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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.
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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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3What do you have so far?

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1r = 6cm h = 12cm r = 3cm^3/min \[v = \pi r^2h\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3`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

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3also, you have the wrong volume formula

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3volume of a cone \[\Large V = \frac{1}{3}\pi*r^2*h\]

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1Oh... So then you take the derivative, yes?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3yeah 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

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3a picture might help dw:1437352686967:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3dw:1437352732409:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3yeah that looks good, r = h/2

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3plug that in, derive, then isolate dh/dt

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ 2 }{ 3 } \pi (\frac{ h^2 }{ 2 })\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3When you plugged r = h/2, and simplified, did you get \[\Large V = \frac{1}{12}\pi h^3\] ??

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1So you get \[\frac{ dv }{ dt } = \frac{ 1 }{ 4 } \pi h^2 \frac{ dh }{ dt }\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3dv/dt = 3 h = 9 solve for dh/dt

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1\[3 = \frac{ 1 }{ 4 } \pi (9)^2 \frac{ dh }{ dt }\] \[\frac{ 12 }{ 81 \pi } = \frac{ dh }{ dt }\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3make sure you reduce the fraction as much as possible

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ 4 }{ 27 \pi } = \frac{ dh }{ dt }\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3yep \[\Large \frac{dh}{dt} = \frac{4}{27\pi} \approx 0.047157\] 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

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3glad to be of help
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