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Math Help - related rates help

  1. #1
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    related rates help

    water is poured into a conical funnel at a constant rate of 1 in^3/sec and flows out at the rate of 1/2in^3/sec. the funnel is a right circular cone with height of 4in and radius of 2in at the base. how fast is the water level changing when the water is 2in high? (v=1/3 pi r^2 h)
    Last edited by korky0429; December 5th 2010 at 12:08 PM.
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  2. #2
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    To start you off:

    By using similar triangles it is easy to see that r/h = 2/4 where r is the radius at height h.

    Use this to write the volume of the cone as a function as h (instead of r and h)

    Then differentiate, and substitute in the information given.
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  3. #3
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    yea i got that much but would the water be flowing in at 1/2 in^3/sec since its flowing in at 1 and flowing out at 1/2?
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    Yep - that's correct. dV/dt will be 1/2, and you are looking for dh/dt.
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  5. #5
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    so far i have

    (dv/dt)=2/3pi r (dr/dt) (dh/dt)
    1/2=2/3pi 1 (dr/dt) (dh/dt)

    i dont kno dr/dt or dh/dt but im trying to find dh/dt
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  6. #6
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    Quote Originally Posted by korky0429 View Post
    so far i have

    (dv/dt)=2/3pi r (dr/dt) (dh/dt) ... this is incorrect
    1/2=2/3pi 1 (dr/dt) (dh/dt)

    i dont kno dr/dt or dh/dt but im trying to find dh/dt
    since \frac{r}{h} = \frac{1}{2} , r = \frac{h}{2}

    as stated earlier by DrSteve, sub this value in for r in the volume formula.

    V = \frac{\pi}{3} \left(\frac{h}{2}\right)^2 \cdot h

    simplify , take the time derivative, then sub in your given values.
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  7. #7
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    Korky, you took the derivative incorrectly. If you leave V in terms of both h and r, then you need to use a product rule. Although you can do it this way, skeeter's method is more efficient.
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  8. #8
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    is 2pi the right answer? does anyone kno or figure it out?
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  9. #9
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    Quote Originally Posted by korky0429 View Post
    is 2pi the right answer? does anyone kno or figure it out?
    2pi is incorrect ... recheck your arithmetic.
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