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$begingroup$

This question comes from section 3.11 of the textbook mentioned in the title.

A hemispherical tank with a radius of $10$m is filled from an inflow pipe at a rate of $3$ cubic meters per minute. How fast is the water level rising when the water level is $5$m from the bottom of the tank? (Hint: the volume of a cap of thickness $h$ sliced from a sphere of radius $r$ is $frac{pi}{3}h^2(3r-h)$).

There's a second part of the question that is stated as follows:

What is the rate of change of the area of the exposed surface of the water when the water is $5$m deep?

I have the answer for the second part from an official solution guide. It is as follows:

Set up a triangle with a base of $r$ (representing the radius of the surface of the water) and a height of $10-h$ (where $h$ is the height of the water). Then the hypotenuse is the radius of the sphere: $10$. By application of the Pythagorean theorem, we can find that $r=5sqrt{3}$ when $h=5$. Then, differentiating $r^2+(10-h)^2+10^2$, we get $rfrac{dr}{dt}+hfrac{dh}{dt}=10frac{dh}{dt}$. Then we need to solve for $frac{dr}{dt}$ to plug in for $frac{dA}{dt}=2pi rfrac{dr}{dt}$.

My issue arises at this point in the solution. The book claims that $frac{dh}{dt}=frac{1}{25pi}$. It is clear to me that this should be solved for in the first part of the problem and that this solution comes from differentiating $V(t)=frac{pi}{3}(h(t))^2(3r-h(t))$. What I don't understand is where this formula comes from. They seems to be using $h$ to mean two different things. First, it is defined to be the thickness of the tank. But then it is used as a function of time representing the height of the water. This makes me think that the hint is just poorly worded, and that this is simply a formula for the volume of a hemisphere. But this also doesn't make sense, as this volume would be given by $frac{frac{4}{3}pi r^3}{2}$. These don't appear to be equivalent formulas. To further add to the confusion, the hint would seem to imply that the radius from the center to the outside of the tank is $10$m, but a figure included in the book shows this measurement as being from the center to the inside of the tank.

If there's anything I can do to clarify my issue, let me know. Thanks in advance.

calculus geometry proof-explanation

share|cite|improve this question

asked yesterday

AtsinaAtsina

825117

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In that formula $h$ is the height of the spherical cap, so the word "thickness" is not really appropriate.
  $endgroup$
  – Aretino
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aretino Excellent, thank you.
  $endgroup$
  – Atsina
  yesterday
  

  
  
  
  

add a comment | 

1

0

$begingroup$

This question comes from section 3.11 of the textbook mentioned in the title.

A hemispherical tank with a radius of $10$m is filled from an inflow pipe at a rate of $3$ cubic meters per minute. How fast is the water level rising when the water level is $5$m from the bottom of the tank? (Hint: the volume of a cap of thickness $h$ sliced from a sphere of radius $r$ is $frac{pi}{3}h^2(3r-h)$).

There's a second part of the question that is stated as follows:

What is the rate of change of the area of the exposed surface of the water when the water is $5$m deep?

I have the answer for the second part from an official solution guide. It is as follows:

Set up a triangle with a base of $r$ (representing the radius of the surface of the water) and a height of $10-h$ (where $h$ is the height of the water). Then the hypotenuse is the radius of the sphere: $10$. By application of the Pythagorean theorem, we can find that $r=5sqrt{3}$ when $h=5$. Then, differentiating $r^2+(10-h)^2+10^2$, we get $rfrac{dr}{dt}+hfrac{dh}{dt}=10frac{dh}{dt}$. Then we need to solve for $frac{dr}{dt}$ to plug in for $frac{dA}{dt}=2pi rfrac{dr}{dt}$.

My issue arises at this point in the solution. The book claims that $frac{dh}{dt}=frac{1}{25pi}$. It is clear to me that this should be solved for in the first part of the problem and that this solution comes from differentiating $V(t)=frac{pi}{3}(h(t))^2(3r-h(t))$. What I don't understand is where this formula comes from. They seems to be using $h$ to mean two different things. First, it is defined to be the thickness of the tank. But then it is used as a function of time representing the height of the water. This makes me think that the hint is just poorly worded, and that this is simply a formula for the volume of a hemisphere. But this also doesn't make sense, as this volume would be given by $frac{frac{4}{3}pi r^3}{2}$. These don't appear to be equivalent formulas. To further add to the confusion, the hint would seem to imply that the radius from the center to the outside of the tank is $10$m, but a figure included in the book shows this measurement as being from the center to the inside of the tank.

If there's anything I can do to clarify my issue, let me know. Thanks in advance.

calculus geometry proof-explanation

share|cite|improve this question

asked yesterday

AtsinaAtsina

825117

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In that formula $h$ is the height of the spherical cap, so the word "thickness" is not really appropriate.
  $endgroup$
  – Aretino
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aretino Excellent, thank you.
  $endgroup$
  – Atsina
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

This question comes from section 3.11 of the textbook mentioned in the title.

A hemispherical tank with a radius of $10$m is filled from an inflow pipe at a rate of $3$ cubic meters per minute. How fast is the water level rising when the water level is $5$m from the bottom of the tank? (Hint: the volume of a cap of thickness $h$ sliced from a sphere of radius $r$ is $frac{pi}{3}h^2(3r-h)$).

There's a second part of the question that is stated as follows:

What is the rate of change of the area of the exposed surface of the water when the water is $5$m deep?

I have the answer for the second part from an official solution guide. It is as follows:

Set up a triangle with a base of $r$ (representing the radius of the surface of the water) and a height of $10-h$ (where $h$ is the height of the water). Then the hypotenuse is the radius of the sphere: $10$. By application of the Pythagorean theorem, we can find that $r=5sqrt{3}$ when $h=5$. Then, differentiating $r^2+(10-h)^2+10^2$, we get $rfrac{dr}{dt}+hfrac{dh}{dt}=10frac{dh}{dt}$. Then we need to solve for $frac{dr}{dt}$ to plug in for $frac{dA}{dt}=2pi rfrac{dr}{dt}$.

My issue arises at this point in the solution. The book claims that $frac{dh}{dt}=frac{1}{25pi}$. It is clear to me that this should be solved for in the first part of the problem and that this solution comes from differentiating $V(t)=frac{pi}{3}(h(t))^2(3r-h(t))$. What I don't understand is where this formula comes from. They seems to be using $h$ to mean two different things. First, it is defined to be the thickness of the tank. But then it is used as a function of time representing the height of the water. This makes me think that the hint is just poorly worded, and that this is simply a formula for the volume of a hemisphere. But this also doesn't make sense, as this volume would be given by $frac{frac{4}{3}pi r^3}{2}$. These don't appear to be equivalent formulas. To further add to the confusion, the hint would seem to imply that the radius from the center to the outside of the tank is $10$m, but a figure included in the book shows this measurement as being from the center to the inside of the tank.

If there's anything I can do to clarify my issue, let me know. Thanks in advance.

calculus geometry proof-explanation

share|cite|improve this question

asked yesterday

AtsinaAtsina

825117

$endgroup$

share|cite|improve this question

asked yesterday

AtsinaAtsina

825117

share|cite|improve this question

asked yesterday

AtsinaAtsina

825117

asked yesterday

AtsinaAtsina

825117

asked yesterday

AtsinaAtsina

825117