# Calculus III boat problem

A boat, initially at rest at what we'll consider the coordinate origin, is pointed north (the $+y$ direction). A water-skier is a distance $L$ away due east $+x$ direction) at $(L,0)$, holding onto a rope of length $L$.

$\text{i.)}\,\,$ As the boat moves, assume that the rope stays taut, and the water-skier is always aimed at the current location of the boat, representing a tangent line to the path he is tracing out. If we assume his position can be expressed $(x,f(x))$, show that $$f'(x)=-\dfrac{\sqrt{L^2-x^2}}{x}$$ Hint: draw a triangle, and deduce stuff. See figure on next page, which is very much not to scale.

$\text{ii.)}\,\,$ Nothing that $f(L)=0$, integrate the equation above to determine $f(x)$. Can anybody help me on coming up with the answer and how to get the answer for this problem? Very tricky and confusing.

I also put a awesome photo below the picture to give a better image of the question.

THanks for the help guys!

• Are you having trouble with (i) or (ii)? What have you tried (such as the position of the boat, or the slope of the line)? – Henry Sep 13 '13 at 7:17

My solution assumes that $f$ is $C^1$.

Note that implicit in the problem description is the fact that $x,y$ are smooth (or sufficiently smooth, at least) functions of time, and that the skier's $y$ position can be written as a function of the $x$ coordinate.

(i) We are given that the boat at $(0,y)$ and the skier at $(x,f(x))$ are a distance $L$ apart, this gives the constraint $x^2+(y-f(x))^2 = L^2$. The boat travels 'north' (presumably in a straight line, although the question did not make this explicit) starting from the origin, hence we have $y \ge 0$. The skier starts at $(L,0)$, hence we have $f(L) = 0$.

We are also given that the skier points towards the boat. The skier's direction is presumably the tangent to movement, we have $\frac{d}{dx} (x,f(x)) = \lambda (-x,y-f(x))$, which gives $f'(x) = -\frac{1}{x}(y-f(x))$.

Substituting this into the constraint gives $x^2(1+(f'(x))^2) = L^2$, from which we get $f'(x) = 0$ iff $|x|=L$, in particular, $f'(x)$ does not change sign on $(0,L)$ (here I am relying on $f'$ being continuous).

Suppose $f'(x) >0$ on $(0,L)$, then since $f(L) = 0$, we have $f(x) < 0$ for $x \in (0,L)$, and since $y \ge 0$, this gives $f'(x) = -\frac{1}{x}(y-f(x)) < 0$, a contradiction. Hence $f'(x) < 0$ on $(0,L)$, and then we have $y > f(x)$.

Then we can solve the constraint for $y-f(x)$ to get $y-f(x) = +\sqrt{L^2-x^2}$, from which we obtain $f'(x) = -\frac{\sqrt{L^2-x^2}}{x}$, as required. (ii) We have $f'(x) = -\frac{\sqrt{L^2-x^2}}{x}$. To reduce clutter, let $\phi(t) = f(Lt)$, then $\phi'(t) = Lf'(Lt) = -L \frac{\sqrt{1-t^2}}{t}$. Then we have $\phi(1)-\phi(t) = \int_t^L \phi'(\tau) d\tau$, or $\phi(t) = -\int_t^L \phi'(\tau) d\tau$, since a boundary condition is $\phi(1) = 0$. It is straightforward to compute $\int_t^1 \frac{\sqrt{1-\tau^2}}{\tau} d \tau = \frac{1}{2}\log \frac{1+\sqrt{1-t^2}}{1-\sqrt{1-t^2}} - \sqrt{1-t^2}$, from which we get $f(x) = \phi(\frac{x}{L})$, or

$$f(x) = \frac{L}{2}\log \left( \frac{L+\sqrt{L^2-x^2}}{L-\sqrt{L^2-x^2}} \right) - \sqrt{L^2-x^2}$$

• Thank you so much cooper.hat! – Christopher Sep 18 '13 at 4:24

To think about this think about what the derivative of a function really tells us. It tells us how the slope of the curve is changing over time. The slope at any point is just the $\frac{Rise}{Run}$. The $Run$ is given and the $Rise$ can be found geometrically by using the Pythagorean theorem. The curve in the image is decreasing hence the derivative will also be decreasing.

(i) Note that, the derivative of a function $f(x)$ equals to the $\tan(\theta)$ where $\theta$ is the angle that the tangent line makes with the $x$-axis. Now, to find $\tan(\theta)$, notice that, from the given triangle $\theta=\pi-\alpha$, where $\alpha$ is the angle between the two line segments $x$ and $L$ which implies

$$f'(x)=\tan(\theta)=\tan(\pi-\alpha)=-\tan(\alpha)=-\frac{\sqrt{L^2-x^2}}{x}.$$

• You are missing a minus sign. – copper.hat Sep 14 '13 at 3:41
• @copper.hat: Thanks for the comment. – Mhenni Benghorbal Sep 19 '13 at 10:39