Chapter 11, Problem 41RE

To determine

To calculate: The maximum speed of the particle.

Answer to Problem 41RE

The required maximum speed of the particle is 1.

Explanation of Solution

Given information:

The position of the particle: $r=\langle \frac{1}{\sqrt{1+t^2}},\frac{t}{\sqrt{1+t^2}}\rangle$

Calculation:

The given position of the particle is $r=\langle \frac{1}{\sqrt{1+t^2}},\frac{t}{\sqrt{1+t^2}}\rangle$ .

It is known that if $r\left(t\right)$ is the position vector of a particle, then the derivative of $r\left(t\right)$ is the velocity vector $v\left(t\right)$ of the particle.

So, $v\left(t\right)=\frac{d\left(r\left(t\right)\right)}{dt}$

Substitute $\langle \frac{1}{\sqrt{1+t^2}},\frac{t}{\sqrt{1+t^2}}\rangle$ for $r\left(t\right)$ in the above expression.

  $\begin{array}{c}v\left(t\right)=\frac{d\left(r\left(t\right)\right)}{dt}\\ =\frac{d}{dt}\langle \frac{1}{\sqrt{1+t^2}},\frac{t}{\sqrt{1+t^2}}\rangle \\ =\langle \frac{d}{dt}\left(\frac{1}{1+t^2}\right),\frac{d}{dt}\left(\frac{t}{\sqrt{1+t^2}}\right)\rangle \\ =\langle \frac{-t}{{\left(1+t^2\right)}^{\frac{3}{2}}},\frac{1}{{\left(1+t^2\right)}^{\frac{3}{2}}}\rangle \end{array}$

Also, the magnitude of the velocity vector is the speed and the magnitude of a vector $\langle a,b\rangle$ is $\left|\langle a,b\rangle \right|=\sqrt{a^2+b^2}$ .

The speed of the particle is:

  $\sqrt{{\left(\frac{-t}{{\left(1+t^2\right)}^{\frac{3}{2}}}\right)}^2+{\left(\frac{1}{{\left(1+t^2\right)}^{\frac{3}{2}}}\right)}^2}=\frac{1}{1+t^2}$

Since, the value of $t^2$ is positive; so the value of $\frac{1}{1+t^2}$ is maximum and $t^2$ is minimum at $t=0$ .

Thus, $\begin{array}{c}\frac{1}{1+t^2}=\frac{1}{1+{\left(0\right)}^2}\\ =1\end{array}$

Hence, the required maximum speed of the particle is 1.