Please help me in this maths question and give full working.?

Question

A spherical balloon is expanding and its volume is increasing at the rate 288cm^3/s when the side is 6cm. Find the rate at which the:

a) side is increasing when the side is 6cm.

b) surface area is increasing when the side is 6cm.

v= 4/3 pie*r^3 A=4pie*r^2

Answer 1

Given:
dV/dt=288 cm^3/s
ill assume the "side" is radius. then r=6cm

a) Find: dr/dt=?
V= 4/3 x pi x r^3 [take the derivative of it]
dV/dt= 4 x pi x r^2 x dr/dt [solve for dr/dt]
dr/dt= [dV/dt] / [4 x pi x r^2] [plug in the numbers]
dr/dt = 288 / [4 x pi x (6)^2] [simplify]
dr/dt = 2/pi cm/s

b)Find: dSA/dt
SA = 4 pi r^2 [take derivative of it]
dSA/dt= 8 pi x r x dr/dt [in part a we got dr/dt so just plug in]
dSA/dt = 8 x pi x 6 x 2/pi [simplify]
dSA/dt= 96 cm^2/s

i hope im correct and if im not then it might be because 6cm is diameter not radius. Just in case 6 was a diameter ill do it as diameter not radius.

Given:
dV/dt=288 cm^3/s
ill assume the "side" is diameter on this one so r = 6/2=3cm

a) Find: dr/dt=?
V= 4/3 x pi x r^3 [take the derivative of it]
dV/dt= 4 x pi x r^2 x dr/dt [solve for dr/dt]
dr/dt= [dV/dt] / [4 x pi x r^2] [plug in the numbers]
dr/dt = 288 / [4 x pi x (3)^2] [simplify]
dr/dt = 8/pi cm/s

b)Find: dSA/dt
SA = 4 x pi x r^2 [take derivative of it]
dSA/dt= 8 x pi x r x dr/dt [in part a we got dr/dt so just plug in]
dSA/dt = 8 x pi x 3 x 8/pi [simplify]
dSA/dt= 192 cm^2/s

hope im right about this one to :p

Answer 2

Just like cattbarf said, there is no side for a sphere since it is a circular solid. Can you please be a specific whether 6cm is a radius or a diameter?

Answer 3

spheres don't have "sides"; rather diameters or radii. Which is a side here?