Question

The current speed ${\mathbf{v}}_{\mathbf{r}}$of a straight river such as that in Problem 28 is usually not a constant. Rather, an approximation to the current speed (measured in miles per hour) could be a function such as ${\mathbf{v}}_{\mathbf{r}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{30}\mathbf{x}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{,}\mathbf{0}⩽\mathbf{x}⩽\mathbf{1}$, whose values are small at the shores (in this case, ${\mathbf{v}}_{\mathbf{r}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$and localid="1663928505474" ${\mathbf{v}}_{\mathbf{r}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$) and largest in the middle of the river. Solve the DE in Problem 30 subject to $\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$, where ${\mathbf{v}}_{\mathbf{s}}\mathbf{=}\mathbf{2}\mathbf{mi}\mathbf{/}\mathbf{h}$and ${\mathbf{v}}_{\mathbf{r}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is as given. When the swimmer makes it across the river, how far will he have to walk along the beach to reach the point (0, 0)?

Short answer

The swimmer needs to walk 2.5 miles south along the west beach to reach the origin.

Step-by-step solution

Definition of Integration

Integration is a technique of finding a function g(x) the derivative of which, Dg(x), is equal to a given function f(x).

Given Data

We are given the differential equation

$\frac{dy}{dx}=-\frac{v_r}{v_x}$

where

role="math" localid="1663928879167" $\begin{array}{rcl}{v}_r& =& -30x(1-x)\\ & =& -30\left(x-x^2\right)\end{array}$

And $v_s=2$.

We are also given the initial condition $y\left(1\right)=0$.

Substitute and integrate

We begin by substituting$v_{\mathbf{r}}$and$v_s$into the differential equation and integrating.

$\begin{array}{rcl}\frac{dy}{dx}& =& -\frac{30\left(x-x^2\right)}{2}\\ \int dy& =& \int -15\left(x-x^2\right)dx\\ y& =& -15\left(\frac{1}{2}{x}^2-\frac{1}{3}{x}^3\right)+e\end{array}$

Solve for constant

Now we substitute the initial condition to solve for constant c.

$\begin{array}{rcl}0& =& -15\left(\begin{array}{cc}1& 1\\ 2& 3\end{array}\right)+c\\ 0& =& -\frac{5}{2}+c\\ \frac{5}{2}& =& c\end{array}$

The solution to the differential equation becomes$y=-15\left(\frac{1}{2}{x}^2-\frac{1}{3}{x}^3\right)+\frac{5}{2}$.

In order to find where on the west beach the swimmer lands we need to find$y\left(0\right)$.

$\begin{array}{rcl}y\left(0\right)& =& -15\left(\frac{1}{2}{0}^2-\frac{1}{3}{0}^3\right)+\frac{5}{2}\\ & =& \frac{5}{2}\end{array}$

So, the required solution is 2.5 km.