Question

A spherical balloon is being filled with air at the rate of 25
cm^3/s. How fast is the radius increasing when the balloon is
20 cm in diameter?​

Answer 1

Answer:

Let r be the radius v be the volume and S be the surface area of a sphere at any instant t

we know volume of sphere

v=

3

4

πr

3

dt

dv

=

3

4

π3r

2

.

dt

dr

given

dt

dv

=25 and r=5

so, 25=

3

4

π×3.5

2

.

dt

dr

or,

dt

dr

=

4×25π

25

=

4π

1

Now surface area

s=4πr

2

dt

ds

=8πr

dt

dr

or,

dt

ds

=8×π×5×

4π

1

[asr=5

dt

dr

=

4n

1

]

dt

ds

=10cm

2

/sec.

so the rate change of surface area 10cm

2

/sec