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Rate of Change of Radius of a Cone

Date: 03/07/99 at 00:08:32
From: Andrew Lesnewski
Subject: Rate of change of radius of a cone

My question is:

Use the measured rate of change in volume of water in the cone to 
calculate the rate at which the circular area of water is reducing 
when the radius is r centimetres.

I had to find a funnel and measure its dimensions and measure the rate 
at which water escapes (i.e., dv/dt) and then apply all the data to 
solving the above question. I have gotten as far as finding dv/dr and 
instead of using h in the volume formula, I related h to the radius, 
and it came out to about 1.2r. Is what I am doing right so far? Where 
do I go now? 

Date: 03/07/99 at 03:38:48
From: Doctor Pat
Subject: Re: Rate of change of radius of a cone

It sounds right. Since volume = 1/3 pi r^2 h and in the cone h and r 
are constantly related, if h = 1.2 r then you can write the equation 
totally in terms of r, i.e. v = 1/3 Pi * 1.2 r^3. If 1.2 is exact, 
then you can simplify the two fractions to get v = 2/5 pi 8 r^3. When 
you plug in the rate of change of the volume and the radius you will 
be left with dr/dt as a variable to solve. There does have to be some 
handwaving here about the rate of change of volume being constant 
(which is not technically true, as the head plays a part in the rate, 
but pretend I never told you that). The difference is probably VERY 
small in a common funnel.

 take the derivative of both sides to get the derivative form.

- Doctor Pat, The Math Forum   
Associated Topics:
High School Calculus

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