The surface area of spherical balloon is increasing at the rate of 2 cm3/sec. At what
rate is the volume of the balloon is increasing when the radius of the balloon is 6cm?
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Answer:
Hence, the volume of the spherical balloon is increasing at the rate of 6 cm3/sec.
Step-by-step explanation:
Let r,V and S be respectively the radius, volume and surface area of the spherical balloon at time t.
∴V=
3
4
πr
3
and S=4πr
2
Rate of change of surface area with respect to t
⇒
dt
dS
=2
⇒
dt
d
(4πr
2
)=2
⇒8πr
dt
dr
=24
⇒
dt
dr
=
4πr
1
.... (i)
Rate of change of volume of the balloon with respect to t
=
dt
dV
=
dt
d
(
3
4
πr
3
)
=
3
4
π⋅3r
2
dt
dr
=4πr
2
⋅
dt
dr
=4πr
2
⋅
4πr
1
....[From (i)]
=r
=6
Hence, the volume of the spherical balloon is increasing at the rate of 6 cm
3
/sec.
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