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anonymous
 3 years ago
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)
anonymous
 3 years ago
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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amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1is there an equation for the volume of a sphere?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1... 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 :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha i meant there's none that came with the problem! Not that none exist :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1;) 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
 3 years ago
Best ResponseYou've already chosen the best response.0it 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!

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1thats 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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
 3 years ago
Best ResponseYou've already chosen the best response.1as long as you know what the power rule for derivatives is, that would help out alot

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok the radius is half the diameter right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i can find the derivative pretty easily but i'm just unsure what the radius would be

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok let me get you the derivative real quick

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the derivative is 4pir^2 so that would be 4pi(7^2) and that would give me 615.44 about

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1very good, but there is one other part that needs to be addressed; the chain rule pops out an r'

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1since 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
 3 years ago
Best ResponseYou've already chosen the best response.0so that is 923.16 or so?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac d{dt}(V=\frac43\pi~r^3)\] \[\frac {dV}{dt}=\frac43\pi~3r^2*\frac{dr}{dt}\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.14pi(7^2)*.3/2 = about 92.36

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol alright i calculated wrong and thanks a million!

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1with a little practice, these should make more sense :) good luck

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks :) 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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