Chapter 5, Problem 23P

a.

To determine

To show: that if the path followed by the boat is the graph of the function y = f(x), then

  $f\text{'}(x)=\frac{dy}{dx}=\frac{-\sqrt{L^2-x^2}}{x}$ .

Explanation of Solution

Given information: A man initially standing at the point O walks along a pier pulling a

rowboat by a rope length L. The man keeps the rope straight and taut. The path followed by the boat is a curve called a tractrix and it has the property that the rope is always tangent to the curve.

Calculation:

Focus on the rope segment. The length of this rope is always fixed to L .Now draw two lines from this segment, so that the rope becomes the hypotenuse and the two drawn lines are legs of a right triangle.

Since the rope is always tangent to the curve f(x), the slop of this segment is its derivative, f’(x). The dx is obviously x . The dy is not so obvious but can be easily obtained using Pythagoras Theorem since know the lengths of the hypotenuse and another leg.

Hence:

  $\begin{array}{l}{x}^2+d{y}^2=L^2\\ dy=\sqrt{L^2-x^2}.\end{array}$

Since the slope is negative, finally have the following differential function

  $f\text{'}(x)=\frac{dy}{dx}=\frac{-\sqrt{L^2-x^2}}{x}$ .

Hence proved.

b.

To determine

To find: the function y=f(x).

Answer to Problem 23P

  $f(x)=\ln (\frac{L}{x}+\frac{\sqrt{L^2-x^2}}{x})-\frac{\sqrt{L^2-x^2}}{L}$ .

Explanation of Solution

Given information: Given $f\text{'}(x)=\frac{-\sqrt{L^2-x^2}}{x}$

Calculation:

  $\begin{array}{c}f\text{'}(x)=\frac{-\sqrt{L^2-x^2}}{x}.\\ f(x)={\displaystyle \int f\text{'}(x)dx}\\ =-{\displaystyle \int \frac{\sqrt{L^2-x^2}}{x}}dx\\ \text{Using}\text{the}\text{substitution}\\ x=L\sin u\\ dx=L\cos udu\\ -{\displaystyle \int \frac{\sqrt{L^2-L^2{\sin }^{2}u}}{L\sin u}}L\cos udu=-{\displaystyle \int \frac{L\cos u}{L\sin u}}L\cos udu\\ =-L{\displaystyle \int \frac{{\cos }^{2}u}{\sin u}}du\\ =-L{\displaystyle \int \frac{1-{\sin }^{2}u}{\sin u}}du\\ =-L{\displaystyle \int (\csc u-\sin u)du}\\ =-L\ln (\csc u+\cot u)-L\cos u\\ =\ln (\frac{L}{x}+\frac{\sqrt{L^2-x^2}}{x})-\frac{\sqrt{L^2-x^2}}{L}\\ f(x)=\ln (\frac{L}{x}+\frac{\sqrt{L^2-x^2}}{x})-\frac{\sqrt{L^2-x^2}}{L}\end{array}$