# Thread: Relate Rates - Vertical Gain

1. ## Relate Rates - Vertical Gain

A waterskier skis over a ramp with a length of 15' and a height of 4' at a constant speed of 30 ft/second. How fast is she rising as she leaves the ramp?

Let x = the horizontal base of the ramp, y = the vertical wall, and z = the hypotenuse with theta as the angle in between x and z.

I see the following facts:

$\displaystyle tan \theta = \frac{4}{15}$
$\displaystyle \frac{dz}{dt} = 30 ft/s$
$\displaystyle z = \sqrt{241}$

Obviously, we are trying to solve for $\displaystyle \frac{dy}{dt}$

I tried setting the problem up like this:

$\displaystyle y = z sin\theta$
$\displaystyle \frac{dy}{dt} = \frac{dz}{dt} sin\theta$
$\displaystyle \frac{dy}{dt} = 30(0.2606)$
$\displaystyle \frac{dy}{dt} = 7.818 ft/sec$

I don't think this is the right answer. Does anybody see where I am going wrong?

2. A waterskier skis over a ramp with a length of 15' and a height of 4' at a constant speed of 30 ft/second. How fast is she rising as she leaves the ramp?
To me, this says that the hypotenuse (i.e., z), not the horizontal base of the ramp, is 30'. But in any case, the answer is $\displaystyle 30\sin\theta$. If $\displaystyle \sin\theta=4/15$, as I think, then the answer is 8 ft/second.

3. The picture in the book shows the base at 15'. The only reference to 30' is the speed that the skier is moving when she hits the end of the ramp. I had to re-write somewhat since I didn't have the picture, but that is definitely how the problem is laid out.

At any rate, your answer is consistent with my answer of 7.8 ft/second. Maybe I got it right after all???

4. If the ramp base is 15', then the ramp angle is

$\displaystyle \theta = \arctan\left(\dfrac{4}{15}\right)$

$\displaystyle \dfrac{dy}{dt} = 30\sin{\theta} = 30 \cdot \dfrac{4}{\sqrt{241}} \approx 7.7 \, ft/s$

5. Originally Posted by emakarov
To me, this says that the hypotenuse (i.e., z), not the horizontal base of the ramp, is 30'.
I meant, 15', of course. However, if 15' is the length of the horizontal base, then according to my calculations, $\displaystyle \sin\theta=4/\sqrt{241}=0.2577$, not 0.2606. Correspondingly, the vertical velocity is $\displaystyle 30\sin\theta=7.73$ ft/second.

The reason this answer is close to 8 is that $\displaystyle \theta$ is relatively small (around $\displaystyle 15^\circ$), so x is pretty close to z. The answer is 30y/z. I thought that z = 15, and while the books says that x = 15, the value of z is still pretty close to 15.

6. Thanks guys.

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### a water skier skis over the ramp shown in the figure at a speed of 30 ft /s

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