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marigirl

  • one year ago

Could someone please help me understand this question and specifically what it means by proportional: Water leaks from the bottom of a tank at a rate proportional to the depth h of the water in the tank. Write down an equation describing this if V is the volume of the water in the tank after time 't' and if the tank is a cylinder show that dh/dt is proportional to h.

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  1. marigirl
    • one year ago
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    @IrishBoy123

  2. marigirl
    • one year ago
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    @AlexandervonHumboldt2

  3. marigirl
    • one year ago
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    so far i understand that \[\frac{ dV }{ dt }\] which is the water leaking is proportional to the depth...How can i mathematically express this?

  4. IrishBoy123
    • one year ago
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    \[\frac{ dV }{ dt } \propto h\] means \[\frac{ dV }{ dt } = k\; h\] where k is some constant now start processing this with \(V = \pi r^2 h\) in mind

  5. marigirl
    • one year ago
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    \[\frac{ dV }{ dt }=\pi r^2 \frac{ dh }{ dt }\] so then right side will equal kh

  6. IrishBoy123
    • one year ago
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    no, LHS will be kh because \(\frac{ dV }{ dt } = k\; h\)

  7. marigirl
    • one year ago
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    \[so \frac{ dh }{ dt }=\frac{ kh }{ \pi r^2 }\]

  8. IrishBoy123
    • one year ago
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    yes \[so \frac{ dh }{ dt }=\frac{ kh }{ \pi r^2 } = const \times h\]

  9. IrishBoy123
    • one year ago
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    or \[\frac{dh}{dt} \propto h\]

  10. marigirl
    • one year ago
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    ah yes! which is what we discussed at the beginning!! thanks! in a nutshell - can you explain what it means when they say proportional

  11. IrishBoy123
    • one year ago
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    for proportional think straight lines, eg \(y = mx + x\), which means \(y \propto x\) the relationship is linear, as x increases y increases. contrast with inversely proportional. here think \(y = \frac{1}{x}\), so \(y \propto \frac{1}{x}\), as y is "inversely proportional" to x

  12. marigirl
    • one year ago
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    Great ! thanks heaps!

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