calculus
A child walks due east on the deck of a ship at 1 miles per hour.
The ship is moving north at a speed of 9 miles per hour.
Find the speed and direction of the child relative to the surface of the water.
Speed = __ mph
The angle of the direction from the north = ___ (radians)
The last time I solved a problem like this was last summer, it looks like a "Related Rates" problem; so, I just need a refresher if anyone could care to help. Thank you very much.
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
$$ f(x) = \frac{L}{2}\log \left( \frac{L+\sqrt{L^2-x^2}}{L-\sqrt{L^2-x^2}} \right) - \sqrt{L^2-x^2} $$
Here is a diagram for your situation. The first ship is at point $A$ at noon and at point $A'$ one hour later. The second ship is at point $B$ at noon and at point $B'$ one hour later.
I teach my calculus class that in related rates problems you should separate the "general" information, which is always true, from the "snapshot" information, which is true only at the relevant moment in time. In your case we have (leaving out the units):
GENERAL INFO:
The first ship is at position $(0,y)$ while the second is at position $(x,0)$.
The distance between them is $z=\sqrt{x^2+y^2}$.
The ship's speeds are given by
$$\frac{dy}{dt}=15$$
$$\frac{dx}{dt}=20$$
SNAPSHOT INFO:
At the relevant time 1:00 p.m.,
$$x=20$$ $$y=-15$$
SOLUTION:
Differentiate the expression for $z$ then substitute the given values for $x,\ y,\ \dfrac{dx}{dt},\ \dfrac{dy}{dt}$.
NOTES:
I use those particular coordinates for the ships in my diagram to get a simple expression for $z$. It should now be clear where your analysis went wrong, but ask if you need details.
Best Answer
Hint: You have a right angle triangle, 1 mi/hr in the easterly direction and 9 mi/hr in the northerly direction. The total speed comes from Pythagoras and the angle from trigonometry.