# Thread: Related rates....a little harder

1. ## Related rates....a little harder

Water is leaking out from a spherical tank of radius 13 feet such that the water level in the tank is falling at a constant rate of 2 ft/hr. At what rate is the radius of the top surface of the water decreasing when the depth of the water in the tank is 8 feet?

Thanks!

Let R=13 ft be the radius of the spherical tank, h be the depth of the water in the tank, and r be the radius of the (circular) surface of the water. Drawing a vertical cross-section of the tank, we see that the distance from the surface to the center of the sphere is R-h, and that a radius r of the surface and the center of the sphere form a right triangle with legs R-h, r, and hypoteneuse R
Thus $r^2 + (R-h)^2 = R^2 \implies r = \sqrt{R^2 - (R-h)^2}$

We find $\frac{dr}{dt}$ using the chain rule:

$\frac{dr}{dt} = \frac{dr}{dh}\frac{dh}{dt}$

We know $\frac{dh}{dt}$ is -2 ft/hr, so we can obtain $\frac{dr}{dt}$ in ft/hr as a function of h from the above. Then just put in R=13 ft and h=8 ft to get the answer.

--Kevin C.