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Math Help - another rate of change problem.

  1. #1
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    another rate of change problem.

    A hallow rigght circular cone has height 18cm and base radius 12cm. It is held vertex downwards beneath a tap leaking at the rate of  2 cm^{3} s^{-1}

    Find the rate of rise of water level when the depth is 6cm.
    I have no idea how to start this, I know that  V = \frac{1}{3} \pi r^{2} h

    any help appreciated,

    thanks
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  2. #2
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    Well, you are looking for an expression for dh/dt...once you have this you can substitute 6 for h for your answer.

    You have V in terms of r and h. First use the fact that the ratio of r to h stays constant to find a formula for V in terms of h only. Then you can calculate dV/dh.

    You are given dV/dt, so no you can use the chain rule to find dh/dt.
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  3. #3
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    Quote Originally Posted by mrbohn1 View Post
    Well, you are looking for an expression for dh/dt...once you have this you can substitute 6 for h for your answer.

    You have V in terms of h, so you can calculate dV/dh. And you are given dV/dt.

    Now use the chain rule to find dh/dt.

    thanks,

    The thing I am getting confused with is that we are not given dv/dt. the  2cm^{3} s^{-1} does not represent the depth of the cone, that's just the rate that the water is dripping from the tap.

    so how can you usethe value for dv/dt?

    thanks
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  4. #4
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    is a measure of volume per second, which is the rate of change of volume over time, ie dV/dt=2.
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  5. #5
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Tweety View Post
    I have no idea how to start this, I know that  V = \frac{1}{3} \pi r^{2} h

    any help appreciated,

    thanks
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