Question

1. you can run at a speed of 6 mph and swim at a speed of 3 mph and are located on the shore, 4 miles east of an island that is 1 mile north of the shoreline. how far should you run west to minimize the time to reach the island

Explanation:

Step1: Set up the time function

Let \(x\) be the distance (in miles) you run west. Then the distance you run is \(x\) miles, and the distance you swim is \(\sqrt{(4 - x)^{2}+1^{2}}\) miles.
The time \(T(x)\) is the sum of the running time and the swimming time. Using the formula \(t=\frac{d}{v}\) (where \(t\) is time, \(d\) is distance, and \(v\) is speed), we have \(T(x)=\frac{x}{6}+\frac{\sqrt{(4 - x)^{2}+1}}{3}\).

Step2: Differentiate the time function

First, rewrite \(T(x)\) as \(T(x)=\frac{x}{6}+\frac{1}{3}((4 - x)^{2}+1)^{\frac{1}{2}}\).
Using the chain - rule, \(T^{\prime}(x)=\frac{1}{6}+\frac{1}{3}\times\frac{1}{2}((4 - x)^{2}+1)^{-\frac{1}{2}}\times2(4 - x)(- 1)\).
Simplify \(T^{\prime}(x)=\frac{1}{6}-\frac{4 - x}{3\sqrt{(4 - x)^{2}+1}}\).

Step3: Set the derivative equal to zero

\(\frac{1}{6}-\frac{4 - x}{3\sqrt{(4 - x)^{2}+1}} = 0\).
Multiply through by \(6\sqrt{(4 - x)^{2}+1}\) to get \(\sqrt{(4 - x)^{2}+1}-2(4 - x)=0\).
Let \(y = 4 - x\), then \(\sqrt{y^{2}+1}-2y = 0\), \(\sqrt{y^{2}+1}=2y\).
Square both sides: \(y^{2}+1 = 4y^{2}\), \(3y^{2}=1\), \(y=\frac{1}{\sqrt{3}}\) (since \(y>0\) as it represents a length).
Since \(y = 4 - x\), then \(x = 4-\frac{1}{\sqrt{3}}\approx4 - 0.577 = 3.423\) miles.

Answer:

You should run approximately \(4-\frac{\sqrt{3}}{3}\approx3.42\) miles west.