Chapter 11, Problem 39RE

a

To determine

To find the velocity and acceleration vector of the particle moving in plane at time $t$

Answer to Problem 39RE

  $v(t)=\langle -4\sin t,\sqrt{2}\cos t\rangle$ , $a(t)=\langle -4\cos t,-\sqrt{2}\sin t\rangle$

Explanation of Solution

Given:

The position vector of a particle moving in plane at time t is $r(t)=\langle 4\cos t,\sqrt{2}\sin t\rangle ,t=\frac{\pi }{4}$

Calculation:

The velocity vector is given by-

  $\begin{array}{l}v(t)=\frac{dr}{dt}=\langle \frac{d\left(4\cos t\right)}{dt},\frac{d\left(\sqrt{2}\sin t\right)}{dt}\rangle ,t=\frac{\pi }{4} \\ v(t)=\langle -4\sin t,\sqrt{2}\cos t\rangle \end{array}$

The acceleration vector is given by-

  $a(t)=\frac{dv}{dt}=\langle -4\cos t,-\sqrt{2}\sin t\rangle$

b

To determine

To find the speed at the given value of $t$ .

Answer to Problem 39RE

  $\sqrt{18}\text{unit/sec}$

Explanation of Solution

Given:

The position vector of a particle moving in plane at time t is $r(t)=\langle 4\cos t,\sqrt{2}\sin t\rangle ,t=\frac{\pi }{4}$

Calculation:

The speed at given value of $t$ is

  $\begin{array}{l}v(t)=-4\sin t+\sqrt{2}\cos t\\ {\left|v(t)\right|}_{\frac{\pi }{4}}=\sqrt{{\left(-4\right)}^2+{(\sqrt{2})}^2}\\ {\left|v(t)\right|}_{\frac{\pi }{4}}=\sqrt{16+2}\\ {\left|v(t)\right|}_{\frac{\pi }{4}}=\sqrt{18}\text{unit/sec}\end{array}$