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shann803
Group Title
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)
 2 years ago
 2 years ago
shann803 Group Title
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)
 2 years ago
 2 years ago

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amistre64 Group TitleBest ResponseYou've already chosen the best response.1
is there an equation for the volume of a sphere?
 2 years ago

amistre64 Group TitleBest 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 :)
 2 years ago

shann803 Group TitleBest ResponseYou've already chosen the best response.0
haha i meant there's none that came with the problem! Not that none exist :)
 2 years ago

amistre64 Group TitleBest 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
 2 years ago

shann803 Group TitleBest ResponseYou've already chosen the best response.0
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!
 2 years ago

shann803 Group TitleBest ResponseYou've already chosen the best response.0
the n is a pi
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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 \[V=\frac43\pi~r^3\]
 2 years ago

shann803 Group TitleBest ResponseYou've already chosen the best response.0
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 :(
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
as long as you know what the power rule for derivatives is, that would help out alot
 2 years ago

shann803 Group TitleBest ResponseYou've already chosen the best response.0
ok the radius is half the diameter right?
 2 years ago

shann803 Group TitleBest ResponseYou've already chosen the best response.0
i can find the derivative pretty easily but i'm just unsure what the radius would be
 2 years ago

shann803 Group TitleBest ResponseYou've already chosen the best response.0
ok let me get you the derivative real quick
 2 years ago

shann803 Group TitleBest ResponseYou've already chosen the best response.0
the derivative is 4pir^2 so that would be 4pi(7^2) and that would give me 615.44 about
 2 years ago

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

shann803 Group TitleBest ResponseYou've already chosen the best response.0
ok... im lost
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

shann803 Group TitleBest ResponseYou've already chosen the best response.0
so that is 923.16 or so?
 2 years ago

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
4pi(7^2)*.3/2 = about 92.36
 2 years ago

shann803 Group TitleBest ResponseYou've already chosen the best response.0
lol alright i calculated wrong and thanks a million!
 2 years ago

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

shann803 Group TitleBest ResponseYou've already chosen the best response.0
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!
 2 years ago
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