Chapter 4.6, Problem 38E

To determine

To calculate: A woman at a point $A$ on the shore of a circular lake with radius $2\text{mi }$ wants to arrive at the point $C$ diametrically opposite $A$ on the other side of the lake in the shortest possible time (see the figure). She can walk at the rate of $4\text{mi}/\text{h}$ and row a boat at $2\text{mi}/\text{h}$ . How should she proceed?

Answer to Problem 38E

The least time is take when $\theta =\frac{\pi }{2}$ means she walk along the circumference to reach the point $C$

Explanation of Solution

Given information:

A woman at a point $A$ on the shore of a circular lake with radius $2\text{mi }$ wants to arrive at the point $C$ diametrically opposite $A$ on the other side of the lake in the shortest possible time (see the figure). She can walk at the rate of $4\text{mi}/\text{h}$ and row a boat at $2\text{mi}/\text{h}$ .

Formula used:

Cosine rule: a, b and c are the sides. C is the angle opposite to the side c. The law of cosine says:

  

  $c=\sqrt{a^2+b^2+2ab\cos C}$

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changessign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

• If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
• If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

Draw the diagram of a woman at a point $A$ on the shore of a circular lake with radius $2\text{mi }$ wants to arrive at the point $C$ diametrically opposite $A$ on the other side of the lake in the shortest possible time (see the figure). She can walk at the rate of $4\text{mi}/\text{h}$ and row a boat at $2\text{mi}/\text{h}$ .

  

Recall that, Cosine rule: a, b and c are the sides. C is the angle opposite to the side c. The law of cosine says:

  $c=\sqrt{a^2+b^2+2ab\cos C}$

In $△AOB$

  $\begin{array}{l}AB=\sqrt{2^2+2^2-2\times 2\times 2\cos \left(\pi -2\theta \right)}\\ =\sqrt{4+4+8\cos 2\theta }\left(\text{Use cos}\left(\pi -\theta \right)=-\cos \theta \right)\\ =\sqrt{8+8\cos 2\theta }\\ =\sqrt{8+8\cos 2\theta }\end{array}$

Now use $\cos 2\theta ={\cos }^2\theta -1$ and solve

  $\begin{array}{l}=\sqrt{8+8\left(2{\cos }^2\theta -1\right)}\\ =\sqrt{8+16{\cos }^2\theta -8}\\ =\sqrt{16{\cos }^2\theta }\\ =4\cos \theta \end{array}$

She row the boat at $2\text{mi}/\text{h}$

Therefore,

The time taken to travel $AB$ is

  $\begin{array}{l}=\frac{\text{distance}}{\text{time}}\\ =\frac{4\cos \theta }{2}\\ =2\cos \theta \end{array}$

Length of the arc is

  $\begin{array}{l}BC=\text{radius}\times \text{central angle}\\ \text{=2}\times 2\theta \\ =4\theta \end{array}$

She walk at the speed of $4\text{mi}/\text{h}$

The time taken to travel $BC$ is

  $\begin{array}{l}=\frac{\text{distance}}{\text{time}}\\ =\frac{4\theta }{4}\\ =\theta \end{array}$

Therefore, the total time

  $T\left(\theta \right)=\theta +2\cos \theta ,0\le \theta \le \frac{\pi }{2}$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and

     $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.

3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

• If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
• If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate on both sides,

  $T^{\prime }\left(\theta \right)=1-2\sin \theta \mathrm{......}\left(1\right)$

Solve for $T^{\prime }\left(\theta \right)=0$ , and simplified

  $\begin{array}{l}1-2\sin \theta =0\\ \Rightarrow \sin \theta =\frac{1}{2}\\ \Rightarrow \sin \theta =\sin \left(\frac{\pi }{6}\right)\\ \Rightarrow \theta =\frac{\pi }{6}\end{array}$

Evaluate the $T\left(\theta \right)$ at the critical point and end point

At $\theta =0$

  $\begin{array}{l}T\left(0\right)=0+2\cos 0\\ =2\end{array}$

At $\theta =\frac{\pi }{6}$

  $\begin{array}{l}T\left(\frac{\pi }{6}\right)=\frac{\pi }{6}+2\cos \frac{\pi }{6}\\ \approx 2.6\end{array}$

At $\theta =\frac{\pi }{2}$

  $\begin{array}{l}T\left(\frac{\pi }{2}\right)=\frac{\pi }{2}+2\cos \frac{\pi }{2}\\ \approx 1.57\end{array}$

Differentiate equation $\left(1\right)$ with respect to $x$

  $T^{″}\left(\theta \right)=-2\cos \theta$

Therefore,

  $T^{″}\left(\theta \right)=-2\cos \theta <0$

By second derivate test gives us maximum.

Conclusion:

Thus the least time is take when $\theta =\frac{\pi }{2}$ means she walk along the circumference to reach the point $C$