Calculus problem, help!

Question

Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 6cm and a height of 12cm, at the rate of 3 cm³/min. At what rate is the depth of the water changing at the instant when the water in the tank is 9 cm deep? Include all the units and work please. Thank you!.

Answer 1

What do we know?

r=6cm

h=12cm

dv/dt = -3cm³/min (this is negative because it's being drained. I also know it's dv/dt because of the units. cm³ measure volume.)

What do we want to know?

dh/dt when h=9

We were told the shape is a right circular cone. Since we were given information about the volume, we're going to use the volume equation: V=pir²(h/3)

There are too many values to differentiate right now. We want V and h, but we need to replace r. We can do this because we know the relationship between h and r

h/r = 12/6

h/r = 2

h = 2r

r = h/2

So, V = pi (h/2)²(h/3)

Simplify:

V = pi (h²/4)(h/3)

V = pi h³/12

Differentiate both sides of the equation. Remember, in a related rates problem, whenever you differentiate a variable, you multiply by d/dt.

dv/dt = pi/4h² (dh/dt)

Plug in what you were given in the original problem:

-3cm³/min = pi/4 (9cm)² (dh/dt)

-3cm³/min = 81pi cm² /4 (dh/dt)

-12cm³/min = 81pi cm² (dh/dt)

-12cm³/min / 81pi cm² = (dh/dt)

dh/dt = -4/27pi cm

Answer 2

Probably the hardest part of this problem is the mathematical modeling stage. It is going to be based on the formula for volume of a cone, which is V = ¹/₃ πr²h. One key factor is that the radius is 6 an the height 12. As this inverted cone fills with water, the shape of the volume of water is a smaller cone that is increasing in volume and is similar to the larger cone that is the container. The relevant fact about this container's dimensions is that the radius is half of the height. So at a given water depth, the radius will be half of that. This will enable us to modify the volume formula to have just one variable, and we want that variable to be h.

r = ^h/₂

V = ¹/₃ π(^h/₂)² h

V = ¹/₁₂ πh³

We then perform implicit differentiation on this function with respect to time t.

^dV/_dt = ¹/₄ πh² (^dh/_dt)

Now all that remains is to substitute all the known values, and solve for the unknown rate of change of the height.

3 = ¹/₄ π(9²)(^dh/_dt)

3 = ^81π/₄ (^dh/_dt)

(^dh/_dt) ≈ 0.0472 cm/min