How Do You Calculate the Rate of Change in a Hemispherical Tank?

[AndersCarlos]

Homework Statement

Apostol, Vol 1: Section 4.12 Problem 26

Water flows into a hemispherical tank of radius 10 feet (flat side up). At any instant, let h denote the depth of the water, measured from the bottom, r the radius of the surface of the water, and V the volume of the water in the tank. Compute dV/dh at the instant when h=5 feet. If the water flows in at a constant rate of 5√3 cubic feet per second, compute dr/dt, the rate at which r is changing, at the instant t when h=5 feet.

Homework Equations

Hemispherical Volume: [2π(r^3)]/3

The Attempt at a Solution

I tried to rewrite the volume in terms of h. If we consider the circle equation, its possible to say that the radius of the water volume is: r' = √[100 - (h^2)], where 100 is the radius of the tank squared and h is the height. So, we can rearrange the volume formula to: V = {2π{[100 - (h^2)]^3/2}}/3, if we derive this with respect to h, we can find: dV/dh = -2πh√[100 - (h^2)], if we change h for 5, it gives that dV/dh = -10π√75. However, according to the answers at the end, the correct answer would be 75. Sorry if I made any mistakes. I would really appreciate any tip, correction or solution. Thank you.

 

[SammyS]

Homework Statement

Apostol, Vol 1: Section 4.12 Problem 26

Water flows into a hemispherical tank of radius 10 feet (flat side up). At any instant, let h denote the depth of the water, measured from the bottom, r the radius of the surface of the water, and V the volume of the water in the tank. Compute dV/dh at the instant when h=5 feet. If the water flows in at a constant rate of 5√3 cubic feet per second, compute dr/dt, the rate at which r is changing, at the instant t when h=5 feet.

Homework Equations

Hemispherical Volume: [2π(r^3)]/3

The Attempt at a Solution

I tried to rewrite the volume in terms of h. If we consider the circle equation, its possible to say that the radius of the water volume is: r' = √[100 - (h^2)], where 100 is the radius of the tank squared and h is the height. So, we can rearrange the volume formula to: V = {2π{[100 - (h^2)]^3/2}}/3, if we derive this with respect to h, we can find: dV/dh = -2πh√[100 - (h^2)], if we change h for 5, it gives that dV/dh = -10π√75. However, according to the answers at the end, the correct answer would be 75. Sorry if I made any mistakes. I would really appreciate any tip, correction or solution. Thank you.

..."If we consider the circle equation, its possible to say that the radius of the water volume is: r' = √[100 - (h^2)], where 100 is the radius of the tank squared and h is the height."​

This is not a hemisphere when r' < r .

 

[AndersCarlos]

So I would have to consider the spherical cap volume formula instead of hemispherical one?

 

[SammyS]

So I would have to consider the spherical cap volume formula instead of hemispherical one?

Yes.

 

[AndersCarlos]

I was able to achieve the correct result in first part. Thank you very much. Also, I made a little mistake, and r and r' are the same thing. Although I got the right answer using spherical cap formulas.
However, when I got into the second part, I've got another problems. It asks for dr/dt. It says that dV/dt equals 5√3. I take dV/dt and rewrite it as (dV/dr)*(dr/dt). V = (π/6)*h*(3(r^2) + h^2). So I try to rewrite it in terms of 'r', using the formula: 10 = (r^2 + h^2)/2h, where 10 is the radius of the tank. I solve this equation for 'h', using quadratic formula e substitute 'h' in the volume formula for 'r'. The derivative was very long and even solving it, the result was not correct.

 

[AndersCarlos]

SammyS:
After some searching, looking for spherical cap's formulas, I was able to derive the correct answer. Thank you very much again.

 

[SammyS]

SammyS:
After some searching, looking for spherical cap's formulas, I was able to derive the correct answer. Thank you very much again.

You're welcome.

I'm glad you found the answer.