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Related Rates, why doesn't my solution work?

  1. Sep 28, 2009 #1
    Conical water tank, vertex down. Radius of 10ft, 24 ft high. Water flows into tank at 20 cubic ft/ min. How fast is the depth of the water increasing at 16ft?

    r=10ft
    y=16ft
    dV/dt=20 cubic ft/min
    dy/dt=?

    So here was my attempt:

    V=1/3 (pi r^2) y
    I treated r as a constant so I could find dv/dt in terms of dy/dt

    dV/dt= 1/3 pi r^2 dy/dt
    dy/dt= (dV/dt)3/(pi r^2)

    I found t by taking similar triangles
    10/24=r/16 so r = 10*16/24 = 20/3

    I plugged in the data and got 27/20pi

    The answer is: 9/20pi

    Here's the actual solution:
    they changed r in terms of y
    r=(2/3)y

    V=(1/3) (pi r^2) y substituting r gives V=(1/3) pi(2/3)^2 (y^3)=(4pi/27) y^3
    dV/dt= (4pi/27)(3y^2)dy/dt

    dy/dt= dV/dt * 27/[4pi(3y^2)]
    plugging in 16ft for y yields: 9/20pi, the right answer

    What went wrong with my solution? Both solutions seem mutually correct. They're not, but I fail to recognize what's wrong with it.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 28, 2009 #2

    Dick

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    Your solution assumes r is a constant. It's not. As the level of the water decreases r changes as well as y.
     
  4. Sep 28, 2009 #3
    That makes sense. I think given a similar problem, I would make the same mistake.

    So, next time I get a similar problem, I'll do implicit differentiation, and get the term dr/dt, I would look at my givens and see that I have no info on dr/dt, then I would go back and get rid of r (by substituting it with y) so I don't have to deal with it...

    I appreciate it, thanks.
     
  5. Sep 28, 2009 #4
    Another question: I'm working on a problem involving piling sand in the shape of a cone. The cone's height and diameter are always equal. The information gave that it's height is increasing at a rate of 5 ft /min at the instant when it's 10ft, I'm supposed to find the rate at which the sand is pouring onto the pile.
    V=1/3 pi r^3
    so dV/dt= pi (r^2)dr/dt
    When I worked through it, I was stumped as to what to put down for dr/dt

    it's DIAMETER is increasing at 5 ft/min, If at that moment, it were to stay at 5 ft/min, a minute later the new diameter would be 15ft. the radius went from 5ft to 7.5 ft. So It appears that the radius is increasing at 2.5 ft/min.

    When I plugged in that information, I got the wrong answer. The books answer matches if you assume dr/dt=5ft/min

    What's wrong with me now?
     
  6. Sep 28, 2009 #5

    Dick

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    The volume is V=(1/3)*pi*r^2*h. If h is equal the diameter, then h=2r. That makes the volume (2/3)*pi*r^3. Not (1/3)*pi*r^3. I think what's wrong with you is that you may need some sleep. You just aren't thinking about this very clearly.
     
  7. Sep 28, 2009 #6
    Yeah I think you're right, I worked the problem again given your new information and I got the wrong answer...because in whatever screwed up mathematical system I'm using, 5^3 =625.

    And I have another mystery, but I have a feeling it's not the mystery I think it is. I should look at it tomorrow after I've lain in bed with my eyes shut for 8 hours.

    Thanks for your insight. ZZZ
     
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