The volume of a sphere is increasing at the rate of 3 cubic cent

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410.

The volume of a sphere is increasing at the rate of 3 cubic centimetres per second. Find the rate of increase of its surface area, when the radius is 2 cm.


straight V space equals 4 over 3 πr cubed
rightwards double arrow space dV over dt space equals space 4 over 3 straight pi.3 straight r squared dr over dt
rightwards double arrow space dV over dt space equals space 4 πr squared dr over dt
rightwards double arrow dr over dt space equals space fraction numerator 1 over denominator 4 πr squared end fraction dV over dt
rightwards double arrow dr over dt space equals space fraction numerator 3 over denominator 4 straight pi left parenthesis 2 right parenthesis squared end fraction
open square brackets straight r equals space 2 space cm space and space dV over dt space equals space 3 space cm squared divided by sec close square brackets

rightwards double arrow dr over dt space space equals fraction numerator 3 over denominator 16 space straight pi end fraction cm divided by sec

Now, let S be the surface area of the sphere at any time t. then,
S = 4πr2
rightwards double arrow space dS over dt space equals space 8 πr dr over dt
rightwards double arrow space dS over dt space equals space 8 straight pi space left parenthesis 2 right parenthesis space straight x fraction numerator 3 over denominator 16 straight pi end fraction
open square brackets straight r space equals 2 space cm space and space dr over dt space equals space fraction numerator 3 over denominator 16 straight pi end fraction space cm divided by sec close square brackets
rightwards double arrow space dS over dt space equals space 3 space cm squared divided by sec



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