Question

The top half of a storage tank is a circular cylinder that is 5 meters tall and has a diameter 2 meters. The bottom half of the tank is shaped like an 8-meter inverted cone (pointed down). Let h represent the depth of the tank's contents.
At t = 0 minutes, a release valve at the bottom of the tank is opened and its contents flow out at a rate of 0.5 cubic meters per minute. Assuming the tank is completely full when the release valve is opened, answer the following:
a) Find the value of dh/dt when t = 30 minutes.
b) Find the value of dh/dt when h = 6 meters

Answer 1

The volume of the cylinder is 25pi = 75.54 m^3

At .5 m^3/min that will take 157.08 min to drain.

So, after 30 min, we are still draining the cylinder. Since the cylinder volume is

v = pi r^2h
dv = pi r^2 dh
-.5 = pi dh
dh = -1/2pi m/min

Now, when h = 6, we are draining the cone.

v = 1/3 pi r^2 h

when h=8, r=1, so
when h = 6, r = 3/4
Also, since h = 8r, dh = 8 dr

dv = pi/3 (2rh dr + r^2 dh)
-1/2 = 9pi/16 dh
dh = -8/9pi m/min

better check the math. a lot of fractions went by there.

Answer 2

Oops. Volume of cylinder is just 5pi = 15.71, which will take 31.42 min. to drain. So, even though my volume was way off, it still takes more than 30 min to drain it, and the dh/dt is still correct.

Answer 3

The math on this is way off. He goes from dv=pir^2 dh to-.5=pi dh. the r^2 just disappeared.and then he multiplies the -.5 by pi instead of dividing. -.5 by pi that's not even taking into account the missing r^2. I don't really know how to solve this problem and i would like help on it but i know the math in this problem is not correct and I do believe the logic is correct though. this is my teachers solution.
Step 1) The volume of the cylindrical portion is V = πr2h; = π(1)2(5) = 5πm3. Since is given as -0.5 m3/min, it will take minutes for that portion of the tank to drain.
Step 2) For the cylindrical portion, V = πr2h;. Since r is constant at 1 m, V = πh. Therefore,
Step 3) For the conical portion, V = (π/3)r2h. Since r/h = 1/8, V = (π/192)h3. Therefore,

Answer for Part a) At t = 30, the top of the contents is still in the cylindrical portion of the tank.

The depth is decreasing at a rate of 0.159 m/min.

Answer for Part b) When h = 6, the top of the contents is in the conical portion of the tank.

The depth is decreasing at a rate of 0.283 m/min.

Answer 4

The math would be considered correct, in the problem, R^2 would be a constant and when you derive a constant it would cancel out