# The spherical balloon is inflated at the rate of 10 m³/sec. find the rate at which the surface area is increasing when the radius of the sphere is 3m?

the balloon has a volume [tex]v[/tex] dependent on its radius [tex]r[/tex]:

[tex]v(r)=\dfrac43\pi r^3[/tex]

differentiating with respect to time [tex]t[/tex] gives

[tex]\dfrac{\mathrm dv}{\mathrm dt}=4\pi r^2\dfrac{\mathrm dr}{\mathrm dt}[/tex]

if the volume is increasing at a rate of 10 cubic m/s, then at the moment the radius is 3 m, it is increasing at a rate of

[tex]10\dfrac{\mathrm m^3}{\mathrm s}=4\pi (3\,\mathrm m)^2\dfrac{\mathrm dr}{\mathrm dt}\implies\dfrac{\mathrm dr}{\mathrm dt}=\dfrac5{18\pi}\dfrac{\rm m}{\rm s}[/tex]

the surface area of the balloon is

[tex]s(r)=4\pi r^2[/tex]

and differentiating gives

[tex]\dfrac{\mathrm ds}{\mathrm dt}=8\pi r\dfrac{\mathrm dr}{\mathrm dt}[/tex]

so that at the moment the radius is 3 m, its area is increasing at a rate of

[tex]\dfrac{\mathrm ds}{\mathrm dt}=8\pi(3\,\mathrm m)\left(\dfrac5{18\pi}\dfrac{\rm m}{\rm s}\right)=\dfrac{20}3\dfrac{\mathrm m^2}{\rm s}[/tex]

48/40 = 1 8/40

1 8/40 Simplify to 1 1/5

Hope this helps!