Helium is injected into a balloon sphere at a constant rate of 80 (cm^3)/s...

Question

Helium is injected into a balloon sphere at a constant rate of 80 (cm^3)/s. Find the rate of change of its radius and how fast the surface of the balloon change at the instant when the radius is 15 centimeters?

Answer 1

The volume of a sphere is (4/3)pir^3

If you differentiate this with respect to time you get

(dv/dt)= (4pi/3)(3r^2dr/dt)= 4pir^2 dr/dt

dv/dt =80 and r=15 and you want to find dr/dti

80= 4pi(15)^2 dr/dt

dr/dt= 80/900pi= .0283cm/sec rounded to nearest thousandth

Surface area of a sphere is 4pir^2

Differentiate both sides with respect to time

da/dt= 4pi(2r)dr/dt =(8pi)r dr/dt

r=15 and we just calculated dr/dt so

da/dt= (8pi)(15)(.0283)=10 2/3 cm^2/sec