A spherical balloon is being inflated. Given that the volume of a sphere in terms of its radius is v(r)=(4/3)(pi(r^3))

Question

and the surface area of a sphere in terms of its radius is s(r)=4pi(r^2), estimate the rate at which the volume of the balloon is changing with respect to its surface area when the surface area measures 50 cm^2.

Answer 1

using the area, r = √(s/4π)
so, v = 4/3 π (s/4π)^(3/2) = 1/(6√π) s^(3/2)
dv/ds =1/(4√π) s^(1/2) = 1/4 √(s/π)
at s=50, that would be 5/4 √(2/π)