Apostol Related Rates Problem

In summary: Just a reminder, please use the hemispherical volume formula for this problem, not the spherical cap volume formula.
  • #1
AndersCarlos
31
0

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.
 
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  • #2
AndersCarlos said:

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 .
 
  • #3
So I would have to consider the spherical cap volume formula instead of hemispherical one?
 
  • #4
AndersCarlos said:
So I would have to consider the spherical cap volume formula instead of hemispherical one?
Yes.
 
  • #5
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.
 
  • #6
SammyS:
After some searching, looking for spherical cap's formulas, I was able to derive the correct answer. Thank you very much again.
 
  • #7
AndersCarlos said:
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.
 

1. What is the Apostol Related Rates Problem?

The Apostol Related Rates Problem is a mathematical problem that involves finding the rate of change of one quantity with respect to another related quantity. It typically involves solving a system of equations using calculus to find the derivative of the related rates.

2. How is the Apostol Related Rates Problem used in real life?

The Apostol Related Rates Problem has many applications in real life, such as in physics, engineering, and economics. For example, it can be used to calculate the rate at which a population is growing, the speed of a moving object, or the change in volume of a container.

3. What are the key steps to solving an Apostol Related Rates Problem?

The key steps to solving an Apostol Related Rates Problem are: 1) identify the related rates and their rates of change, 2) write an equation relating the rates, 3) differentiate both sides of the equation with respect to time, 4) substitute in known values and solve for the unknown rate, and 5) check the solution for reasonableness.

4. What are some common mistakes to avoid when solving an Apostol Related Rates Problem?

Some common mistakes to avoid when solving an Apostol Related Rates Problem include: not correctly identifying the related rates, not differentiating both sides of the equation, making errors in algebraic manipulation, and not checking the solution for reasonableness.

5. Are there any tips for successfully solving an Apostol Related Rates Problem?

Yes, some tips for successfully solving an Apostol Related Rates Problem include: carefully reading and understanding the problem, drawing a diagram to visualize the situation, labeling all known and unknown rates, using appropriate units, and double-checking the solution for accuracy.

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