Relate Rates - Vertical Gain

**joatmon** · March 8th 2011, 08:44 PM

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:

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

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

I tried setting the problem up like this:

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

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

**emakarov** · March 9th 2011, 02:42 AM

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 $30\sin\theta$ . If $\sin\theta=4/15$ , as I think, then the answer is 8 ft/second.

**joatmon** · March 9th 2011, 06:01 AM

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???

**skeeter** · March 9th 2011, 07:40 AM

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

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

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

**emakarov** · March 9th 2011, 07:46 AM

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, $\sin\theta=4/\sqrt{241}=0.2577$ , not 0.2606. Correspondingly, the vertical velocity is $30\sin\theta=7.73$ ft/second.

The reason this answer is close to 8 is that $\theta$ is relatively small (around $15^\circ$ ), 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.

**joatmon** · March 9th 2011, 08:14 AM

Thanks guys.

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