related rates waterskier rising off ramp I feel this should be an easy one, but something is tripping me up. here is the question:
A waterskier skis over the ramp at a speed of 30ft/s. How fast is she rising as she leaves the ramp? 
(the ramp is a right triangle with adjacent 15ft and opposite 4ft.) 
I want to set it up as a $x^2+y^2=z^2$
so I think I need to differentiate first before plugging values in.
$2x \frac {dx}{dt} + 2y \frac {dy}{dt} = 2z \frac {dz}{dt}$
divide both sides by 2 and plug the values I have in, that being x=15, y=4, dz/dt = 30ft/s, z = $sqrt241$ (because $15^2+4^2 = 241$)
so this leaves me with:
$15 \frac {dx}{dt} + 4 \frac {dy}{dt} = sqrt241*30$
how should I go about finding dx/dt? where did I go wrong? I get really stuck on related rates so any help in the right direction or approach is greatly appreciated. 
 A: There is a relation between $x$ and $y$ from similar triangles, namely $x/y=15/4.$ From this, $x=(15/4)y$ so that $dx/dt=(15/4)dy/dt$ which you may now place in what you have found already, and only $dy/dt$ will remain unfound, which can therefore be determined.
Another perhaps faster way, since you know $dz/dt$ and are only interested in finding $dy/dt,$ is to use the relation $y/z=4/\sqrt{241},$ so that $y=(4/\sqrt{241})z$ and then $dy/dt=(4/\sqrt{241})dz/dt,$ and you only have to fill in the known $dz/dt=30$ ft/sec.
A: My son (and I) struggled with this problem in his calculus class.  Though the other answers satisfactorily solve the problem, I thought I might give a more lengthy explanation for those who have had trouble with this problem (as we did).
The following discussion shows (1) how to arrive at a quick "guesstimate" so that you will know if your ultimate answer is "in the ballpark"; (2) a non-calculus solution (like stretch's answer, just less offensive); and (3) two alternative calculus solutions. 
For the solutions below, assume that x = 15 (the length of the base of the ramp), y = 4 (the height at the end of the ramp), z = 15.524175 (the length of the ramp/hypotenuse), and ${dz\over dt} = 30$ ft/sec.
(1) A quick guesstimate
You know that you are rising approximately 4 feet in approximately 1/2 second, therefore you know that you are rising roughly 8 feet / second.  If your ultimate answer is not in this "ballpark" then you know you did something wrong.
(2) Non-calculus solution
We are traveling 30 feet per second up the ramp.  How long does it take us to reach the end of the ramp?  That is 15.524175 / 30 which is 0.5176725 seconds.  Therefore, at the end of the ramp, we rose 4 feet in 0.5176725 seconds, or 7.7298795 feet per second.  (4/0.5176725 = 7.7298795).  That is close to 8 feet per second.  We are close to our ballpark figure.  What we have calculated is the average rate of change of the height.  But, the rate of change in height is not changing.  It is constant.  Therefore the average rate of change is also the instantaneous rate of change.  So we have our answer.
(3) Calculus solutions
Each calculus solution should give us the same answer we got above, namely, 7.7298795.
Because we are dealing with a (right) triangular ramp, the pythagorean theorem presents itself as a vehicle leading to a solution.  We want to differentiate 
$z^2 = x^2 + y^2$ 
with respect to time.  That will give us
$2z {dz\over dt} = 2x {dx\over dt} + 2y {dy\over dt}$
But we have a problem.  We are given ${dz\over dt}$, and we want to find ${dy\over dt}$, but we are not given, and do not know $dx\over dt$.  We have one equation and two unknowns.  What do we do?  We either have to eliminate $dx\over dt$ entirely from the equation, or we have to figure out what it is.  The first solution below eliminates $dx\over dt$ from the equation; the second solution figures out the value of $dx\over dt$.
(3a) Calculus solution - similar triangles
As @coffeemath pointed out above, we can define x in terms of y using similar triangles.  ${x\over y} = {15\over 4}$ so $x = {15y\over4}$.
Substituting into our formula,
$z^2 = {({15y\over4})}^2 + y^2 $
$z^2 = {241y^2\over16}$
Differentiating both sides with respect to time gives us
${32z\over482y}*30 = {dy\over dt}$
${dy\over dt} = 7.7298795$
(3b) Calculus solution - finding ${dx\over dt}$
When we first differentiated we had a ${dx\over dt}$ term that was unknown:
$2z {dz\over dt} = 2x {dx\over dt} + 2y {dy\over dt}$
How do we find that value?  We know that
${dx\over dt} = {dx\over dz} * {dz\over dt}$
We know the value of ${dz\over dt}$, so if we can find a relation between x and z we can find ${dx\over dz}$.  They are related by $cos(a)$ where a is the acute angle at the base of the ramp.
$cos(a) = {x\over z}$
$x = zcos(a)$
Differentiating both sides, we find that ${dx\over dz} = cos(a)$.  Therefore, ${dx\over dt} = 30cos(a)$
Now we can substitute values into the equation:
$2z {dz\over dt} = 2x {dx\over dt} + 2y {dy\over dt}$
${{(2 * 15.524175 * 30) - (2 * 15 * 30 * {15\over 15.524175})} \over {2 * 4}} = 7.7298795$
A: It doesn't require trigonometry or calculus. It really seems more like a grade school problem.
  Calculate ramp length using Pythagorean Theorem.
  Calculate time on the ramp using t=d/r formula.
  Calculate vertical speed using r=d/t (same formula).
For every 15.52 feet (the ramp length) of travel on the ramp, the skier rises 4 feet (and travels 15 feet over the water surface). Time on the ramp is 15.52/30 or 0.5173 seconds. Vertical rate of rise (at all points on the ramp, and leaving it) is 4/.5173 or 7.732 ft/sec.
A: The previous answers are incorrect because they all assume that the water skier's speed along the ramp is 30 ft/s, but that is the horizontal speed of the boat pulling her and thus her horizontal speed as well (assuming the tow rope does not stretch). Her speed relative to the surface of the ramp will be greater than 30 ft/s. But what we are after is her vertical speed.
If we let $x$ and $y$ represent her horizontal and vertical position respectively and let $s$ represent her position along the ramp then we can relate these changing quantities via the Pythagorean theorem:
$$
s^2 = x^2 + y^2
$$
Implicitly differentiating with respect to time, $t$, yields:
$$
2s\cdot s' = 2x\cdot x' + 2y\cdot y' \qquad \Longrightarrow \qquad
s\cdot s' = x\cdot x' + y\cdot y'
$$
We don't know $s$ nor $s'$, but we can use similar triangles to express them in terms of $x$ and $x'$ respectively,
$$
\frac{s}{x} = \frac{\sqrt{241}}{15} \quad \Longrightarrow \quad
s = \frac{\sqrt{241}}{15} x \quad \Longrightarrow \quad s' = \frac{\sqrt{241}}{15} x'
$$
We can now replace $s$ and $s'$ in our previous equation above:
$$
\frac{\sqrt{241}}{15} x \cdot \frac{\sqrt{241}}{15} x' = x\cdot x' + y\cdot y'
\quad \Longrightarrow \quad
y' = \left( \frac{4}{15} \right)^{2} \frac{x\cdot x'}{y}
$$
Finally, substituting $x=15$ ft, $x'=30$ ft/s and $y=4$ ft into the formula for $y'$ above yields:
$$y'=8 \text{ ft/s.}$$
