Chapter 10.2, Problem 31E

$\text{(a)}$

To determine

To find: The velocity vector $v(t)$ .

Answer to Problem 31E

The answer:

  $v(t)=(2t+2\cos (2t),2t+2\sin (2t))$

Explanation of Solution

Given information:

In the plane with position vector $r(t)=(t^2+\sin 2t,t^2-\cos 2t)$

Calculation:

Know that if the position vector of a particle is equal to:

  $r(t)=\left(x(t),y(t)\right)$

Then the velocity vector of a particle is equal to:

  $v(t)=\left(\frac{dx}{dt},\frac{dy}{dt}\right)$

In this exercise, the position vector of the particle is equal to $r(t)=(t^2+\sin 2t,t^2-\cos 2t)$

Then,

  $\begin{array}{c}x(t)=t^2+\sin (2t)\\ y(t)=t^2-\cos (2t)\end{array}$

Which means that,

  $\begin{array}{c}\frac{dx}{dt}=2t+\cos (2t)\cdot 2\\ =2t+2\cos (2t)\\ \frac{dy}{dt}=2t-(-\sin (2t))\cdot 2\\ =2t+2\sin (2t)\end{array}$

The previous results show that the particle's velocity vector is equal to:

  $v(t)=(2t+2\cos (2t),2t+2\sin (2t))$

Therefore, the required velocity vector $v(t)$ is $v(t)=(2t+2\cos (2t),2t+2\sin (2t))$ .

$\text{(b)}$

To determine

To find: The acceleration vector $a(t)$ .

Answer to Problem 31E

The answer:

  $a(t)=(2-4\sin (2t),2+4\cos (2t))$

Explanation of Solution

Given information:

In the plane with position vector $r(t)=(t^2+\sin 2t,t^2-\cos 2t)$

Calculation:

Know that if the position vector of a particle is equal to:

  $r(t)=\left(x(t),y(t)\right)$

Then the acceleration vector of a particle is equal to:

  $a(t)=\left(\frac{d^{2}x}{d{t}^2},\frac{d^{2}y}{d{t}^2}\right)$

Hence, from part $\text{(a)}$ :

  $\begin{array}{c}dx/dt=2t+2\cos (2t)\\ dy/dt=2t+2\sin (2t)\end{array}$

Then,

  $\begin{array}{c}\frac{d^{2}x}{d{t}^2}=\frac{d(dx/dt)}{dt}\\ =2\cdot 1+2\cdot (-\sin (2t))\cdot 2\\ =2-4\sin (2t)\\ \frac{d^{2}y}{d{t}^2}=\frac{d(dy/dt)}{dt}\\ =2\cdot 1+2\cdot \cos (2t)\cdot 2\\ =2+4\cos (2t)\end{array}$

The previous results show that the particle's acceleration vector is equal to:

  $a(t)=(2-4\sin (2t),2+4\cos (2t))$

Therefore, the required acceleration vector $a(t)$ is $a(t)=(2-4\sin (2t),2+4\cos (2t))$ .