Calculating Depth of Water in Hemispherical Bowl with Related Rates

[ƒ(x)]

Homework Statement

Water is dripping into a hemispherical bowl with a radius of 8 cm at a rate of 1 cubic cm per minute. At what rate is the depth increasing when it is 4 cm.

A) 1/96π
B) 1/48π
C) 1/24π
D) 1/16π
E) 1/8π

Homework Equations

V = (4πr³)/3
and any other relevant equations, but that's the only one I know of that will help.

The Attempt at a Solution

V = (4πr³)/3
and...that's about it. B is the correct answer btw.

 

[ƒ(x)]

dV/dt = 4πr²(dr/dt)
-1/(4π) = r²(dr/dt)

when h = 4, then r = 4

-1/(4π) = 16(dr/dt)
-1/(64π) = dr/dt

But...that can't be correct.

Since dr/dt is decreasing, dh/dt has to be increasing.

 

[ƒ(x)]

Anyone?

 

[lanedance]

hmmm... haven't checked your working, but as a short cut without driving the full volume of the filled region with depth, for the given depth, work out the radius of the cirlce at the top at relate the total volume change to the volume change of a circluar prism, hieght dz...

 

[lanedance]

dV/dt = 4πr²(dr/dt)
-1/(4π) = r²(dr/dt)

when h = 4, then r = 4

-1/(4π) = 16(dr/dt)
-1/(64π) = dr/dt

But...that can't be correct.

Since dr/dt is decreasing, dh/dt has to be increasing.

so what i think you did here is differentiate the volume of a sphere, which doens;t work for this problem,

it only applies if you are expanding shperically, (ie effectively addind or removing spherical shells of volume

 

[ƒ(x)]

Ok, I tried a different approach to this, but still didnt get the right answer.

 

[ƒ(x)]

Sorry for the large image.

 

[ƒ(x)]

Oh...

Ok...

So if I look at it as being composed of tiny slices, each of Δh height and of ΔV volume I get:

ΔV = (π)(r²)Δh

1 = π(16h-h²)Δh
1 = 48πΔh
1/48π = Δh

But, why doesn't the other method (the one in the image) work?

 

[ƒ(x)]

I guess its because V doesn't equal (2π/3)(r²)h

V = ∫ πr² dh and then the derivative of that is just dV/dt = πr²(dh/dt)

Any ideas?