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Math Help - Cone Calculus Problem

  1. #1
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    Cone Calculus Problem [ Detailed workings and help needed ]

    Last edited by Punch; March 4th 2010 at 03:00 AM.
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  2. #2
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    Is there no reply because none can see the image I posted?
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  3. #3
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    After 3 minutes the volume is 390 - 3 \cdot 60~cm^3 = 210~cm^3.
    The volume of a cone is V = \frac{\pi}{3}r^2 h.
    The 45 degree angle gives r = h~\Rightarrow~V = \frac{\pi}{3}h^3~\Leftrightarrow~h = \left(\frac{3 V}{\pi}\right)^{1/3}.
    \frac{dh}{dt} = \frac{dh}{dV}\cdot\frac{dV}{dt} = \frac{1}{\left(9\pi V^2\right)^{1/3}}\frac{dV}{dt} \approx 0.028~cm/s.
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  4. #4
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    Quote Originally Posted by sitho View Post
    After 3 minutes the volume is 390 - 3 \cdot 60~cm^3 = 210~cm^3.
    The volume of a cone is V = \frac{\pi}{3}r^2 h.
    The 45 degree angle gives r = h~\Rightarrow~V = \frac{\pi}{3}h^3~\Leftrightarrow~h = \left(\frac{3 V}{\pi}\right)^{1/3}.
    \frac{dh}{dt} = \frac{dh}{dV}\cdot\frac{dV}{dt} = \frac{1}{\left(9\pi V^2\right)^{1/3}}\frac{dV}{dt} \approx 0.028~cm/s.

    Correction: After 3mins, the volume is 390-(3\cdot60\cdot3)
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  5. #5
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    Also the answer is -0.0680 cm^3/s
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  6. #6
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    Quote Originally Posted by Punch View Post
    Correction: After 3mins, the volume is 390-(3\cdot60\cdot3)
    Ah, that's true. My mistake. But that means the cone is emptied after 3 mins, or more precisely after 2 mins and 10 secs.
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  7. #7
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    I guess there was a typo in the question. It should be \frac{dV}{dt} = -2~cm^3/s. Then after 3 mins the volume is 390 - 2 \cdot 180~cm^3 = 30~cm^3~\Rightarrow
    \frac{dh}{dt} = -0.0680~cm/s
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  8. #8
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    Did you mean negative 2 or negative 3?
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  9. #9
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    Quote Originally Posted by Punch View Post
    Did you mean negative 2 or negative 3?
    Negative 2. Hence the volume decreases 2~cm^3/s, because if it would decrease 3~cm^3/s as it says in the question it would give a negative volume after 3 mins. Replacing 3 with 2 also gives the correct answer.
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  10. #10
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    However, I really doubt that there is any mistake with the question.... This is because my teacher said she have already done this question in the past years with the past batches and said this question is relatively common.
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  11. #11
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    Quote Originally Posted by Punch View Post
    However, I really doubt that there is any mistake with the question.... This is because my teacher said she have already done this question in the past years with the past batches and said this question is relatively common.
    Yeah, that's a bit strange. Try showing her the expression

    \frac{dh}{dt} = \frac{1}{\left(9\pi\left(V_0 + t\cdot \frac{dV}{dt}\right)^2\right)^{1/3}}\cdot\frac{dV}{dt}

    where V_0 is the initial volume in cm^3, t is the time in seconds and \frac{dV}{dt} is the change in volume w.r.t. time [cm^3/s], and ask her if she gets the same. Because if it's true, then you get the correct answer when inserting 3 mins and 2~cm^3/s.
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  12. #12
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    I guess I will require a detailed working...
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  13. #13
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    Although my teacher have not confirmed this, I conclude that the question has a typo mistake in it and the rate of volume flowing out should instead be 2cm^3/s
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