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Rate of change: Height of a conical pile

  1. Dec 4, 2006 #1
    The problem states: sand falls onto a conical pile at a gravel yard at a rare of 10 cubic feet per minute. The base of the pile is approximately three times the altitude. How fast is the pile getting taller when the pile is 15 feet tall?


    Volume = πr² h/3

    dV = 10 h = 15
    dt

    When I use implicit differentiation, I get

    d [V] = π/3(r²dh/dt + 2rh dr/dt)
    dt
    solving for dh/dt I get dV/dt(3/πr² - 2rh dr/dt

    The answer they give is 8/405πm however, without knowing what the dr/dt is, I'm not sure how to solve this. The closest I cpme is 8/15π, bur r=t this is without a quantity for dr/dt.
     
    Last edited: Dec 4, 2006
  2. jcsd
  3. Dec 4, 2006 #2

    AlephZero

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    The base of the pile is approximately three times the altitude.
     
  4. Dec 4, 2006 #3
    I have that. r Approximatelly equals 1.5h, however, I do not know how to find the rate of change for the radius with respect to time (dr/dt) in order to plug that into my equation and solve for the change in height with respect to time t (dh/dt). Since both radius and height change as the volume changes, how do I find the dr/dt to solve for dh/dt?
     
  5. Dec 5, 2006 #4

    HallsofIvy

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    Then use it! replace r with 1.5h so that you have only one variable.

    If [itex]V= (1/3)\pi r^2 h[/itex] and r= 1.5h then [itex]V= (1/3)\pi (1.5)^2h^3[/itex]. Differentiate that!
     
  6. Dec 5, 2006 #5
    Thanks! I see wehere I made my mistake. I went with [tex]2/9\pi\[/tex]r^3

    Iwas eliminatimg the wrong variable.

    I appreciate the help.
     
    Last edited: Dec 5, 2006
  7. Dec 5, 2006 #6
    Since I am making the height as a function of the radius, I don't think I need to squlare the 1.5. Instead, if i just write [(2/3)πh^3]/3, or (2/9)πh^3, I can differentiate thar and hopefully I will get thte corect answer.
     
  8. Dec 6, 2006 #7
    By any chance, has any0ne else worked this problem out and come with an answer of 8/405[itex]pi[/itex]?

    When I put the height as a function of the radius, I get h = 2/3[itex]r^2[/itex]

    so the volume = [itex]2/9\pi h^3[/itex]

    when I differentiate both sides of the equation with respect to t, I get [itex]V\prime = 2/9\pi h^2 h\prime[/itex]

    solving for [itex]h\prime[/itex] I get [itex]h\prime = 3V\prime/2\pi h^2[/itex]

    When I plug in the values for h and [itex]V\prime[/itex] I get an answer of [itex]1/15\pi[/itex]

    Ant suggestions on where I went wrong?
     
  9. Dec 7, 2006 #8

    AlephZero

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    I don't understand where h = 2/3 r^2 came from.

    h = 2/3 r (not squared). But r = 3/2 h is more relevant.

    Volume = 1/3 pi r^2 h = 1/3 pi 9/4 h^2 h = 3/4 pi h^3

    .... which leads to 8/405 pi
     
  10. Dec 7, 2006 #9

    HallsofIvy

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    Several posts back you said "I have that. r Approximatelly equals 1.5h"
    How did r become r2? Was that a typo? If r= (3/2)h then r2= (9/4) h^2.

    No, V= (1/3)\pi r^2 h= (1/3)\pi (9/4)h^2 h= (3/4)\pi h^3
    No, [itex]V'= (3/4)\pi (3h^2 h')= (9/4)\pi h^2 h'[/itex]
    h'= (4V')/(9\pi h^2)[/itex].

    V'= 10, h= 15 so [itex]h'= (40)/(9\pi (225))= (40)/(2025 \pi)= 8/(405\pi) ft per minute is the correct answer for dh/dt.
     
  11. Dec 7, 2006 #10
    thanks for the help. Abput 15 minutes ago. i saw where i made my mistake and finally came up with the corredt answer.
     
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