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zeesbrat3

  • one year ago

A water filter has the shape of an inverted right circular cone with base radius 3 meters and height 9 meters. Water is being pumped into the filter at the rate of 10 meters3/sec. Find the rate, in meters/sec, at which the water level is rising when the water is 3 meters deep.

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  1. zeesbrat3
    • one year ago
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    Hey

  2. ganeshie8
    • one year ago
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    Hey! related rates problem

  3. ganeshie8
    • one year ago
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    |dw:1436729337203:dw|

  4. zeesbrat3
    • one year ago
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    Ya, my favorite...

  5. ganeshie8
    • one year ago
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    |dw:1436729438553:dw|

  6. zeesbrat3
    • one year ago
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    What's the next step?

  7. Astrophysics
    • one year ago
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    I miss these problems, so fun. What we're given is \[\frac{ dV }{ dt } = 10\frac{ m^3 }{ s }\] and you have to find \[\frac{ dh }{ dt }\] ganeshie used similar triangles above to find r so we can use the volume formula \[V = \frac{ 1 }{ 3 }\pi \left( \frac{ h }{ 3 } \right)^2h = \frac{ 1 }{ 9 }\pi h^3\] now we differentiate!!

  8. zeesbrat3
    • one year ago
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    So.. \[\frac{ dv }{dt? } = \frac{ 1 }{ 3 }pih^2\]

  9. Astrophysics
    • one year ago
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    \[\frac{ dV }{ dt } = \frac{ \pi }{ 3 }h^2 \frac{ dh }{ dt }\] notice we differentiate respect to t

  10. Astrophysics
    • one year ago
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    Remember we are using the chain rule here, sorry @ganeshie8 for intruding hehe I just like these problems :D

  11. Astrophysics
    • one year ago
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    Now we solve for \[\frac{ dh }{ dt }\] can you go ahead and try that please

  12. Michele_Laino
    • one year ago
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    here is my reasoning: |dw:1436729996057:dw| we can write: \[\Large \begin{gathered} dV = \pi {r^2}dh \hfill \\ \hfill \\ \frac{{dV}}{{dt}} = \pi {r^2}\frac{{dh}}{{dt}} \hfill \\ \hfill \\ \frac{{dV}}{{dt}} = \pi {\left( {\frac{h}{3}} \right)^2}\frac{{dh}}{{dt}} \hfill \\ \hfill \\ \frac{{dV}}{{dt}} = \frac{{\pi {h^2}}}{9}\frac{{dh}}{{dt}} \hfill \\ \hfill \\ \frac{{dV}}{{dt}} = \frac{\pi }{{27}}\frac{{d\left( {{h^3}} \right)}}{{dt}} \hfill \\ \end{gathered} \]

  13. zeesbrat3
    • one year ago
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    \[10 = \frac{ 1 }{ 3 }\pi(9)^2\frac{ dh }{ dt }\]

  14. Astrophysics
    • one year ago
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    \[\frac{ dh }{ dt } = \frac{ 3 }{ \pi h ^2 } \frac{ dV }{ dt }\]

  15. Astrophysics
    • one year ago
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    \[\frac{ dh }{ dt } = \frac{ 3 }{ \pi(3)^2 } \times 10 = \frac{ 30 }{ 9 \pi ^2 }\]

  16. Astrophysics
    • one year ago
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    That is the rate of water rising

  17. zeesbrat3
    • one year ago
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    So \[\frac{ 30 }{ 88.7364 }\]

  18. zeesbrat3
    • one year ago
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    Or .33807

  19. Astrophysics
    • one year ago
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    The second one

  20. anonymous
    • one year ago
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    This is wrong

  21. anonymous
    • one year ago
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    The answer is 3.18. Astrophysics is wrong on so many levels.

  22. Astrophysics
    • one year ago
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    @dahmanman show your work and I might believe you.

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