Application of Derivatives

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Question

The radius of a spherical soap bubble is increasing at the rate of 0.2 cms^–1. Find the rate of change of its (i) volume (ii) surface area, when the radius is 4 cm.

Solution

Let r be radius of spherical soap bubble.

therefore space space space space space space space space space space space space space dr over dt space equals space 0.2 space space cms to the power of negative 1 end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

(i)

straight V space equals space 4 over 3 πr cubed space space space where space straight V space is space volume.

therefore space space space space space space space space space dV over dt space equals space fraction numerator 4 straight pi over denominator 3 end fraction cross times 3 straight r squared space dr over dt space equals space 4 space straight pi space straight r squared space cross times space 0.2 space space space space space space space space space space space space space space space space space open square brackets because space space space of space left parenthesis 1 right parenthesis close square brackets space space space space space space space space space space space space space space space space space space space space space space space equals space 0.8 space straight pi space straight r squared

When

straight r space equals space 4 comma space space dV over dt space equals space 0.8 space straight pi space cross times space 16 space equals space 12.8 space straight pi space cubic space cm divided by sec

(ii)

straight S space equals space 4 πr squared space space space where space straight S space is space surface space area

therefore space space space space space dS over dt space equals space 8 πr dr over dt space equals space 8 πr cross times 0.2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets space space space space space space space space space space space space space space space space space space space equals space 1.6 space straight pi space straight r when space straight r space equals 4 comma space space space space dS over dt space equals space 1.6 space straight pi space cross times space 4 space equals space 6.4 space straight pi space sq. space cm divided by sec