Rate of Change of Radius of a ConeDate: 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 http://mathforum.org/dr.math/ |
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