G
Guest
Guest
Water is poured at a constant rate of 20 cm^3/s into a container in the shape of an inverted cone whose semi-verticle angle is 45 degrees. What is the rate at which the level of water is rising when the depth is 50cm?
skeeter
Elite Member
- Joined
- Dec 15, 2005
- Messages
- 3,216
I'm assuming the cone's vertex angle is 90 degrees since its "semi-verticle" (?) angle is 45 degrees.
If so, then height of water in the cone = radius of water in the cone
\(\displaystyle \L V = \frac{\pi}{3}h^3\)
take the derivative of the volume formula w/r to time ... you were given dV/dt, solve for dh/dt when h = 50.
If so, then height of water in the cone = radius of water in the cone
\(\displaystyle \L V = \frac{\pi}{3}h^3\)
take the derivative of the volume formula w/r to time ... you were given dV/dt, solve for dh/dt when h = 50.