Chapter 7.4, Problem 58E

a.

To determine

Show that the function , $v\left(t\right)=\sqrt{\frac{mg}{k}}\frac{e^{at}-e^{-at}}{e^{at}+e^{-at}}$ .

Explanation of Solution

Given information: If a body of mass m falling from rest under the action of gravity encounters an air resistance proportional to the square of the velocity, then the body velocity v(t) is modelled by the initial value problem.

Differential equation: $m\frac{dv}{dt}=mg-k{v}^2$ .

Initial condition: $v\left(0\right)=0$ ,

Proof:

To verify that the differential equation $m\frac{dv}{dt}=mg-k{v}^2$ is satisfied by the function . $v\left(t\right)=\sqrt{\frac{mg}{k}}\frac{e^{at}-e^{-at}}{e^{at}+e^{-at}}$ , To simplify the calculation, first rewrite v by multiplying the numerator and denominator by $e^{at}$ ,

  $\begin{array}{l}v\left(t\right)=\sqrt{\frac{mg}{k}}\frac{e^{at}-e^{-at}}{e^{at}+e^{-at}}\\ =\sqrt{\frac{mg}{k}}\frac{e^{2at}-1}{e^{2at}+1}\end{array}$

Now find its derivative,

  $\begin{array}{l}\frac{dv}{dt}=\sqrt{\frac{mg}{k}}\frac{2a{e}^{2at}\left(e^{2at}+1\right)-2a{e}^{2at}\left(e^{2at}-1\right)}{{\left(e^{2at}+1\right)}^2}\\ \text{ =}\sqrt{\frac{mg}{k}}\frac{2a{e}^{4at}+2a{e}^{2at}-2a{e}^{4at}+2a{e}^{2at}}{{\left(e^{2at}+1\right)}^2}\\ =\sqrt{\frac{mg}{k}}\frac{4a{e}^{4at}}{{\left(e^{2at}+1\right)}^2}\end{array}$

Given that $\text{a =}\sqrt{\frac{gk}{m}}$ , Substitute in $\sqrt{\frac{gk}{m}}$ for a and then simplify,

  $\begin{array}{l}\frac{dv}{dt}=a\sqrt{\frac{mg}{k}}\frac{4a{e}^{4at}}{{\left(e^{2at}+1\right)}^2}\\ \text{ =}\sqrt{\frac{gk}{m}}\sqrt{\frac{mg}{k}}\frac{4a{e}^{4at}}{{\left(e^{2at}+1\right)}^2}\\ \text{ =}\sqrt{g^2}\frac{4a{e}^{4at}}{{\left(e^{2at}+1\right)}^2}\end{array}$

Substitute this into $m\frac{dv}{dt}$ to find an expression for the left side of the differential equation:

  $m\frac{dv}{dt}=m\left[\frac{4g{e}^{2at}}{{\left(e^{2at}+1\right)}^2}\right]=\frac{4mg{e}^{2at}}{{\left(e^{2at}+1\right)}^2}$

Substitute $v=\sqrt{\frac{mg}{k}}\frac{e^{2at}-1}{e^{2at}+1}$ into $mg-k{v}^2$ and simplify,

  $\begin{array}{l}mg-k{v}^2=mg-k{\left(\sqrt{\frac{mg}{k}}\frac{e^{2at}-1}{e^{2at}+1}\right)}^2\\ =mg-k\left(\frac{mg}{k}\frac{{\left(e^{2at}-1\right)}^2}{{\left(e^{2at}+1\right)}^2}\right)\\ =mg-mg\frac{{\left(e^{2at}-1\right)}^2}{{\left(e^{2at}+1\right)}^2}\\ =mg\left(1-\frac{{\left(e^{2at}-1\right)}^2}{{\left(e^{2at}+1\right)}^2}\right)\\ =mg\left(\frac{{\left(e^{2at}+1\right)}^2-{\left(e^{2at}-1\right)}^2}{{\left(e^{2at}+1\right)}^2}\right)\\ =mg\left(\frac{\left(e^{4at}+2e^{2at}+1\right)-\left(e^{4at}-2e^{2at}+1\right)}{{\left(e^{2at}+1\right)}^2}\right)\\ =mg\left(\frac{4e^{2at}}{{\left(e^{2at}+1\right)}^2}\right)\\ =\frac{4mg{e}^{2at}}{\left(e^{2at}+1\right)}\end{array}$

Since $m\frac{dv}{dt}=\frac{4mg{e}^{2at}}{\left(e^{2at}+1\right)}=mg-k{v}^2$ , the differential equation is satisfied by $v\left(t\right)=\sqrt{\frac{mg}{k}}\frac{e^{at}-e^{-at}}{e^{at}+e^{-at}}$ .

The last thing to check is that the initial condition of $v\left(0\right)=0$ is satisfied. Substitute t = 0 into $v\left(t\right)$ to show its equal to 0,

  $\begin{array}{l}v\left(t\right)=\sqrt{\frac{mg}{k}}\frac{e^{at}-e^{-at}}{e^{at}+e^{-at}}\\ =\sqrt{\frac{mg}{k}}\frac{e^{a\left(0\right)}-e^{-a\left(0\right)}}{e^{a\left(0\right)}+e^{-a\left(0\right)}}\\ =\sqrt{\frac{mg}{k}}\frac{e^0-e^0}{e^0+e^0}\\ =\sqrt{\frac{mg}{k}}\frac{1-1}{1+1}\\ =\sqrt{\frac{mg}{k}}\left(0\right)\\ =0\end{array}$

Hence, proved.

b.

To determine

To find the body’s limiting velocity, ${\lim }_t\to \infty v\left(t\right)$ .

Answer to Problem 58E

  $\sqrt{\frac{mg}{k}}$ is the body’s limiting velocity.

Explanation of Solution

Given Information: : If a body of mass m falling from rest under the action of gravity encounters an air resistance proportional to the square of the velocity, then the body velocity v(t) is modelled by the initial value problem.

Differential equation: $m\frac{dv}{dt}=mg-k{v}^2$ .

Initial condition: $v\left(0\right)=0$ ,

Calculation:

As per the given problem,

  $\begin{array}{l}\underset{t\to \infty }{\lim }v\left(t\right)=\sqrt{\frac{mg}{k}}\frac{e^{a\times \infty }-e^{-a\times \infty }}{e^{a\times \infty }+e^{-a\times \infty }}\\ =\sqrt{\frac{mg}{k}}\frac{e^{a\times \infty }}{e^{a\times \infty }}\\ =\sqrt{\frac{mg}{k}}\end{array}$

Hence, $\sqrt{\frac{mg}{k}}$ is the body’s limiting velocity.

c.

To determine

To find the driver’s limiting velocity in feet per second? In miles per hours.

Answer to Problem 58E

The driver’s limiting velocity is 178.9 feet per second and 122 miles per hours.

Explanation of Solution

Given Information: If a body of mass m falling from rest under the action of gravity encounters an air resistance proportional to the square of the velocity, then the body velocity v(t) is modelled by the initial value problem.

Differential equation: $m\frac{dv}{dt}=mg-k{v}^2$ .

Initial condition: $v\left(0\right)=0$ ,

For a 160-lb skydiver (mg=160), and with time in seconds and distance in feet, a typical value for k is 0.005.

Calculation:

As per the given problem,

  $\sqrt{\frac{mg}{k}}=178.9$

178.9 feet per second.

122 miles per hours.

The driver’s limiting velocity is 178.9 feet per second and 122 miles per hours.