# Homework Help: Derivatives and rates

1. Nov 23, 2004

### SyntheticVisions

I'm having some trouble with a few problems involving derivatives and rates. Here's one of the ones that is confusing me:

A waterskier skis over the ramp shown in the figure at a speed of 30ft/s. How fast is she rising as she leaves the ramp?

The ramp is a right triangle facing the waterskier so that they can go up it, with sides of (assume y is the vertical axis, x is horizontal, z is hypotenuse) x = 15ft, y = 4ft, and by the pythagorean theorem, z = $$\sqrt{241}$$

The first problem I was having here, is whether or not the speed given is relative to the x axis, or the z axis? I would think generally the x axis, but the problem states "over the ramp". Also, when doing the problem for the speed given on the x axis I got a negative number for dy/dt. When doing it given on the z axis I get about 116ft/s, which seems too fast.

Assuming it's on the x I set it up like so (I wasn't sure if I should use z^2, but I don't know what dz/dt is so I didn't...):

$$x^2 + y^2 = \sqrt{241}$$

$$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$$

$$900 + 8 \frac{dy}{dt} = 0$$

$$\frac{dy}{dt} = \frac{-225}{2}$$

Which doesn't make any sense. Where am I going wrong here?

Last edited: Nov 23, 2004
2. Nov 24, 2004

### Astronuc

Staff Emeritus
Assume that the boat pulling the skier is traveling at 30 ft/sec as the skier starts up the ramp and the boat is traveling a constant 30 ft/sec. Then the skier's speed must be higher in the along the hypotenuse to compensate for the longer distance.

However, consider v_x = 30 ft/sec. Then the skier travels the length of the ramp (15 ft) in 15/30 = 0.5 sec.

In the same amount of time, the skier has risen the height.

Now figure out the vertical speed.

3. Nov 24, 2004

### SyntheticVisions

Makes sense, thanks.

4. Nov 24, 2004

### Leo32

By the way, if you approach this problem in a vectorial way, the solution is much more elegant than using derivatives and rates.

Just assume either the 30m/s velocity vector to be either along the hypothenuse or the x-axis, depending on your interpretation of the question (make sure to clearly mention the option you choose, as this will show your scientific approach).
Just split it then according to the other direction and a vertical component in the right way. The speed of the vertical component is then your vertical velocity component.

Regards,
Leo