Chapter 10.2, Problem 45E

$\text{(a)}$

To determine

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

Answer to Problem 45E

The answer: $\approx 2.399$

Explanation of Solution

Given information:

  $\begin{array}{c}x=5\cos (\pi t/6)\\ y=3\sin (\pi t/6)\end{array}$

Calculation:

Know that,

  

Then,

  

In this exercise,

  

Then,

  $\begin{array}{c}\frac{dx}{dt}=5\cdot (-\sin (\pi t/6))\cdot \frac{\pi }{6}\\ =-\frac{5\pi \sin (\pi t/6)}{6}\\ \frac{dy}{dt}=3\cos (\pi t/6)\cdot \frac{\pi }{6}\\ =\frac{\pi \cos (\pi t/6)}{2}\end{array}$

To use the previous result,

For $t=2$ ,

Then,

  $\begin{array}{c}\frac{dx}{dt}(2)=-\frac{5\pi \sin (\pi \cdot 2/6)}{6}\\ =-\frac{5\pi \sin (\pi /3)}{6}\\ =-\frac{5\pi \cdot (\sqrt{3}/2)}{6}\\ =-\frac{5\sqrt{3}\pi /2}{6}\\ =-\frac{5\sqrt{3}\pi }{12}\\ \frac{dy}{dt}(2)=\frac{\pi \cos (\pi \cdot 2/6)}{2}\\ =\frac{\pi \cos (\pi /3)}{2}\\ =\frac{\pi \cdot (1/2)}{2}\\ =\frac{\pi /2}{2}\\ =\frac{\pi }{4}\end{array}$

According to the previous result, the velocity of the vector for $t=2$ :

  

Know that the magnitude of a vector is equal to .

Then,

  

Therefore, the required speed of the point at $t=2$ is

  $\approx 2.399$ .

$\text{(b)}$

To determine

To find: The acceleration vector at $t=2$ .

Answer to Problem 45E

The answer:

Explanation of Solution

Given information:

  $\begin{array}{c}x=5\cos (\pi t/6)\\ y=3\sin (\pi t/6)\end{array}$

Calculation:

Know that,

  

Then,

  

From part $(a)$ :

  $\begin{array}{c}dx/dt=-(5\pi \sin (\pi t/6))/6\\ dy/dt=(\pi \cos (\pi t/6))/2\end{array}$

Then,

  $\begin{array}{c}\frac{d^{2}x}{d{t}^2}=\frac{d(dx/dt)}{dt}\\ =-\frac{5\pi \cos (\pi t/6)\cdot (\pi /6)}{6}\\ =-\frac{\left(5{\pi }^2\cos (\pi t/6)\right)/6}{6}\\ =-\frac{5{\pi }^2\cos (\pi t/6)}{36}\\ \frac{d^{2}y}{d{t}^2}=\frac{d(dy/dt)}{dt}\\ =\frac{\pi \cdot (-\sin (\pi t/6))\cdot (\pi /6)}{2}\\ =\frac{\left(-{\pi }^2\sin (\pi t/6)\right)/6}{2}\\ =-\frac{{\pi }^2\sin (\pi t/6)}{12}\end{array}$

To use the previous result,

  $\begin{array}{c}\frac{d^{2}x}{d{t}^2}(2)=-\frac{5{\pi }^2\cos (\pi \cdot 2/6)}{36}\\ =-\frac{5{\pi }^2\cos (\pi /3)}{36}\\ =-\frac{5{\pi }^2\cdot (1/2)}{36}\\ =-\frac{5{\pi }^2/2}{36}\\ =-\frac{5{\pi }^2}{72}\\ \frac{d^{2}y}{d{t}^2}(2)=-\frac{{\pi }^2\sin (\pi \cdot 2/6)}{12}\\ =-\frac{{\pi }^2\sin (\pi /3)}{12}\\ =-\frac{{\pi }^2\cdot (\sqrt{3}/2)}{12}\\ =-\frac{\sqrt{3}{\pi }^2/2}{12}\\ =-\frac{\sqrt{3}{\pi }^2}{24}\end{array}$

Therefore, the required acceleration vector at $t=2$ is:

   .

$(c)$

To determine

To find: The estimate parameter and equation an $x$ and $y$ .

Answer to Problem 45E

The answer: The curve is eclipse.

The equation is ${\left(\frac{x}{5}\right)}^2+{\left(\frac{y}{3}\right)}^2=1$

Explanation of Solution

Given information:

  $\begin{array}{c}x=5\cos (\pi t/6)\\ y=3\sin (\pi t/6)\end{array}$

Calculation:

To have,

  $\begin{array}{c}{\left(\frac{x}{5}\right)}^2={\left(\cos \frac{\pi t}{6}\right)}^2\\ {\left(\frac{y}{3}\right)}^2={\left(\sin \frac{\pi t}{6}\right)}^2\end{array}$

Then,

  ${\left(\frac{x}{5}\right)}^2+{\left(\frac{y}{3}\right)}^2=1$

Therefore, the required curve is eclipse and the equation is ${\left(\frac{x}{5}\right)}^2+{\left(\frac{y}{3}\right)}^2=1$ .