Question

surface area of a sphere increases at the rate of 0.6 cm sq per sec. The rate at which is volume is increasing when it's radius is 10 cm is​

Answer 1

Answer:

A≈1256.64cm²

r Radius

10

cm

Solution

A=4πr2=4·π·102≈1256.63706cm 2

Answer 2

Therefore the rate of increase in the volume of the sphere is 3 cm³/sec.

Given:

Rate of increase of the surface area of a sphere = 0.6 cm²/sec

The radius of the sphere = 10 cm

To Find:

The rate at which volume is increasing.

Solution:

The given question can be solved as shown below.

Given that,

Rate of increase of the surface area of a sphere = d( SA ) / dt = 0.6 cm²/sec

The radius of the sphere = r = 10 cm

Surface area of a sphere = SA = 4πr²

Differentiating concerning time on both sides.

⇒ d ( SA) / dt = d ( 4πr² ) / dt

⇒ d ( SA) / dt = 4π × dr²/dt

⇒ d ( SA) / dt = 4π × 2r dr/dt = 0.6     [ ∵ d( SA ) / dt = 0.6 ]

⇒ dr/dt = 0.6 / ( 4π × 2 × 10 ) = 0.03/4π

Now Volume of the sphere = V = ( 4/3 ) × πr³

Now differentiating both sides to get a rate of change of volume,

⇒ dV/dt = ( 4/3 ) × d( πr³ ) / dt

⇒ dV/dt = ( 4/3 ) × π × 3r² dr/dt

⇒ dV/dt = ( 4/3 ) × π × 3 × 10²× ( 0.03/4π )     [ ∵ r = 10 and dr/dt = 0.03/4π ]

⇒ dV/dt = 100 × 0.03 = 3

Therefore the rate of increase in the volume of the sphere is 3 cm³/sec.

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