Chapter 10, Problem 41RE

To determine

To find: The maximum speed of particle.

Answer to Problem 41RE

The maximum speed of particle is 1.

Explanation of Solution

Given information:

The position of the particle in the plane at time $t$ is .

Formula used:

If the position vector of particle is then the particle’s velocity is and its speed is magnitude of velocity.

Calculation:

Consider the position vector .

Substitute $\frac{1}{\sqrt{1+t^2}}$ for $x$ and $\frac{t}{\sqrt{1+t^2}}$ for $y$ in the formula .

   …… (1)

Differentiate $\frac{1}{\sqrt{1+t^2}}$ with respect to $t$ .

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

Differentiate $\frac{1}{\sqrt{1+t^2}}$ with respect to $t$ .

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

Substituting the values from equations (2) and (3) into equation (1) to find the velocity of the particle.

  

Calculate the speed of the particle.

  

Thus, the speed of particle is .

From the above expression it can be observed that as the time $t$ increases, the speed of the particle decreases. Thus, the particle has maximum speed at $t=0$ .

Substitute $0$ for $t$ in to find the maximum speed.

  

Thus, the maximum speed of the particle is $1$ when time $t$ is zero.