Question

a man launches his boat from point a on a bank of a straight river, 4 km wide, and wants to reach point b, 4 km downstream on the opposite bank, as quickly as possible (see the figure below). he could row his boat directly across the river to point c and then run to b, or he could row to some point d between c and b and then run to b. if he can row 6 km/h and run 8 km/h, how far (in km) downstream from c should he land to reach b as soon as possible? (we assume that the speed of the water is negligible compared to the speed at which the man rows. hint: this question is based on example 4 in section 4.7 of the textbook. however, for this question, the textbook has added a challenge which may require an unexpected solution. look for it!)

Explanation:

Step1: Set up variables

Let \(x\) be the distance from \(C\) to \(D\). Then the distance from \(A\) to \(D\) (rowing distance) is \(\sqrt{4^{2}+x^{2}}\) and the distance from \(D\) to \(B\) (running distance) is \(4 - x\). The time function \(T(x)\) is the sum of the rowing time and running time. The rowing speed \(v_{1}=6\) km/h and the running speed \(v_{2} = 8\) km/h. So \(T(x)=\frac{\sqrt{16 + x^{2}}}{6}+\frac{4 - x}{8}\), where \(0\leq x\leq4\).

Step2: Differentiate the time function

First, use the chain - rule to differentiate \(\frac{\sqrt{16 + x^{2}}}{6}\). Let \(u = 16+x^{2}\), then \(\frac{d}{dx}(\frac{\sqrt{u}}{6})=\frac{1}{6}\times\frac{1}{2\sqrt{u}}\times2x=\frac{x}{6\sqrt{16 + x^{2}}}\). The derivative of \(\frac{4 - x}{8}\) is \(-\frac{1}{8}\). So \(T^{\prime}(x)=\frac{x}{6\sqrt{16 + x^{2}}}-\frac{1}{8}\).

Step3: Set the derivative equal to zero and solve for \(x\)

Set \(T^{\prime}(x) = 0\), then \(\frac{x}{6\sqrt{16 + x^{2}}}-\frac{1}{8}=0\). Cross - multiply to get \(8x=6\sqrt{16 + x^{2}}\). Square both sides: \(64x^{2}=36(16 + x^{2})\). Expand: \(64x^{2}=576+36x^{2}\). Subtract \(36x^{2}\) from both sides: \(28x^{2}=576\). Then \(x^{2}=\frac{576}{28}=\frac{144}{7}\), and \(x=\frac{12}{\sqrt{7}}\approx4.54\) (but \(x\leq4\), so we made a mistake. Let's start again.

Another way: \(T(x)=\frac{(16 + x^{2})^{\frac{1}{2}}}{6}+\frac{4 - x}{8}\)

\(T^{\prime}(x)=\frac{1}{6}\times\frac{1}{2}(16 + x^{2})^{-\frac{1}{2}}\times2x-\frac{1}{8}=\frac{x}{6\sqrt{16 + x^{2}}}-\frac{1}{8}\)

Set \(T^{\prime}(x)=0\)

\(\frac{x}{6\sqrt{16 + x^{2}}}=\frac{1}{8}\)

\(8x = 6\sqrt{16+x^{2}}\)

\(64x^{2}=36(16 + x^{2})\)

\(64x^{2}-36x^{2}=576\)

\(28x^{2}=576\)

\(x^{2}=\frac{144}{7}\)

Since \(x\in[0,4]\), we check the endpoints:

When \(x = 0\), \(T(0)=\frac{\sqrt{16}}{6}+\frac{4}{8}=\frac{4}{6}+\frac{1}{2}=\frac{4 + 3}{6}=\frac{7}{6}\approx1.17\)

When \(x = 4\), \(T(4)=\frac{\sqrt{16 + 16}}{6}+\frac{4-4}{8}=\frac{\sqrt{32}}{6}=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3}\approx0.94\)

Answer:

\(\frac{2\sqrt{2}}{3}\approx0.94\) km from \(C\)