Chapter 10, Problem 40RE

a)

To determine

To find:The velocity and acceleration vectors for the position vector, $r\left(t\right)=〈4\cos t,\sqrt{2}\sin t〉$ of a particle moving in a plane at time $t$ .

Answer to Problem 40RE

For a particle with position vector, $r\left(t\right)=〈4\cos t,\sqrt{2}\sin t〉$ moving in a plane at time $t$ :

The velocity vector is $v\left(t\right)=〈-4\sin t,\sqrt{2}\cos t〉$ .

The acceleration vector is $a\left(t\right)=〈-4\cos t,-\sqrt{2}\sin t〉$ .

Explanation of Solution

Given information:

The position vector of the particle moving in a plane at time $t$ :

  $r\left(t\right)=〈4\cos t,\sqrt{2}\sin t〉$ .

Formula used:

If a particle moves along a smooth curve in the plane so that its position at any time $t$ is $\left(x\left(t\right),y\left(t\right)\right)$ , given by the position vector, $r\left(t\right)=〈x\left(t\right),y\left(t\right)〉$ where $x$ and $y$ are differentiable functions of $t$ , then,

1. The Velocity vector of the particle is $v\left(t\right)=〈\frac{dx}{dt},\frac{dy}{dt}〉$ .
2. The Acceleration vector of the particle is $a\left(t\right)=〈\frac{d^{2}x}{d{t}^2},\frac{d^{2}y}{d{t}^2}〉$ .

Calculation:

Compare the given position vector $r\left(t\right)=〈4\cos t,\sqrt{2}\sin t〉$ with the standard form of position vectors, $r\left(t\right)=〈x\left(t\right),y\left(t\right)〉$ to obtain:

  $x\left(t\right)=4\cos t$

  $y\left(t\right)=\sqrt{2}\sin t$

Differentiate both sides of the equations with respect to $t$ :

  $\begin{array}{l}\frac{dx}{dt}=\frac{d}{dt}\left(4\cos t\right)\\ \frac{dx}{dt}=4\frac{d}{dt}\left(\cos t\right)\\ \frac{dx}{dt}=4\left(-\sin t\right)\\ \frac{dx}{dt}=-4\sin t\end{array}$

And,

  $\begin{array}{l}\frac{dy}{dt}=\frac{d}{dt}\left(\sqrt{2}\sin t\right)\\ \frac{dy}{dt}=\sqrt{2}\frac{d}{dt}\left(\sin t\right)\\ \frac{dy}{dt}=\sqrt{2}\cos t\end{array}$

Substitute $\frac{dx}{dt}=-4\sin t$ and $\frac{dy}{dt}=\sqrt{2}\cos t$ in the formula for the Velocity vector of a particle moving in a plane at time $t$ to obtain the velocity vector, $v\left(t\right)$ :

  $v\left(t\right)=〈-4\sin t,\sqrt{2}\cos t〉$

Thus, the velocity vector of the particle moving in a plane at time $t$ with position vector, $r\left(t\right)=〈4\cos t,\sqrt{2}\sin t〉$ is $v\left(t\right)=〈-4\sin t,\sqrt{2}\cos t〉$ .

Evaluate the second-order derivatives $\frac{d^{2}x}{d{t}^2}$ and $\frac{d^{2}y}{d{t}^2}$ by differentiating $\frac{dx}{dt}=-4\sin t$ and $\frac{dy}{dt}=\sqrt{2}\cos t$ with respect to time $t$ :

  $\begin{array}{l}\frac{dx}{dt}=-4\sin t\\ \frac{d^{2}x}{d{t}^2}=\frac{d}{dt}\left(-4\sin t\right)\\ \frac{d^{2}x}{d{t}^2}=-4\frac{d}{dt}\left(\sin t\right)\\ \frac{d^{2}x}{d{t}^2}=-4\cos t\end{array}$

And,

  $\begin{array}{l}\frac{dy}{dt}=\sqrt{2}\cos t\\ \frac{d^{2}y}{d{t}^2}=\frac{d}{dt}\left(\sqrt{2}\cos t\right)\\ \frac{d^{2}y}{d{t}^2}=\sqrt{2}\frac{d}{dt}\left(\cos t\right)\\ \frac{d^{2}y}{d{t}^2}=-\sqrt{2}\sin t\end{array}$

Substitute $\frac{d^{2}x}{d{t}^2}=-4\cos t$ and $\frac{d^{2}y}{d{t}^2}=-\sqrt{2}\sin t$ in the formula for the Acceleration vector of a particle moving in a plane at time $t$ to obtain the Acceleration vector, $a\left(t\right)$ :

  $a\left(t\right)=〈-4\cos t,-\sqrt{2}\sin t〉$

Thus,

The Acceleration vectorof the particle moving in a plane at time $t$ with position vector, $r\left(t\right)=〈4\cos t,\sqrt{2}\sin t〉$ is $a\left(t\right)=〈-4\cos t,-\sqrt{2}\sin t〉$ .

b)

To determine

To find:The speedof the particle with position vector, $r\left(t\right)=〈4\cos t,\sqrt{2}\sin t〉$ , moving in the plane at time, $t=\frac{\pi }{4}$ .

Answer to Problem 40RE

The speed of the particle with position vector, $r\left(t\right)=〈4\cos t,\sqrt{2}\sin t〉$ , moving in the plane at time, $t=\frac{\pi }{4}$ is: 9 units.

Explanation of Solution

Given information:

The position vector of the particle moving in a plane at time $t$ :

  $r\left(t\right)=〈4\cos t,\sqrt{2}\sin t〉$ .

Formula used:

If a particle moves along a smooth curve in the plane so that its position at any time $t$ is $\left(x\left(t\right),y\left(t\right)\right)$ , given by the position vector, $r\left(t\right)=〈x\left(t\right),y\left(t\right)〉$ where $x$ and $y$ are differentiable functions of $t$ , then,

1. The Velocity vector of the particle is $v\left(t\right)=〈\frac{dx}{dt},\frac{dy}{dt}〉$ .

The speed of the particle is the magnitude of the vector $v\left(t\right)$ , denoted by $\left|v\right|$ . Speed is a scalar quantity.

Magnitude of a vector:

The magnitude of the vector $〈a,b〉$ is the non-negative real number:

  $\left|〈a,b〉\right|=\sqrt{a^2+b^2}$ .

Calculation:

Compare the given position vector $r\left(t\right)=〈4\cos t,\sqrt{2}\sin t〉$ with the standard form of position vectors, $r\left(t\right)=〈x\left(t\right),y\left(t\right)〉$ to obtain:

  $x\left(t\right)=4\cos t$

  $y\left(t\right)=\sqrt{2}\sin t$

Differentiate both sides of the equations with respect to $t$ :

  $\begin{array}{l}\frac{dx}{dt}=\frac{d}{dt}\left(4\cos t\right)\\ \frac{dx}{dt}=4\frac{d}{dt}\left(\cos t\right)\\ \frac{dx}{dt}=4\left(-\sin t\right)\\ \frac{dx}{dt}=-4\sin t\end{array}$

And,

  $\begin{array}{l}\frac{dy}{dt}=\frac{d}{dt}\left(\sqrt{2}\sin t\right)\\ \frac{dy}{dt}=\sqrt{2}\frac{d}{dt}\left(\sin t\right)\\ \frac{dy}{dt}=\sqrt{2}\cos t\end{array}$

Substitute $\frac{dx}{dt}=-4\sin t$ and $\frac{dy}{dt}=\sqrt{2}\cos t$ in the formula for the Velocity vector of a particle moving in a plane at time $t$ to obtain the velocity vector, $v\left(t\right)$ :

  $v\left(t\right)=〈-4\sin t,\sqrt{2}\cos t〉$

Let $\left|v\right|$ be the magnitude of the vector $v\left(t\right)=〈-4\sin t,\sqrt{2}\cos t〉$ .

Apply the formula for Magnitude of vectors to calculate, the speed of the particle, $\left|v\right|$ at time $t$ :

  $\begin{array}{l}\left|v\right|=\sqrt{{\left(-4\sin t\right)}^2+{\left(\sqrt{2}\cos t\right)}^2}\\ \left|v\right|=\sqrt{16{\sin }^{2}t+2{\cos }^{2}t}\end{array}$

Substitute $t=\frac{\pi }{4}$ to evaluate the speed of the particle at time, $t=\frac{\pi }{4}$:

  $\begin{array}{l}\left|v\right|=\sqrt{16{\sin }^2\frac{\pi }{4}+2{\cos }^2\frac{\pi }{4}}\\ \left|v\right|=\sqrt{16{\left(\frac{1}{\sqrt{2}}\right)}^2+2{\left(\frac{1}{\sqrt{2}}\right)}^2}\\ \left|v\right|=\sqrt{16\times \frac{1}{2}+2\times \frac{1}{2}}\\ \left|v\right|=9\end{array}$

Thus,

The speed of the particle with position vector, $r\left(t\right)=〈4\cos t,\sqrt{2}\sin t〉$ , moving in the plane at time, $t=\frac{\pi }{4}$ is: 9 units.