Calculus Related Rates Problem Help?

Question

Water is draining through a cone into a cylinder. The cone is 9 cm tall, with a radius of 3 cm. The pot is 15 cm tall with a radius of 4 cm. How fast is the water level in the cylinder changing when the cylinder has 42 cm^3 of water in it, the water level in the cone is 4 cm deep, and the water level in the cone is dropping at a rate of 1.6 cm/min?

I've drawn out a diagram to represent the situation and defined my variables like this:
h = height of cone (9)
r = radius of cone (3)
H = height of cylinder (15)
R = radius of cylinder (4)
t = time
v = volume of cone
V = volume of cylinder

I know that I must find:
dH/dt

When:
V = 42 and h = 3 and dh/dt = 1.6

I'm at the point where I'm writing equations to relate the variables but I'm super stuck. So far I only have:

H = .8356 when V =42 and R = 4

I'm having trouble conceptualizing how these rates and numbers are related to each other.

RealPro · Answer

Assuming the cone is indeed apex-down

Let's fix up a bit
h = height of cone (9 cm) 
r = radius of cone (3 cm) 
R = radius of cylinder (4 cm) 
t = time
volume of cone = 1/3 pi r^2 h (better without evaluation)
v = volume of _water in cone_ (don't care what it is, only need the variable)
V = volume of _water in cylinder_ (don't care what it is, only need the variable)

Additionally, let
y = height of _water in cone_ = 4 cm
x = height of _water in cylinder_ (don't care what it is, only need the variable)

volume and height of cylinder completely irrelevant as long as it can store the water.

So let's go first part:

The water in the cone forms another cone (of height y) that's similar to the first one.
Their volumes are in the ratio of (y / h)^3. So to solve for volume of water in cone...
v = (1/3 pi r^2 h) * (y/h)^3
v = (pi/3)(r / h)^2 * y^3
Differentiate with respect to time on both sides; have in mind r and h are constant
dv/dt = (pi/3)(r / h)^2 * 3y^2 * dy/dt
dv/dt = pi(r / h)^2 * y^2 * dy/dt
Plug in with dy/dt = -1.6 cm/s (don't forget the minus)
dv/dt = pi(3/9)^2 * (4 cm)^2 * (-1.6 cm/s)
dv/dt = -8.936 cm^3/s
(Quick brute force calculation for a change in time of 0.001 s shows difference in volumes is -0.008933 cm^3 so we're close enough)

2nd part:

What flows out of the cone, flows into the cylinder.
dV/dt = 8.936 cm^3 / s
Since V = pi R^2 * x, dV/dt = pi R^2 * dx/dt
dx / dt = 8.936 cm^3/s / (pi R^2) = 8/45 cm/s = 0.1777... cm/s
Common sense shortcut: Each second, the water falling into the pot adds a thin disk of volume 8.936 cm^3 and base area piR^2, so height = volume / base.

? · Answer

The point would be facing downward since the water is draining from the cone into the cylinder

Pope · Answer

We would need more information, particularly the orientations of the cone and cylinder.

I would suppose that the cone had a vertical axis, although even that should have been given. Even given that, we must know whether it is point-up or point-down.

A vertical axis seems likely for the cylinder too. However, you gave the volume of the water in the cylinder, which would be irrelevant if the axis were vertical.

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Calculus Related Rates Problem Help?

3 Answers