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zeesbrat3

  • one year ago

Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 6cm and a height of 12cm, at the rate of 3 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 9 cm deep? Give an exact answer showing all work and include units in your answer.

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  1. zeesbrat3
    • one year ago
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    @jim_thompson5910

  2. jim_thompson5910
    • one year ago
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    What do you have so far?

  3. zeesbrat3
    • one year ago
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    r = 6cm h = 12cm r = 3cm^3/min \[v = \pi r^2h\]

  4. jim_thompson5910
    • one year ago
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    `r = 3cm^3/min` is incorrect I think you meant to say dV/dt = -3 to represent the fact that the volume is decreasing by 3 cubic cm

  5. jim_thompson5910
    • one year ago
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    also, you have the wrong volume formula

  6. jim_thompson5910
    • one year ago
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    volume of a cone \[\Large V = \frac{1}{3}\pi*r^2*h\]

  7. zeesbrat3
    • one year ago
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    Oh... So then you take the derivative, yes?

  8. jim_thompson5910
    • one year ago
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    yeah but first you have to somehow get everything in terms of h you have to find a connection between r and h and do a substitution

  9. jim_thompson5910
    • one year ago
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    a picture might help |dw:1437352686967:dw|

  10. zeesbrat3
    • one year ago
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    r = 1/2h?

  11. jim_thompson5910
    • one year ago
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    |dw:1437352732409:dw|

  12. jim_thompson5910
    • one year ago
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    yeah that looks good, r = h/2

  13. jim_thompson5910
    • one year ago
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    plug that in, derive, then isolate dh/dt

  14. zeesbrat3
    • one year ago
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    \[\frac{ 2 }{ 3 } \pi (\frac{ h^2 }{ 2 })\]

  15. jim_thompson5910
    • one year ago
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    When you plugged r = h/2, and simplified, did you get \[\Large V = \frac{1}{12}\pi h^3\] ??

  16. zeesbrat3
    • one year ago
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    So you get \[\frac{ dv }{ dt } = \frac{ 1 }{ 4 } \pi h^2 \frac{ dh }{ dt }\]

  17. jim_thompson5910
    • one year ago
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    good

  18. jim_thompson5910
    • one year ago
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    dv/dt = -3 h = 9 solve for dh/dt

  19. zeesbrat3
    • one year ago
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    \[-3 = \frac{ 1 }{ 4 } \pi (9)^2 \frac{ dh }{ dt }\] \[\frac{ -12 }{ 81 \pi } = \frac{ dh }{ dt }\]

  20. jim_thompson5910
    • one year ago
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    make sure you reduce the fraction as much as possible

  21. zeesbrat3
    • one year ago
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    \[\frac{ -4 }{ 27 \pi } = \frac{ dh }{ dt }\]

  22. jim_thompson5910
    • one year ago
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    yep \[\Large \frac{dh}{dt} = -\frac{4}{27\pi} \approx -0.047157\] the units for dh/dt are cm/min the rate of the depth of the water is changing at roughly -0.047157 cm/min at exactly the depth of h = 9 cm

  23. zeesbrat3
    • one year ago
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    Thank you!!

  24. jim_thompson5910
    • one year ago
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    glad to be of help

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