Chapter 11.2, Problem 51E

(a)

To determine

To find: The $x$ coordinates of the position of the object at time $t=4$ .

Answer to Problem 51E

The x coordinate of the new position is that is 3.942.

Explanation of Solution

Given Data:

The object moves along the curve in the xy plane has the position $\left(x\left(t\right),y\left(t\right)\right)$ at time $t\ge 0$ with position $\left(x\left(t\right),y\left(t\right)\right)$ at time $t\ge 0$ with $\frac{dx}{dt}=2+\sin \left(t^2\right)$ .

The position of the object at $\left(3,5\right)$ .

Calculation:

Consider the displacement of the object in the $x$ direction for the time $t=1$ to $t=4$ is ${\displaystyle {\int }_2^4\left(2+\sin \left(t^2\right)\right)dt}=0.942$

Since, at $t=2$ the object is at the $\left(3,5\right)$ and for $t=2$ the displacement of the object in the $x$ direction is $0.942$ .

Thus, the x coordinate of the new position is $3+0.942$ that is 3.942.

(b)

To determine

To find: The equation for the tangent line to the curve at the point $\left(x\left(2\right),y\left(2\right)\right)$ .

Answer to Problem 51E

The required equation of the tangent line is $y-5=\frac{-6}{2+\sin 4}\left(x-3\right)+5$ .

Explanation of Solution

Given:

At time $t=2$ the value of $\frac{dy}{dt}$ is $-6$ .

Calculation:

Consider that at $t=2$ the object is at $\left(3,5\right)$ and $\frac{dy}{dt}=-6$ then $\frac{dx}{dt}=2+\sin 4$ .

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

  $\begin{array}{c}m=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{-6}{2+\sin 4}\end{array}$

Then the equation of the line tangent to the curve at $\left(3,5\right)$ is,

  $\begin{array}{c}y-5=\frac{-6}{2+\sin 4}\left(x-3\right)\\ =\frac{-6}{2+\sin 4}\left(x-3\right)+5\end{array}$

(c)

To determine

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

Answer to Problem 51E

The speed of the object is $6.127$ .

Explanation of Solution

Consider the speed of the object at time $t=2$ is,

  $\begin{array}{c}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}=\sqrt{{\left(2+\sin 4\right)}^2+{\left(-6\right)}^2}\\ =6.127\end{array}$

(d)

To determine

To find: The acceleration vector of the object at time $t=4$ .

Answer to Problem 51E

The acceleration vector is $a\left(t\right)=\langle -7.661,-50.205\rangle$ .

Explanation of Solution

Consider that for $t\ge 3$ the line tangent to the curve at $\left(x\left(t\right),y\left(t\right)\right)$ has the slope of $2t-1$ .

Then,

  $\begin{array}{c}m=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =2t-1\end{array}$

Here,

  $\begin{array}{c}\frac{dx}{dt}=2+\sin t^2\\ =\left(2t-1\right)\frac{dx}{dt}\\ =\left(2t-1\right)\left(2+\sin t^2\right)\end{array}$

Then, the acceleration vector of the particle is,

  $\begin{array}{c}a\left(t\right)=\frac{d}{dt}\langle \frac{dx}{dt},\frac{dy}{dt}\rangle \\ =\frac{d}{dt}\langle \left(2+\sin t^2\right)\rangle ,\left(2t-1\right)\left(2+\sin t^2\right)\\ =\langle 2t\cos t^2,\left(2t-1\right)\left(2t\cos t^2\right)+2\left(2+\sin t^2\right)\rangle \end{array}$

Then, at time $t=4$ the acceleration vector is,

  $a\left(t\right)=\langle -7.661,-50.205\rangle$ b