Chapter 4.4, Problem 20E

To determine

To find: The distance that Jane should leave her boat to reach the village in the least amount of time.

Answer to Problem 20E

  $\frac{4}{\sqrt{21}}\text{ mi}$

Explanation of Solution

Given information: Jane is 2 mi offshore in the boat and village is 6 mi away from the sea-shore.

Velocity of Jane in water is 2 mph and in land is 5 mph.

Concept used: The maximum or minimum value of the function $f\left(x\right)$ will be at the critical point of that function.

Calculation:

Consider that the distance of Jane’s boat from the point on the shoreline is $x$ .

So, according to question, Jane will row in water for the distance of $\sqrt{4+x^2}\text{ mi}$ and $\left(6-x\right)\text{ mi}$ .

Since the velocity of Jane in water is 2 mph and in land is 5 mph, total time(t) required to reach the destination is $t=\frac{\sqrt{4+x^2}}{2}+\frac{6-x}{5},0<x<6$ .

Differentiate the obtained equation with respect to $x$ .

  $\frac{dt}{dx}=\frac{x}{2\sqrt{4+x^2}}-\frac{1}{5}$

Equate the obtained derivative to 0 and solve for $x$ .

  $\begin{array}{c}\frac{dt}{dx}=0\\ \frac{x}{2\sqrt{4+x^2}}-\frac{1}{5}=0\\ x=\pm \frac{4}{\sqrt{21}}\end{array}$

Since $0<x<6$ , the only value of $x$ is $\frac{4}{\sqrt{21}}$ .

Now, differentiate the equation, $\frac{dt}{dx}=\frac{x}{2\sqrt{4+x^2}}-\frac{1}{5}$ with respect to $r$ .

  $\frac{d^{2}t}{d{x}^2}=\frac{2}{\left(x^2+4\right)\sqrt{x^2+4}}$

Since at $x=\frac{4}{\sqrt{21}}$ , $\frac{d^{2}t}{d{x}^2}>0$ , the distance will be minimum at $x=\frac{4}{\sqrt{21}}$ .

So, the required distance is $\frac{4}{\sqrt{21}}\text{ mi}$.