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Related Rates - cone draining into cylinder

  • Thread starter aFk-Al
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  • #1
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Homework Statement


Water is draining from a conical tank with height 12 feet and diameter 8 feet into a cylindrical tank that has a base with area [tex]400 \pi[/tex] square feet. The depth, h, in feet, of the water in the conical tank is changing at the rate of (h-12) feet per minute.

A) Write an expression for the volume of the water in the conincal tank as a function of h.

B) At what rate is the volume of the water in the conical tank changing when h=3?

C) Let y be the depth, in feet, of the water in the cylindrical tank. At what rate is y changing when h = 3?


Homework Equations


[tex]V = \frac{1}{3} \pi r^2 h[/tex]


The Attempt at a Solution


First I took the derivative of the volume function using the product rule:
[tex]\frac{dv}{dt} = \frac{2}{3} \ast \pi \ast r \ast \frac{dr}{dt} \ast h + \frac{1}{3} \ast \pi \ast r^2 \ast \frac{dh}{dt}[/tex]
I know that dh/dt = (h-12), but I'm not sure how to find dr/dt and I'm also not sure if I use 4 as the radius (b/c the radius of the water in the cone is constantly changing.) I'm guessing there is another relevant equation I'm missing, or something obvious. I know that the ratio of height of the cone to radius is constant, but I'm not sure if that helps or not.

If anyone could help me figure these out, I'm pretty sure I can do B) and C) on my own. Thanks in advance!
 
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Answers and Replies

  • #2
HallsofIvy
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Homework Statement


Water is draining from a conical tank with height 12 feet and diameter 8 feet into a cylindrical tank that has a base with area [tex]400 \pi[/tex] square feet. The depth, h, in feet, of the water in the conical tank is changing at the rate of (h-12) feet per minute.

A) Write an expression for the volume of the water in the conincal tank as a function of h.

B) At what rate is the volume of the water in the conical tank changing when h=3?

C) Let y be the depth, in feet, of the water in the cylindrical tank. At what rate is y changing when h = 3?


Homework Equations


[tex]V = \frac{1}{3} \pi r^2 h[/tex]


The Attempt at a Solution


First I took the derivative of the volume function using the product rule:
[tex]\frac{dv}{dt} = \frac{2}{3} \ast \pi \ast r \ast \frac{dr}{dt} \ast h + \frac{1}{3} \ast \pi \ast r^2 \ast \frac{dh}{dt}[/tex]
I know that dh/dt = (h-12), but I'm not sure how to find dr/dt and I'm also not sure if I use 4 as the radius (b/c the radius of the water in the cone is constantly changing.) I'm guessing there is another relevant equation I'm missing, or something obvious. I know that the ratio of height of the cone to radius is constant, but I'm not sure if that helps or not.

If anyone could help me figure these out, I'm pretty sure I can do B) and C) on my own. Thanks in advance!
That's exactly why that asked you first, in (a), to write the volume of the water in the cone as a function of h only! Do that first. That has nothing to do with differentiating! If you are not sure how to do that, draw a side view of the conical tank. The cone formed by the entire tank has height twelve and radius 8 while the cone formed by the water has height h and radius r. Of course the triangles formed are similar triangles so the sides are in the same proportion. Once you have the volume of water in the conical tank as a function of h only, it should be easy to differentiate.
 
  • #3
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Answer for A) V = (pi/9)h^3
 
  • #4
Would you mind solving c? and as for a, I actually got v=(1/27)∏h^3

Work process:

r=4 and h=12 so r/h=1/3 ; r= (1/3)h

v=(1/3)∏((1/3h)^2)h
v=(1/3)∏(1/9h^2)h
v=((1/3)*(1/9))∏h^3
v=(1/27)∏h^3
 

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