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

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.

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.

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.

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

To solve this problem, we need to find the rate at which the height of the tank's contents is changing over time. We can do this by finding the derivative of h with respect to time (dh/dt).

a) To find dh/dt when t = 30 minutes, we first need to determine the relationship between the volume of liquid in the tank and the height h.

The volume V of a cylinder can be found using the formula V = πr^2h, where r is the radius of the cylinder's base. Since the diameter is 2 meters, the radius is 1 meter.

The volume V of an inverted cone can be found using the formula V = (1/3)πr^2h, where r is the radius of the cone's base. In this case, the base of the cone has a radius of 2 meters (half the diameter).

Let's calculate the volumes of both regions of the tank:

Volume of the cylinder = π(1^2)(h - 8)
Volume of the cone = (1/3)π(2^2)(8)

The total volume of liquid in the tank is the sum of these two volumes:

V = π(1^2)h - π(1^2)8 + (1/3)π(2^2)8
V = πh - 8π + (32/3)π

Now we need to differentiate V with respect to t (since we're looking for dh/dt).

dV/dt = (dV/dh)(dh/dt)

To find dV/dh, we differentiate the equation for V with respect to h:

dV/dh = π - 8π

Now we can use the given flow rate of 0.5 cubic meters per minute to find dh/dt when t = 30 minutes.

0.5 = (π - 8π)(dh/dt)

Now, solve for dh/dt:

dh/dt = 0.5 / (π - 8π)

Evaluate this expression to find the value of dh/dt when t = 30 minutes.

b) To find dh/dt when h = 6 meters, we can use the equation we derived earlier:

dh/dt = 0.5 / (π - 8π)

Plug in h = 6 and evaluate to find the value of dh/dt when h = 6 meters.