Chapter 11.2, Problem 52E

(a)

To determine

To find: The equation for the line tangent tot the curve at $\left(4,5\right)$ .

Answer to Problem 52E

The required equation is $y=\frac{3\cos 4}{\sin 8}\left(x-4\right)+5$ .

Explanation of Solution

Given Data:

The time interval is $0\le t\le 3$ .

The object moves in the x y plane has the position $\left(x\left(t\right),y\left(t\right)\right)$ with the $\frac{dx}{dt}=\sin t^3$ and $\frac{dy}{dt}=3\cos \left(t^2\right)$ .

The time $t=2$ and object is at the position $\left(4,5\right)$ .

Calculation:

Consider that at the time $t=2$ the object is at the point $\left(4,5\right)$ . Then,

  $\begin{array}{l}\frac{dy}{dt}=3\cos 4\\ \frac{dx}{dt}=\sin 8\end{array}$

Then, the slope of the tangent line at $t=2$ is,

  $\begin{array}{c}m=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{3\cos 4}{\sin 8}\end{array}$

Then, the equation of the tangent line is,

  $y=\frac{3\cos 4}{\sin 8}\left(x-4\right)+5$

(b)

To determine

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

Answer to Problem 52E

The speed of the object is $2.196$ .

Explanation of Solution

The speed of the object at the time $t=2$ is obtained as,

  $\sqrt{{\left(\sin 8\right)}^2+{\left(3\cos 4\right)}^2}=2.196$

Thus, the speed of the object is $2.196$ .

(c)

To determine

To find: The total distance travelled by the object over the time interval $0\le t\le 1$ .

Answer to Problem 52E

The distance travelled is $2.741$ .

Explanation of Solution

The velocity components of the particle is,

  $\begin{array}{l}{v}_1\left(t\right)=\sin t^3\\ v_2\left(t\right)=3\cos t^2\end{array}$

Then, the distance travelled by the particle is,

  ${\displaystyle {\int }_0^1\sqrt{{\left(\sin t^3\right)}^2+{\left(3\cos t^2\right)}^2}}dt=2.741$

(d)

To determine

To find: The find the position of the object at time $t=3$ .

Answer to Problem 52E

The new position of the particle is $\left(4.004,5.724\right)$ .

Explanation of Solution

The displacement of the particle is obtained by,

  $\langle {\displaystyle {\int }_2^3\sin \left(t^3\right)dt,{\displaystyle {\int }_2^3\cos \left(t^2\right)dt}}\rangle =\langle 0.004,0.724\rangle$

Then, at the time $t=2$ the particle is at the $\left(4,5\right)$ and for the time $t=2$ to $t=3$ the displacement of the particle is $\langle 0.004,0.724\rangle$ .

Then, the new position of the particle is,

  $\langle 4+0.004,5+0.724\rangle =\left(4.004,5.724\right)$