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MarcLeclair

  • 3 years ago

A water tank is in the shape of a cone with vertical axis and vertex downward. The tank has a radius of 3m and is high 5m. At first the tank is full of water, but at time t=0 (in seconds), a small hole at the vertex is opened and the water begins to drain. When the height of the water in the tank has dropped to , the water is flowing out at a rate of 2m (cube)/s. At what rate, in meters per seconds, is the water level dropping then? I got this: dv/dt = 2m(cube)/s and we're looking for dh/dt. We also know V=Pi(r^5)(h) /3 and r=3h/5. However I don't know what to do afterwards

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  1. MarcLeclair
    • 3 years ago
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    |dw:1354523543605:dw|

  2. Shadowys
    • 3 years ago
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    you can use \(\frac{dh}{dt}=\frac{dh}{dr} \times \frac{dr}{dV}\times \frac{dV}{dt}\) or just simply sub r=...h into V

  3. Shadowys
    • 3 years ago
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    do you need further help?

  4. MarcLeclair
    • 3 years ago
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    Sorry I hadn't seen the response, but after substituting r, i would find its derivative? And I don't get the first equation youve done :/

  5. Shadowys
    • 3 years ago
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    after sub-bing r, you'll get an eq where V is a function of h. the first eq is the chain rule lol for lazy people like me

  6. MarcLeclair
    • 3 years ago
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    how would that give you a function, that's where I always get confuse. I mean its just a volume equation :/ at a certain point in time

  7. Shadowys
    • 3 years ago
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    yup, that's the function, actually, that eq.

  8. Shadowys
    • 3 years ago
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    at that particular time where the time change is dt.

  9. MarcLeclair
    • 3 years ago
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    Ahhhhh its like as if the volume decrease, heigh decreases, so we could map a graph and therefore find its derivative.... ahaha thanks :)

  10. Shadowys
    • 3 years ago
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    lol you're welcome :)

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