calculus

[lilkrazyrae]

Here's another
A container in the shape of an inverted right circular cone has a radius of 6.00 inches at the top and a height of 8.00 inches. At the instant when the water in the container is 7.00 inches deep, the surface level is falling at the rate of -0.800 in/s. Find the rate at which water is being drained.

Any help would be GREATLY appreciated.
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[lilkrazyrae]

HElP

The answer to this problem is -69.3 in^3/s. But i have no idea how to get that please help!!!

 

[tkhunny]

A container in the shape of an inverted right circular cone has a radius of 6.00 inches at the top and a height of 8.00 inches. At the instant when the water in the container is 7.00 inches deep, the surface level is falling at the rate of -0.800 in/s. Find the rate at which water is being drained.

I'll do a "thought process" version. Careful! I might wander down a path I don't like and have to back up a little.

inverted right circular cone

Right circular cone. Volume = (1/3)*pi*r²*h

Note: I don't even know how I'm going to solve this, yet, I'm just writing down stuff I think I might need.

Find the rate at which water is being drained

Oh, well, that's dV/dt, the change in the volume per unit change in time. We'll need the derivative of that volume formula, assuming everything is a function of 't'. I'll use r(t) and r interchangeable, depending on what I want to emphasize.

V(t) = (1/3)*pi*r(t)²*h(t)

dV/dt = (1/3)*pi*(r²*dh + h*2*r*dr)/dt

OK, I'm thinking that's a little more complicated than it has to be. Maybe there is something we can use to simplify things. Besides, we know only how fast the surface level is falling. We DON'T know how fast the radius is shrinking. Maybe we can find a way to express the height in terms of the radius.

a radius of 6.00 inches at the top and a height of 8.00 inches

Aha! A right circular cone is a regular object, so this proportion holds all the way down the inverted cone. Height/8 = Radius/6, or Radius = (6/8)*Height = (3/4)*Height. Substitute this into the volume formula and I expect the derivative will be simpler.

V(t) = (1/3)*pi*r(t)²*h(t)
V(t) = (1/3)*pi*[(3/4)*h(t)]²*h(t)
V(t) = (1/3)*pi*(9/16)*h(t)³
V(t) = (3/16)*pi*h(t)³

dV/dt = (3/16)*pi*3*h²dh/dt
dV/dt = (9/16)*pi*h²dh/dt

At the instant when the water in the container is 7.00 inches deep

We have the 'h' of interest.

the surface level is falling at the rate of -0.800 in/s

We have the dh/dt of interest.

We're all done except for the shouting. You do the rest.

Note: The information about the falling surface level strikes as maybe a little poorly worded. Maybe it's just me.

 

[lilkrazyrae]

sorry

Ok that all works but you inverted the numbers for the Radius = Height it should be 3/4 not 4/3 and that changes the whole problem. So at the end it should be 9/16*pi*3*h^2*dh/dt. This way gives the right answer!!!! Oh and I just wrote the problem straight off the paper. So the wording is weird!

 

[tkhunny]

Good catch.

I thought I didn't like the constants I was getting, but obviously, I didn't think about it enough to go check it.

Sorry about that. Repaired above.