# Rate of Change in the Legs of a Right Triangle

One leg of a right triangle is always $6$ feet long and the other leg is increasing at a rate of $2$ ft/s. Find the rate of change in ft/s of the hypotenuse when it is $10$ feet long.

The answer is $1.6$

So I try the following formula based on the Pythagorean theorem:

$(6^2)^2 + (2t)^2 = 10^2$

Compute for $t$; which is the time elapsed as the hypotenuse got to $10$. I get $t = 4$. So I divide $10/4$ and I get $2.5$. Am I doing something wrong? Any Hint?

Let $h$ be the length of the hypotenuse and $x$ be the other leg which is varying, then at an instant of time $t$, we have $$x^2+6^2=h^2.$$ Now take the derivative with regards to time $t$, to get $$2x\frac{dx}{dt}=2h\frac{dh}{dt}.$$ Now use the fact that $h=10$ and $\frac{dx}{dt}=2$.

• ok i get [ 2x(dx/dt) = 2h(dh/dt) ] Then substitute your variables; I get... [ (dh/dt) = (1/4)x ] I then substitute the x with the earlier pythagorean relation... getting (1/4)sqrt(10^2 - 6^2)... = 2. It is among the choices! But maybe the answer key maybe wrong? – james Jun 13 '15 at 6:41
• @james You should have $\frac{\text{d}h}{\text{d}t} = \frac{1}{4}x\frac{\text{d}x}{\text{d}t}$. – N. F. Taussig Jun 13 '15 at 9:38

The answer key is correct, the answer is 1.6. We are asked to find the dh/dt not dx/dt.

dy/dt (2y)=dh/dt(2h)

Rearranging we get

dh/dt= [dy/dt(2y)]/2h

Subtitute the given h=10, dy/dt=2, and find y using pythagorean theorem (√10^2 - 6^2 ) = 8

Then we get 1.6

Try it..