SOLUTION: A spherical balloons is released from rest and expands as it rises After rising for t seconds its radius is r cm and its surface area is A cm square where A = 4 pi r^2.The initial

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Question 1150211: A spherical balloons is released from rest and expands as it rises After rising for t seconds its radius is r cm and its surface area is A cm square where A = 4 pi r^2.The initial radius of the balloon is 16cm given that the rate of increases of the radius is constant and has the value 0.8 cm s^-1 find the rate of increase of A when t=5.
Answer by greenestamps(13215) About Me  (Show Source):
You can put this solution on YOUR website!


[1] dr%2Fdt+=+0.8

[2] r+=+16%2B0.8t

[3] A+=+4%28pi%29%28r%5E2%29

Find the derivative with respect to time

[4] dA%2Fdt+=+4%28pi%29%282r%2A%28dr%2Fdt%29%29

Evaluate dA/dt at t=5 using dr/dt (from [1]) and r(5) (from[2]).