A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.3cm/min. At what rate is the volume of the snowball decreasing when the diameter is 14 cm? (Note the answer is a positive number)

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- anonymous

- schrodinger

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- amistre64

is there an equation for the volume of a sphere?

- anonymous

no :(

- amistre64

... im pretty sure there is. spheres have been studied now for quite some time, and some ancient mathmatikers are bound to have come up with some sort of formula that tells them how to determine the volume of a sphere. If not, i know a way that we can become very very rich :)

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- anonymous

haha i meant there's none that came with the problem! Not that none exist :)

- amistre64

;) you hunt down a formula for it, and then we can see what can be done to find an answer to this question then

- anonymous

it says online that it is 4/3nr^3 so i'm not really sure what that means but thats what i got :) I hope it helps!

- anonymous

the n is a pi

- amistre64

thats the one i was thinking of
to find out how fast the volume is changing, we need to take its derivative; and we are in luck that we only have 1 variable and a power rule to deal with
\[V=\frac43\pi~r^3\]

- anonymous

ok sounds good but i have a class right now :( so I dont know how you wanna do this because I'll be in class untill 12 :(

- amistre64

as long as you know what the power rule for derivatives is, that would help out alot

- anonymous

ok the radius is half the diameter right?

- amistre64

correct

- anonymous

i can find the derivative pretty easily but i'm just unsure what the radius would be

- anonymous

ok let me get you the derivative real quick

- anonymous

the derivative is 4pir^2 so that would be 4pi(7^2) and that would give me 615.44 about

- amistre64

very good, but there is one other part that needs to be addressed; the chain rule pops out an r'

- anonymous

ok... im lost

- amistre64

since the radius is half the diamter; the rate of change of the radius is half the rate of change of the diamter
so multiply that result by .3/2

- anonymous

so that is 923.16 or so?

- amistre64

\[\frac d{dt}(V=\frac43\pi~r^3)\]
\[\frac {dV}{dt}=\frac43\pi~3r^2*\frac{dr}{dt}\]

- amistre64

4pi(7^2)*.3/2 = about 92.36

- anonymous

lol alright i calculated wrong and thanks a million!

- amistre64

with a little practice, these should make more sense :) good luck

- anonymous

thanks :) ill be posting 2 more on here because I completely suck at this so maybe if you're on later you can help me again :) You actually help you don't just say the answer I like that :) Thank you! I have class now so hopefully I'll talk to you later!

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