Rate of change problem

  • Thread starter Incog
  • Start date
  • #1
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Homework Statement



A water tank is built in the shape of a circular cone with a height of 6 m and a diameter of 10 m at the top. Water is being pumped into the tank at a rate of 2m[tex]^{3}[/tex]
per minute. Find the rate at which the water level is rising when the water is 2 m deep.

Homework Equations



Volume of a cone - [tex]\frac{1}{3}[/tex] [tex]\Pi[/tex] r[tex]^{2}[/tex] h

Surface area of a cone - [tex]\Pi[/tex] r s + [tex]\Pi[/tex] r[tex]^{2}[/tex]

The Attempt at a Solution



[tex]\frac{dV}{dt}[/tex] = 2m[tex]^{3}[/tex]/min

I think I have to find [tex]\frac{dh}{dt}[/tex] but other than that I'm completely lost.
 

Answers and Replies

  • #2
1,707
5
what is h as a function of v?
 
  • #3
17
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I'm not sure I understand.
 
  • #4
TMM
92
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Can you relate r and h with the knowledge that the tank is a cone?
 
  • #5
1,341
3
What both ice109 and TMM are trying to say is that you have two variables since

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

Both the radius and height are effecting the volume. So before you can find dh/dt you need to find a way to relate r in terms of h.
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
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Draw a triangle, vertex at the bottom, base horizontal, with height 6 and base length 10, representing the water tank. Draw a horizontal line representing the water line in the tank, with length 2r (since the diameter is twice the radius) and height above the vertex h. Use "similar triangles" to connect h and r.
 

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