# related rates of change

• October 2nd 2009, 04:00 PM
alfredo
related rates of change
Hi

I am new to this forum so I hope i post this question at the right level.

A spherical vacuole (something in a cell) is decreasing in size so that its volume decreases at a constant rate of 100 cubic micrometres per seconds. At what rate is the radius decreasing at the instant when the radius of 5 micrometre.

This question needs the chain rule and is based on related rates of change.

So, the [change in the radius wrt time] is equal to the [change in the radius wrt to the volume] * [change in the volume wrt time]

dr /dt = dr/dv * dv/dt

The change in volume wrt time is constant at 100, so can substitute dv/dt for 100 (or -100 because its a decrease?)

dr /dt = dr/dv * 100

Now to find how the radiu changes wrt volume

v = 4/3 * pi * r^3

r = (3/4 * 1/pi * v) ^1/3

when the radius is 5, the volume is 523.6

to find dr/dv, use teh chain rule

r = u^1/3 where u = (3/4 * 1/pi * v)

dr/du = 1/3u^-2/3
du/dv = (3/4 * 1/pi )
= 1/3 (3/4 * 1/pi * v)^-2/3 * (3/4 * 1/pi )
= 1/4 * pi * (3/4 * 1/pi * v)^-2/3

so that's how r changes wrt to the volume. I'm sure that is wrong!

If i then substitute the volume for then the radius is 5

= 1/4 * pi * (3/4 * 1/pi * 523.6 )^-2/3

then multiply * 100 i should have the answer.

I have no idea how they get that. All of the previous examples in the book i am using have been based on the approach i've just used.

many thanks for any help
• October 2nd 2009, 04:10 PM
Jameson
Welcome to MHF!

You're making this way harder than it needs to be.

$V=\frac{4}{3} \pi r^3$

$\frac{dV}{dt}=4 \pi r^2 \frac{dr}{dt}$

You are given dV/dt and r, so use that info to solve for dr/dt.