}{6}\right)\\ \frac{x^2}{25}+\frac{y^2}{9}=1\end{array}$

Hence, the required equation is $\frac{x^2}{25}+\frac{y^2}{9}=1$ .