Derivatives/Rates Word Problem: An inverted conical storage

[tarynt1]

An inverted conical storage tank has a radius of 30 meters and a height of 50 meters. It is being drained at a rate of 10 cubic meters per minute. When the height of the liquid reaches 40 meters, find the rate of change of the surface area of the liquid that is exposed to the air.

I've already done the problem, but I'd like to check my answer because I very well could have made an error.

I started out by finding the derivative, with respect to time, of the formula for the volume of a cone, V = (1/3)(pi)(r^2)(5r/3). (I substituted 5r/3 for h in the formula.) After finding the derivative and substituting variables like r (24) and dV/dt (-10), I found dr/dt to be -1/(96pi).

The next part is confusing to me. The problem asks for the rate of change of the surface area of the liquid exposed to the air, but according to the way I've analyzed it, the liquid exposed to the air can be found by using the area of a circle, rather than any formula for surface area. To me, it's a 2-D rather than 3-D function. Am I misinterpreting something?

Taking the derivative of the area of a circle, A = pi(r^2), and then substituting values for r (24) and dr/dt (-1/(96pi)), I get a final answer of dA/dt = -.5 meters squared per minute.

Can someone clarify or explain this? I'd appreciate any help.

 

[tkhunny]

Of course Surface Area is 2-D. Why do you want it to be 3-D?

The dV hint is just to find dr. This does not require that you stay in 3D. Let your model guide you.

I think you have it, except for the part where you are confusing yourself.

Good work.

 

[arthur ohlsten]

I agree with your answer.
We are both right, or both wrong. But seeing I agree with you, I am probably correct.


define a xy coordinate system with the apex of the cone at 0,0 and height along the y axis. [I find it helps to "see" the problem.]

then the equation of the radius of the surface,x, when the water is at a height y is:
x=[ 30/50] y

the area of the surface is :
A=pi x^2 substitute
A= pi [900/2500]y^2 the rate of change of the area is:
dA/dt =[ 9pi/25]2y dy/dt at y=40
dA/dt=[ 720pi/25] dy/dt
dA/dt=[144pi/5] dy/dt
We must now find dy/dt, the rate at which the height is changing

the volume,V, of the water when the water is at height y is:
V=1/3 pi x^2 y
V= pi/3[900/2500]y^3
V= [3pi/25]y^3

the rate of change of the volume is dV/dt= 10 cubic meters a minute
taking the derivative of V
dV/dt = [3pi/25]3y^2 dy/dt or
10=[9pi/25] y^2 dy/dt or the rate of change of y is
dy/dt = 250/[9piy^2] when y = 40 meters the rate of change is
dy/dt = 250/[14400pi]
dy/dt= 25/[1440pi]
dy/dt=5/[288pi] substitute

dA/dt=[144pi/5][5/(288pi)]
dA/dt = 1/2 square meters per minute answer