SOLUTION: A spherical ballon is being filled with air at a constant rate of 2cm^3 per second. By using the chain rule dr/dt equals to 1/2(pi)(r^2) where t is in seconds. The rate at which th

Question 1142720: A spherical ballon is being filled with air at a constant rate of 2cm^3 per second. By using the chain rule dr/dt equals to 1/2(pi)(r^2) where t is in seconds. The rate at which the radius is increasing when r=54 is 1/5832pi. The balloon will burst when the radius reaches 100cm. Find the rate at which the surface area is increasing at that point in time.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A spherical ballon is being filled with air at a constant rate of 2cm^3 per second. By using the chain rule dr/dt equals to 1/2(pi)(r^2) where t is in seconds. The rate at which the radius is increasing when r=54 is 1/5832pi. The balloon will burst when the radius reaches 100cm. Find the rate at which the surface area is increasing at that point in time.
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dV/dt = 4pi*r^2*dr/dt = 2
dr/dt = 1/(2pi*r^2) which is 1/(5832pi) at r = 54
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dr/dt at r=100 is 1/(2pi*10000) = 1/(20000pi)
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SA = 4pi*r^2
dSA/dt = 8pi*r*(dr/dt)
dSA/dt = 8pi*100/20000pi
= 1/25 sq cm/sec
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"...at that point in time." is superfluous.
"at that point." or "at that time." is better and sufficient.
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No need for "at the point in space, or in the space-time continuum" either.

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