the surface area of a sphere increases at 0.6cm 2/sec the rate at which volume increase when its radius is 10
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Answer:
/s. When the radius of the bubble is 6 cm, then the rate by which the volume of the bubble increasing is
Given
dt
dA
=2cm
2
/sec,r=6cm
Surface area of sphere A=4πr
2
dt
dA
=8πr
dt
dr
⇒2=8×π×6×
dt
dr
dt
dr
=
24π
1
cm/sec
Now V=
3
4
πr
3
⇒
dt
dV
=
3
4
π3r
2
dt
dr
⇒
dt
dV
=
24π
4.π.(6)
2
⇒
dt
dV
=6cm
3
/sec.
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