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Math Help - How to calculate wetted surface area?

  1. #1
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    How to calculate wetted surface area?

    A resevoir has the shape of an inverted cone with altitude 6m and diameter at the top 18m. Water runs in at 3 m^3/s. At the instant the maximum depth is 4m, determine the following.
    a) the rate of increase of depth [completed Answer for \frac{dh}{dt} is \frac{1}{12 \pi} when h=4m
    b) the rate of increase of wetted surface area of the reservoir [answer is
    \frac{\sqrt{13}}{2}

    b) since i am not given \frac{ds}{dt} or \frac{dr}{dt} the only way i can solve for surface area is if i put r and s in terms of h then differentiate, plug in values and hopefully get the answer.

    SA = \pi r^2 + \pi rs

    SA = \pi \left(\frac{3}{2}h\right)^2 + \pi \left(\frac{3}{2}h\right) \left(\frac{\sqrt{13}}{2}h\right)

    SA = \frac{9}{4} \pi h^{2} + \frac{3\sqrt{13}}{4} \pi h^{2}

    \frac{dSA}{dt} = \frac{9}{2} \pi h \frac{dh}{dt} + \frac{3 \sqrt{13}}{2} \pi h \frac{dh}{dt}

    \frac{dSA}{dt} = \frac{9}{2} \pi (4)    \left(\frac{1}{12 \pi} \right)  + \frac{3 \sqrt{13}}{2} \pi (4) \left(\frac{1}{12 \pi} \right)

    \frac{dSA}{dt} = \frac{36 \pi}{24 \pi} + \frac{12 \sqrt{13} \pi}{24 \pi}

    \frac{dSA}{dt} = \frac{3}{2} + \frac{ \sqrt{13}}{2}

    \frac{dSA}{dt} = \frac{3 + \sqrt{13}}{2}

    which is definenty not the answer. So what am i doing wrong???? my logic seems fine... and also I even tried direct subsitution for s in the SA formula and still didnt get the right answer. I am not given \frac{dr}{dt} or \frac{ds}{dt} the only thing i know is to write s and r in terms of h and even then as u see is still the incorrect answer. Please help. Im in dire need of assistance.

    And if u are wondering how i got the slant height. Its from using phythagorean theorum and substituting r = \frac{3}{2}h Then using phythagorean theorum:

    \left( \frac{3}{2}h \right)^2 + h^2 = s^2

    \frac{9}{4} h^2 + h^2 = s^2

    \sqrt{\frac{13}{4}h^2} = s

    \frac{\sqrt{13}}{2}h = s

    And also the relation of r:h is from making a diagram. I do not know how to create pictures in LateX but from reading this problem and making a diagram this is what you should get:

    \frac{r}{h} = \frac{9}{6} = \frac{3}{2}

    2r = 3h

    r = \frac{3}{2}h
    Last edited by ghostanime2001; April 10th 2009 at 08:12 PM.
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  2. #2
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    Quote Originally Posted by ghostanime2001 View Post
    A resevoir has the shape of an inverted cone with altitude 6m and diameter at the top 18m. Water runs in at 3 m^3/s. At the instant the maximum depth is 4m, determine the following.
    a) the rate of increase of depth [completed Answer for \frac{dh}{dt} is \frac{1}{12 \pi} when h=4m
    b) the rate of increase of wetted surface area of the reservoir [answer is
    \frac{\sqrt{13}}{2}

    b) since i am not given \frac{ds}{dt} or \frac{dr}{dt} the only way i can solve for surface area is if i put r and s in terms of h then differentiate, plug in values and hopefully get the answer.

    SA = \pi r^2 + \pi rs

    SA = \pi \left(\frac{3}{2}h\right)^2 + \pi \left(\frac{3}{2}h\right) \left(\frac{\sqrt{13}}{2}h\right)

    SA = \frac{9}{4} \pi h^{2} + \frac{3\sqrt{13}}{4} \pi h^{2}

    \frac{dSA}{dt} = \frac{9}{2} \pi h \frac{dh}{dt} + \frac{3 \sqrt{13}}{2} \pi h \frac{dh}{dt}

    \frac{dSA}{dt} = \frac{9}{2} \pi (4) \left(\frac{1}{12 \pi} \right) + \frac{3 \sqrt{13}}{2} \pi (4) \left(\frac{1}{12 \pi} \right)

    \frac{dSA}{dt} = \frac{36 \pi}{24 \pi} + \frac{12 \sqrt{13} \pi}{24 \pi}

    \frac{dSA}{dt} = \frac{3}{2} + \frac{ \sqrt{13}}{2}

    \frac{dSA}{dt} = \frac{3 + \sqrt{13}}{2}

    which is definenty not the answer. So what am i doing wrong???? my logic seems fine... and also I even tried direct subsitution for s in the SA formula and still didnt get the right answer. I am not given \frac{dr}{dt} or \frac{ds}{dt} the only thing i know is to write s and r in terms of h and even then as u see is still the incorrect answer. Please help. Im in dire need of assistance.

    And if u are wondering how i got the slant height. Its from using phythagorean theorum and substituting r = \frac{3}{2}h Then using phythagorean theorum:

    \left( \frac{3}{2}h \right)^2 + h^2 = s^2

    \frac{9}{4} h^2 + h^2 = s^2

    \sqrt{\frac{13}{4}h^2} = s

    \frac{\sqrt{13}}{2}h = s

    And also the relation of r:h is from making a diagram. I do not know how to create pictures in LateX but from reading this problem and making a diagram this is what you should get:

    \frac{r}{h} = \frac{9}{6} = \frac{3}{2}

    2r = 3h

    r = \frac{3}{2}h
    Kudos to you for showing all your work

    Note that \frac{dS}{dt} = \frac{dS}{dh} \cdot \frac{dh}{dt} = \frac{dS}{dh} \cdot \frac{1}{12 \pi}.

    Note that the wetted surface area of the reservoir does not include the flat circular top of the cone. It is only the sloped side. So S = \pi r l = \pi r \sqrt{h^2 + r^2} = \frac{3 \pi \sqrt{13}}{4} \cdot h^2.
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  3. #3
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    So now do i just differentiate with respect to h for this formula?



    S = \pi r l = \pi r \sqrt{h^2 + r^2} = \frac{3 \pi \sqrt{13}}{4} \cdot h^2
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  4. #4
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    Quote Originally Posted by ghostanime2001 View Post
    So now do i just differentiate with respect to h for this formula?



    S = \pi r l = \pi r \sqrt{h^2 + r^2} = \frac{3 \pi \sqrt{13}}{4} \cdot h^2
    Yes. Then substitute h = 4. Then substitute into the expression for dS/dt.
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  5. #5
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    how do i eliminate the \pi
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  6. #6
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    Quote Originally Posted by ghostanime2001 View Post
    how do i eliminate the \pi
    Substitute everything into the expression for dS/dt. The pi's cancel. If you're having trouble doing this you will need to show your working.
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