A spherical balloon is inflated so that its radius (r) increases at a rate of 2/r cm/sec. How fast is the volume of the balloon increasing when the radius is 4 cm?

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Answer 1 · 2017-05-22T19:03:19

#(dv)/dt = 32pi" "("cm"^3)/"s"#

Using the chain rule:

#(dv)/dt = (dv)/(dr)(dr)/dt#

given: #(dr)/dt = 2/r" ""cm"/"s"#

The volume of a sphere is:

#v = 4/3pir^3" cm"^3#

#(dv)/(dr) = 4pir^2" cm"^2#

#(dv)/dt = (4pir^2" cm"^2)(2/r" ""cm"/"s")#

#(dv)/dt = 8pir" "("cm"^3)/"s"" [1]"#

Evaluate equation [1] at #r = 4" cm"#

#(dv)/dt = 32pi" "("cm"^3)/"s"" [2]"#