Answer:
Therefore the surface area of the balloon is increased at 4 cm³/s.
Explanation:
The balloon is being filled with air at a rate of 10 cm³/s
It means the volume of the balloon is increased at a rate 10 cm³/s.
i.e
Consider r be the radius of the balloon.
The volume of of a sphere is
Differentiate with respect to t
The surface of area of the balloon is(S) =
Differentiate with respect to t
Putting the value of
Given that r = 5 cm
=4 cm³/s
Therefore the surface area of the balloon is increased at 4 cm³/s.