Application of Derivatives

The volume of a spherical balloon is increasing at the rate of 25 cm 3 /sec. Find the rate of change of its surface area at the instant when its radius is 5 cm. - Wired Faculty

Question

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The volume of a spherical balloon is increasing at the rate of 25 cm³/sec. Find the rate of change of its surface area at the instant when its radius is 5 cm.

Solution

Let V be volume of sphere of radius r.

therefore space space space space space space space space space space space space straight V space equals space 4 over 3 πr cubed

From given condition,

dV over dt space equals space 25 space cm cubed divided by straight s comma space space space space space space space space space space space space space space space space space space therefore space space space space straight d over dt open parentheses 4 over 3 πr cubed close parentheses space equals space 25 space space space space space rightwards double arrow space space space space 4 over 3 straight pi. space 3 straight r. space dr over dt space equals space 25

therefore space space space space space space space dr over dt space equals space fraction numerator 25 over denominator 4 πr squared end fraction 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 space space space space space space space... left parenthesis 1 right parenthesis

Let S be surface area of balloon.

therefore space space space space space space space space space straight S space equals space 4 πr squared

Rate of increase of surface area =

dS over dt space equals space straight d over dt left parenthesis 4 πr squared right parenthesis

equals space 8 space πr space dr over dt space equals space 8 space πr space cross times space fraction numerator 25 over denominator 4 πr squared end fraction 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 space left parenthesis 1 right parenthesis close square brackets

equals space 50 over straight r

When r =5, rate of increase of surface area =

50 over 5 space equals space 10 space cm squared divided by straight s.

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Application Of Derivatives

The volume of a spherical balloon is increasing at the rate of 25 cm³/sec. Find the rate of change of its surface area at the instant when its radius is 5 cm.

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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For Daily Free Study Material Join wiredfaculty

Application Of Derivatives

The volume of a spherical balloon is increasing at the rate of 25 cm3/sec. Find the rate of change of its surface area at the instant when its radius is 5 cm.

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.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

The volume of a spherical balloon is increasing at the rate of 25 cm³/sec. Find the rate of change of its surface area at the instant when its radius is 5 cm.