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anonymous

  • 5 years ago

a tank shape of inverted cone with diameter 40ft depth of 15 ft being filled at rate of 10cubic ft/min how fast is the depth changing when tank filled of 10 ft?

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  1. anonymous
    • 5 years ago
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  2. dumbcow
    • 5 years ago
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    Volume cone = 1/3*h*pi*r^2 given is rate volume is changing dV/dt = 10 required is rate depth or height is changing dh/dt dV/dt = dh/dt * dV/dh notice how dh will cancel on right side now to find dV/dh we need V in terms of h so find relationship between r and h h of tank = 15, r of tank = 20 r/h = 20/15 --> r = 20h/15 plug this in for r in volume formula V = 1/3pi*h*(20h/15)^2 = 1/3pi*(16/9)h^3 dV/dh = pi*(16/9)h^2 dV/dt = dh/dt * dV/dh 10=dh/dt * (pi*(16/9)h^2) dh/dt = 10/(pi*(16/9)h^2) find dh/dt when h=10 dh/dt = 1/pi*160/9 = .0179

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