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### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# A model for the velocity v at time t of 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

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

See the step by step solution

## Step 1: Solving the given differential equation

$\begin{array}{l}\frac{\mathrm{dv}}{\mathrm{dt}}=1-\frac{\mathrm{v}}{8}\\ \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{8-\mathrm{v}}{8}\\ \frac{8\mathrm{dv}}{8-\mathrm{v}}=\mathrm{dt}\\ -8\int \frac{\mathrm{dv}}{\mathrm{v}-8}=\int \mathrm{dt}\\ -8\mathrm{log}\left|\mathrm{v}-\right8|=\mathrm{t}+\mathrm{c}\\ \mathrm{v}-8={\mathrm{e}}^{\frac{-\mathrm{t}}{8}}{\mathrm{c}}_{1}\\ \mathrm{v}=8+{\mathrm{e}}^{\frac{-\mathrm{t}}{8}}{\mathrm{c}}_{1}\end{array}$

## Step 2: Applying the initial condition v(0) = 5  in the solution found in Step 1.

$\begin{array}{l}5=8+{\mathrm{c}}_{1}\\ {\mathrm{c}}_{1}=-3\\ \mathrm{v}=8-3{\mathrm{e}}^{\frac{-\mathrm{t}}{8}}\end{array}$

## Step 3: Putting the first condition v(0) = 8 in the solution from Step 1.

$\begin{array}{l}8=8+{\mathrm{c}}_{1}\\ {\mathrm{c}}_{1}=0\\ \mathrm{v}=8\end{array}$

## Step 4: Appealing the primary condition v(0) = 15 in the solution developed in Step 1.

$\mathrm{v}=8$

## Step 6: 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.

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