SOLUTION: A balloon is being inflated. If its radius is increasing at a speed of 2cm/s, determine its surface area as a function of time t. Begin with A(x)=4(pi)x^2.
This is the exact qu
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Question 1165072: A balloon is being inflated. If its radius is increasing at a speed of 2cm/s, determine its surface area as a function of time t. Begin with A(x)=4(pi)x^2.
This is the exact question on my assignment and I have no idea how to approach it. I think I am overthinking the question.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ =\ 4\pi{r^2})
But 
is a function of time, )
. We are given the rate of change of , namely 2 cm/s.
\ =\ 2t\ +\ c)
where the constant of integration, 
, represents the initial the initial radius, so 
\ =\ 2t\ +\ r_o)
\ =\ 4\pi(2t\,+\,r_o)^2\,+\,c\ =\ 4\pi\(4t^2\,+\,4tr_o\,+\,r_o^2\))
John
My calculator said it, I believe it, that settles it
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