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Rate of change problem

  1. Apr 19, 2008 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant 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]

    3. 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.
  2. jcsd
  3. Apr 19, 2008 #2
    what is h as a function of v?
  4. Apr 20, 2008 #3
    I'm not sure I understand.
  5. Apr 20, 2008 #4


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    Can you relate r and h with the knowledge that the tank is a cone?
  6. Apr 20, 2008 #5
    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.
  7. Apr 21, 2008 #6


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    Staff Emeritus
    Science Advisor

    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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