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Math Help - Derivative word problem.

  1. #1
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    Derivative word problem.

    No idea what to do here, and all help is appreciated.

    A coffee filter has the shape of an inverted cone. Water drains out of the filter at a rate of 10 cm^3/min. When the depth of the water in the cone is 8 cm, the depth is decreasing at 2 cm/min. What is the ratio of the height of the cone to its radius?

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    Quote Originally Posted by Hibijibi View Post
    No idea what to do here, and all help is appreciated.

    A coffee filter has the shape of an inverted cone. Water drains out of the filter at a rate of 10 cm^3/min. When the depth of the water in the cone is 8 cm, the depth is decreasing at 2 cm/min. What is the ratio of the height of the cone to its radius?

    Help.
    A very similar question was asked here. Read over the reply and then try your question again.
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    That problem kind of confused me. :/
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    Quote Originally Posted by Hibijibi View Post
    That problem kind of confused me. :/
    Then I'm not sure how a solution to your specific problem will be any less confusing for you .....?
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    Well, I understood what was happening, but I don't know how to take what was applied to that problem and use it to find a radius, which was given in that one. Please also bear in mind I am no where near your level of math education, and my perception into all of this is much more limited. I apologize for such things.
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    Quote Originally Posted by Hibijibi View Post
    Well, I understood what was happening, but I don't know how to take what was applied to that problem and use it to find a radius, which was given in that one. Please also bear in mind I am no where near your level of math education, and my perception into all of this is much more limited. I apologize for such things.
    Since you're dealing with a related rates problem, the chain rule is the key:

    \frac{dV}{dt} = \frac{dV}{dh} \, \frac{dh}{dt}.

    When h = 8, \frac{dV}{dt} = -10 \, cubic cm per minute and \frac{dh}{dt} = -2 \, cm per minute.

    Therefore when h = 8: 10 = \frac{dV}{dh} \, 2 \Rightarrow \frac{dV}{dh} = 5.

    Let the radius of the conical volume be x when h = 8. Draw a side view of the filter and use similar triangles:

    \frac{x}{8} =  \frac{r}{h} \Rightarrow r = \frac{hx}{8}.

    Therefore:

    V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left( \frac{hx}{8} \right)^2 h = \pi \, x^2 \, \frac{h^3}{192} \Rightarrow \frac{dV}{dh} = \pi x^2 \frac{h^2}{64}.

    Therefore, when h = 8, \frac{dV}{dh} = 5 \Rightarrow \pi x^2 \frac{8^2}{64} = 5 \Rightarrow \pi x^2 = 5 \Rightarrow x^2 = \frac{5}{\pi}.

    Substitute x^2 = \frac{5}{\pi} into \frac{x}{8} =  \frac{r}{h} \Rightarrow \frac{x^2}{64} = \left( \frac{r}{h} \right)^2 and solve for the required ratio.

    There are probably careless mistakes in the above - check for them. The method is what matters.
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    After looking at it, it seems so easy. Much appreciate the help! I finally get it, sort of!
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