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Application Of Derivatives

Question
CBSEENMA12035004
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The radius of a spherical soap bubble is increasing at the rate of 0.4 cms–1 Find the rate of change of its (i) volume (ii) surface area, when the radius is 10 cm.

Solution

Let r be radius of spherical soap bubble.
therefore space space space space space dr over dt space equals space 0.4 space space cm space straight s 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 space space... left parenthesis 1 right parenthesis
left parenthesis straight i right parenthesis space space space space space space space space space space space space space straight V space equals space fraction numerator 4 straight pi over denominator 3 end fraction straight r cubed space space where space straight V space is space volume.
therefore 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 dr over dt space equals 4 πr squared space cross times space 0.4 space space space space space space space space space space space space space space space space space space space open square brackets because 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 equals space 1.6 space πr squared space When space straight r space equals space 10 comma space space space dV over dt space equals space 1.6 space straight pi space cross times space 100 space equals space 160 space straight pi space cubic space cm divided by sec
(ii) Again,          straight S space equals space 4 πr squared comma space space space space where space straight S space is space surface space area.
therefore space space space space space space space space space space space space space dS over dt space equals space 8 πr dr over dt space equals space 8 space πr space cross times 0.4 space equals space 3.2 space πr space space space space space space space space space space space space space when space straight r space equals 10 comma space space space space space dS over dt space equals space 3.2 space straight pi space cross times space 10 space equals space 32 space straight pi space sq. space cm divided by sec.
           

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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