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

Algebra ->  Test -> 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      Log On


   

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) About Me  (Show Source):
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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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V+=+4%2Api%2Ar%5E3%2F3
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.