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 dependent on its radius :

Differentiating with respect to time gives

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

The surface area of the balloon is

and differentiating gives

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

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x

gmt is 4 hrs ahead of edt so just add four hours to 1: 14: 20 (minutes and seconds stay the same)

gmt is 5: 14: 20

hope this !

~just a girl in love with shawn mendes