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Difficult rates of change question (1 Viewer)

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Hey all,

the question is : A conical container with base angle pie/3 is being filled with water at the rate of 5 litres/sec. if the base is r metres and the height is h metres, find the height at which the rate of change in depth of water in the container is 1.19 metres/sec.

The trouble i'm experiencing is determining the volume of the container. Thx for helping
 

braintic

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Is the container sitting apex-up or apex-down?
 
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The question did not have an image to visualise the problem.... so im not sure.
 

braintic

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Sorry, can't do until I know. The solutions are different.
 
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Could we consider the conical container to be in the shape of a cone? and so use the cone volume to solve this?
 

Pharmerbrah

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Hey all,

the question is : A conical container with base angle pie/3 is being filled with water at the rate of 5 litres/sec. if the base is r metres and the height is h metres, find the height at which the rate of change in depth of water in the container is 1.19 metres/sec.

The trouble i'm experiencing is determining the volume of the container. Thx for helping
Normally when we deal with "filling container questions" - it is apex down.

So...



Basically the question is asking...find h when dh/dt = 1.19.
We know that dV/dt = 5 (as it says litres...and litres is VOLUME)

dh/dt = dv/dt * dh/dV

to find dh/dV we need to find an equation relating V and h. How about the volume equation? V = (1/3)(pi)(r^2)(h).
We need to get rid of the r.

Tan 60 = h/r. therefore r = h/(root 3). Replacing this in the equation....V = (pi/9)(h^3). therefore dV/dh = (pi/3)(h^2)

therefore dh/dV = 3/[(pi)(h^2)]

therefore dh/dt = 5 * (3/[(pi)(h^2)]) = 15/[(pi)*(h^2)]

dh/dt = 1.19

so solve: 1.19 = 15/[(pi)(h^2)

and I believe h = 2.0 m.
 

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