# word problem using implicit differentiation

• Nov 28th 2008, 12:31 PM
LexiRae
word problem using implicit differentiation
fluid is entering a spherical tank with a 6 inch radius at a constant rate of 30 pi cubic inches per hour. when the water is h feet deep, the volume of the water in the tank is given by:
V = [(pi)(h^2)/3] (18-h)
what is the rate at which the depth of the water in the tank is increasing at the moment when the water is 2 inches deep?

thanks so much i really appreciate it! im so confused on this one!
• Nov 28th 2008, 01:01 PM
TheEmptySet
Quote:

Originally Posted by LexiRae
fluid is entering a spherical tank with a 6 inch radius at a constant rate of 30 pi cubic inches per hour. when the water is h feet deep, the volume of the water in the tank is given by:
V = [(pi)(h^2)/3] (18-h)
what is the rate at which the depth of the water in the tank is increasing at the moment when the water is 2 inches deep?

thanks so much i really appreciate it! im so confused on this one!

First lets simplfy the equation a bit

$\displaystyle V=6\pi h^2-\frac{\pi h^3}{3}$

Now taking the derviative with respect to time gives

$\displaystyle \frac{dV}{dt}=12 \pi h \frac{dh}{dt}-\pi h^2 \frac{dh}{dt}=\left(12 \pi h-\pi h^2 \right) \frac{dh}{dt}$

Now so we can solve for $\displaystyle \frac{dh}{dt}$ and plug in what we know

$\displaystyle \frac{dh}{dt}=\frac{\frac{dV}{dt}}{12 \pi h-\pi h^2}=\frac{30}{12\pi (2)-\pi (2)^2 }=\frac{30}{20 \pi}=\frac{3}{2\pi}$