Question

A model for the velocity v at time tof a certain object falling under the influence of gravity in a viscous medium is given by the equation $\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathbf{v}}{\mathbf{8}}$.From the direction field shown in Figure 1.14, sketch the solutions with the initial conditions v(0) = 5, 8, and 15. Why is the value v = 8 called the “terminal velocity”?

Figure 1.14

Short answer

For v = 8, the value of $\frac{\mathrm{dv}}{\mathrm{dt}}=0,$hence, it is the terminal velocity.

Step-by-step solution

Solving the given differential equation

$\begin{array}{l}\frac{\mathrm{dv}}{\mathrm{dt}}=1-\frac{\mathrm{v}}{8}\\ \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{8-\mathrm{v}}{8}\\ \frac{8\mathrm{dv}}{8-\mathrm{v}}=\mathrm{dt}\\ -8\int \frac{\mathrm{dv}}{\mathrm{v}-8}=\int \mathrm{dt}\\ -8\log \left|\mathrm{v}-\right.\left.8\right|=\mathrm{t}+\mathrm{c}\\ \mathrm{v}-8={\mathrm{e}}^{\frac{-\mathrm{t}}{8}}{\mathrm{c}}_1\\ \mathrm{v}=8+{\mathrm{e}}^{\frac{-\mathrm{t}}{8}}{\mathrm{c}}_1\end{array}$

Applying the initial condition v(0) = 5  in the solution found in Step 1.

$\begin{array}{l}5=8+{\mathrm{c}}_1\\ {\mathrm{c}}_1=-3\\ \mathrm{v}=8-3{\mathrm{e}}^{\frac{-\mathrm{t}}{8}}\end{array}$

Putting the first condition v(0) = 8 in the solution from Step 1.

$\begin{array}{l}8=8+{\mathrm{c}}_1\\ {\mathrm{c}}_1=0\\ \mathrm{v}=8\end{array}$

Appealing the primary condition v(0) = 15 in the solution developed in Step 1.

$\mathrm{v}=8$

Draw the Sketch for the solutions found in Steps 2 to 4.

Finding the terminal velocity by substituting dvdt=0.

$\begin{array}{l}1-\frac{\mathrm{v}}{8}=0\\ \mathrm{v}=8\end{array}$

Therefore, v = 8 is the terminal velocity as the slope is 0.