Relatively easy math problem...

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  1. #1
    Rad gcn_zelda's Avatar
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    Relatively easy math problem...

    We're learning related rates, and I'm still trying to completely understand them, so I'm not sure how correct my answer to this question is, and I just wanted some help on it.

    A container has the shape of an open right circular cone. The height of the cone is 10 cm and the diameter of the opening is 10 cm. Water in the container is evaporating so that its depth h is changing at the constant rate of -3/10 cm/hr.

    Find the rate of change of the volume of water in the container with respect to time, when h = 5 cm.
    How I think it should be done:
    Code:
    let h = depth of water in container
    let r = radius of surface of water
    V(cone) = 1/3(pi)(r^2)(h)
    
    h/r = 10/5 = 2, so r = h/2
    
    Since I'm trying to relate dh/dt and dV/dt, I solved for r and will plug that into the formula for volume of a cone.
    
    V = 1/3(pi)[(h^2)/4](h)
    V = [(pi)(h^3)]/12
    
    I'll differentiate that:
    
    dV/dt = [2h^2(pi)/3](dh/dt)
    
    Substitute 5 cm for h and -3/10 cm/hr for dh/dt.
    
    dV/dt = [10(pi)/3](-3/10)
    
    Thus, dV/dt = -(pi)
    Did I do this about right?

  2. #2
    aoeuhtns
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    Quote Originally Posted by gcn_zelda
    Did I do this about right?
    No. Had you differentiated correctly, your answer would be correct. I think you're taking a roundabout solution to the problem, too, but it works.

    Don't forget to tack units onto your answer, or else it's lost all meaning.

  3. #3
    Rad gcn_zelda's Avatar
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    Oh, wow. I suck at basic math.

    I just wasn't paying attention when differentiating.

    Well, I just wanted to make sure I've got the basic idea of it down.

  4. #4
    & the hat of GPL slaying Thantos's Avatar
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    Ok did it a little hasty but I used the Chain Rule for One Independent Variable
    dv/dt = ∂v/∂r * dr/dt + ∂v/∂h * dh/dt
    So I started with:
    v(r,h) = π[r(t)]^2 [h(t)] / 3
    h(t) = (-3/10)t + 10 = (-3t+100)/10
    r(t) = (h/2) = (-3t+100)/20 = (-3/20)t + 5
    Then its a matter of finding the derivatives
    ∂v/∂r = 2πrh/3
    ∂v/∂h = (πr^2)/3
    dr/dt = -3/20
    dh/dt = -3/10
    Plug into the above equation and combine turns / reduce / etc
    dv/dt = (2πrh/3)*(-3/20) + (πr^2)/3 * (-3/10)
    = (-2πrh)/20 - (πr^2)/10
    = (-πr/10)(h+r)
    Here I went back to the shape and used the fact that r=h/2
    dv/dt = (πh/20)(h+h/2)
    =(πh/20)( [3/2]h )
    = (-3πh^2)/40
    Here I went back and plugged in h(t) for h to get the function in terms of t instead of h
    dv/dt = (-3π[3t+100]^2)/(40 * 10^2)
    = (-3π)(3t+100)^2 / 4000
    here I went back to h(t) and solved for t when the result was 5
    5 = (-3t+100)/10
    50 = -3t + 100
    -50 = -3t
    t = 50/3
    Now we plug that t value into the above equation:
    dv/dt = (-3π)(3[50/3]+100)^2 / 4000
    = (-3π)(150)^2 / 4000
    = (-3π)(45/8)
    = (-135/8)π cm^3/h

  5. #5
    aoeuhtns
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    Um.... Here's a solution.

    As Δh converges towards zero, (ΔV - A*Δh) converges towards zero, where A is the surface area, since the change in volume is approximable by a cylinder. So ΔV/Δt converges towards 6.25pi Δh/Δt. Hence dV/dt = 6.25pi dh/dt = 6.25pi * -3/10 = -15pi/8 cm^3/hr.

    Quote Originally Posted by Thantos
    dv/dt = (πh/20)(h+h/2)
    =(πh/20)( [3/2]h )
    = (-3πh^2)/40
    Where'd that negative sign come from?

  6. #6
    & the hat of GPL slaying Thantos's Avatar
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    sorry that was caused by a slight difference in my work vs what I typed. The neg came from:
    = (-πr/10)(h+r)
    then sub in r=h/s


    Edit: Hmmm after reworking it with a less tired mind I found an easier way the came up with same answer as Mr. Nazi. As much as I'd like to not hand wave I really don't want to type it up.

  7. #7
    Rad gcn_zelda's Avatar
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    Okay, I understand now. Thanks for the help, guys!

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