Gasoline is pouring into a cylindrical tank of radius 3 feet. When the depth of the gasoline is 4 feet, the depth is increasing at 0.2 ft/sec. How fast is the volume of gasoline changing at that instant? Include units

Question

anonymous · Answer

i know the goal is to find dv/dt, but im not sure on how to do that
$$v=\pi r^2*h$$

cruffo · Answer

first find the volume when r = 3 and h = 4  ( i got $v=36\pi$)
Thus you know:
r = 3   h = 4   v = 36$\pi$
dr/dt = 0  (radius is not changing)
dh/dt = 1/5

Take derivative of formula with respect to time (must use product rule), substitute known values, and solve for dv/dt:

anonymous · Answer

i'll try it out thanks!

cruffo · Answer

as a check, i got $\large\dfrac{9\pi}{5}$ ft/sec.

anonymous · Answer

$$dv/dt=\pi *r^2*(dh/dt)$$
and once i subsitute, i think i do get 9pi/5

anonymous · Answer

thanks alot!!

cruffo · Answer

np