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Application Of Derivatives

Question

Wired Faculty App

Prove that the radius of the right circular cylinder of greatest curved surface which can be inscribed in a given cone is half of that of the cone.

Solution

Let x be the radius of the base of the cone of height PR = h. Let r be the radius of cylinder of height PQ = y.
Now

increment RQA space space and space space increment RPB

are similar

therefore space space space QA over PB space equals RQ over RP rightwards double arrow space space space space straight r over straight x space equals space fraction numerator straight h minus straight y over denominator straight h end fraction rightwards double arrow space space space space space straight r space equals space fraction numerator left parenthesis straight h minus straight y right parenthesis space straight x over denominator straight h end fraction

Let S be the curved surface of cylinder

therefore space space straight S space equals space 2 πry space equals 2 straight pi fraction numerator left parenthesis straight h minus straight y right parenthesis space straight x over denominator straight h end fraction straight y space equals fraction numerator 2 πx over denominator straight h end fraction left parenthesis hy minus straight y squared right parenthesis therefore space space dS over dv space equals space fraction numerator 2 πx over denominator straight h end fraction left parenthesis straight h minus 2 straight y right parenthesis

For S to be maximum or minimum.

dS over dy space equals space 0

therefore space space space space space space fraction numerator 2 πx over denominator straight h end fraction left parenthesis straight h minus 2 straight y right parenthesis space equals space 0 space space space space space rightwards double arrow space space space space space straight h minus 2 straight y space equals space 0 space space space space space space space rightwards double arrow space space space straight h space equals space 2 straight y Also comma space space fraction numerator straight d squared straight S over denominator dy squared end fraction space equals fraction numerator 2 πx over denominator straight h end fraction left parenthesis negative 2 right parenthesis space equals space fraction numerator negative 4 πx over denominator straight h end fraction When space straight h space equals space 2 straight y comma space space space space space fraction numerator straight d squared straight s over denominator dy squared end fraction space equals negative fraction numerator 4 πx over denominator straight h end fraction less than 0 therefore space space space space space straight S space is space maximum space when space straight h space equals space 2 straight y Now comma space space space space straight r space equals space fraction numerator straight x over denominator 2 straight y end fraction left parenthesis 2 straight y minus straight y right parenthesis space equals space straight x over 2

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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