(I know this is long, but I tried to be as clear and thorough as possible.)

Here's a problem I have. There is a man walking alongside the edge of a long pool when he notices a valuable object floating in the water at point C. (If you have trouble reading the diagram notation, the label on AC is F(t), the label on AB is x(t), and the label on BC is y(t).)

So he wants to get to C as fast as possible. Note that he is starting his journey at a point that is to the left of A, and he runs faster than he swims. We will assume that his running speed, R, and his swimming speed, S, are each constant. At each point in time, he chooses a path the allows him to maximize his speed of approach to point C. Note that as he stays running on land toward B, his speed of approach to C decreases. (If he stayed running all the way from A to B, his running speed of approach to C would reach zero at B.)

It is at point A that his speed of approach to point C at running speed becomes equal to his speed of approach of swimming directly to point C. He will want to jump in and start swimming at point A then to minimize his time spent getting to C.

I want to find F'(t) in terms of x(t), y(t), and R.

F(t) is the distance from A to C, so

$\displaystyle F(t) = \sqrt{x(t)^2 + y(t)^2}$

At this point, I want to point out a few things. Let's see what happens with F(t) and x(t) as he runs so we can better understand the functions and their implications. Consider the following diagram:

He will jump in the water at some point before getting to B, so x(t) will be positive. But notice that as the man (and therefore x(t)) gets closer to B, x(t) is decreasing to zero. Therefore, $\displaystyle \frac{dx}{dt}$ is negative.

We can see that F(t) is also a decreasing function from A to B, so F'(t) will be negative.

Note that y(t) is a constant as he is running alongside the pool. So $\displaystyle \frac{dy}{dt} = 0$.

Okay, so we now derive F(t).

$\displaystyle F'(t) = \frac{2x(t)\frac{dx}{dt} + 2y(t)\frac{dy}{dt}}{2\sqrt{x(t)^2 + y(t)^2}}$

$\displaystyle F'(t) = \frac{x(t)\frac{dx}{dt} + y(t)\frac{dy}{dt}}{\sqrt{x(t)^2 + y(t)^2}}$

and since $\displaystyle \frac{dy}{dt} = 0$,

$\displaystyle F'(t) = \frac{x(t)\frac{dx}{dt}}{\sqrt{x(t)^2 + y(t)^2}}$

Now, here is the one final part that I cannot wrap my head around.

By definition, $\displaystyle \frac{dx}{dt}$ is his speed as he runs alongside the pool, which we designated before as the constant R. But we also noted before that $\displaystyle \frac{dx}{dt}$ is negative.

So

$\displaystyle F'(t) = \frac{x(t)(-R)}{\sqrt{x(t)^2 + y(t)^2}}$

That fits all of the criteria. F'(t) is negative since $\displaystyle \frac{dx}{dt}$ is negative, as we expected.

However, in the text, F'(t) is given as follows:

$\displaystyle F'(t) = \frac{x(t)(R)}{\sqrt{x(t)^2 + y(t)^2}}$, with R as a positive value.

So I have my answer for F'(t) and the text's solution of F'(t). Is one wrong and the other correct? The text was not completely clear on defining x(t), y(t) and F(t), so is it possible that some other approach could give the other answer and it wouldn't be wrong? I have tried a few other ways to set up the problem and just cannot get the text's answer.

Also, I'm having a little bit of trouble understanding why it is -R (if that is indeed correct). Yes, the man is running in such a way that x(t) is decreasing to zero. And that does mean that $\displaystyle \frac{dx}{dt}$ must be negative. But he is running at speed R in the problem, and I'm confused about why he actually seems to be running at speed -R when all is said and done.