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Isosceles Triangular Prism Related Rates Problem

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A trough is 9 ft long and its ends have the shape of isosceles triangles that are 5 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 14 ft3/min, how fast is the water level rising when the water is 7 inches deep?

    I know b, h, l, dv/dt, dl/dt.
    I need to first find db/dt then solve for dh/dt

    2. Relevant equations

    Volume of Iso. triangular prism= 1/2bh*l
    dv/dt=1/2bh(dl/dt)+l(1/2b(dh/dt)+1/2h(db/dt)

    I assume that dl/dt=0, so the new equation for the derivative is equal to.. dv/dt=l(1/2b(dh/dt)+1/2h(db/dt)

    3. The attempt at a solution
    If anyone could just give me some direction where to start in solving for db/dt that would be great!
     
  2. jcsd
  3. Oct 11, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi maladroit! Welcome to PF! :smile:
    oooh, that's far too complicated! :cry:

    There's a relation between b and h, so write b in terms of h, and then differentiate! :wink:
     
  4. Oct 11, 2009 #3
    That seems to be just what my problem is... I can't see the relationship between b and h.

    I tried setting up a triangle and solving for what b is equal to in terms of h but I am still getting the wrong answer.
    I used cos(theta)=(1/2b)/h, differentiated, and solved for what db/dt was equal to.
     
  5. Oct 11, 2009 #4

    tiny-tim

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    uhh? :confused: … it's a triangle … it's the same all the way up … b = 5h.
     
  6. Oct 11, 2009 #5
    Once I replaced the db/dt with the 5dh/dt and solved for the problem I got .4817 ft/min but that was incorrect.

    The equation I used was...
    14=9(1/2*5(dh/dt)+1/2*(7/12)5(dh/dt)
     
  7. Oct 11, 2009 #6

    tiny-tim

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    Sorry, not following that :confused:

    can you write it out properly: what is v in terms of h, then what is dv/dt in terms of h, then what is dv/dt when h = 7/12. :smile:
     
  8. Oct 11, 2009 #7
    Right!
    I used the same equation I put before
    dv/dt=l(1/2b(dh/dt)+1/2h(db/dt))
    and, differentiating b=5h,

    dv/dt=l(1/2b(dh/dt)+1/2h(5dh/dt))

    Also, I really appreciate your help!!!
     
  9. Oct 11, 2009 #8

    tiny-tim

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    hmm … v = (22.5)h2, dv/dt = … ? :wink:
     
  10. Oct 11, 2009 #9
    oh!!! I differentiated much too early... thank you so much for your help!!!
     
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