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Application Of Derivatives

Question
CBSEENMA12035050
Wired Faculty App

Water is running out of a conical funnel at the rate of 5 cm3/sec. If the radius of the base of the funnel is 10 cm and the altitude is 20 cm, find the rate at which the water level is dropping when it is 5 cm from the top.

Solution
Let r be the radius and h the height of the surface of water at time t. Let V be the volume of water in the funnel.
therefore space space space space straight V space equals space 1 third πr squared straight h                                   ...(1)
Also, by similar triangles, we have
          straight r over straight h space equals space 10 over 20 space space space space space space space space rightwards double arrow space space space space space straight r space equals space 1 half straight h
therefore space space space space from space left parenthesis 1 right parenthesis comma space space space straight V space equals space 1 third straight pi open parentheses straight h over 2 close parentheses squared space straight h space equals space πh squared over 12 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 2 right parenthesis
Since water is running out of the funnel at the rate of 5 space cm cubed divided by sec.
therefore space space space space space dV over dt space equals space minus 5 space space space space space space space space space space space space space space space space space space space open parentheses dV over dt space is space minus ve space because space straight V space decreases space as space straight t space increases close parentheses From space left parenthesis 2 right parenthesis comma space space dt over dt space equals straight pi over 12 straight d over dt left parenthesis straight h cubed right parenthesis space equals space fraction numerator 3 πh cubed over denominator 12 end fraction dh over dt therefore space space space minus 5 space equals space fraction numerator straight pi space straight h squared over denominator 4 end fraction dh over dt therefore space space space space dh over dt equals negative fraction numerator 20 over denominator straight pi space straight h squared end fraction therefore space space space rate space of space dropping space of space water space level space left parenthesis straight i. straight e. space of space straight h right parenthesis space straight w. straight r. straight t space time space straight t space equals space dh over dt space equals space fraction numerator negative 20 over denominator πh squared end fraction. When space water space level space is space 5 space cm space from space the space top comma space space straight h space equals space 20 minus 5 space equals space 15. therefore space space space space rate space of space dropping space of space water space level space straight w. straight r. straight t. space apostrophe straight t apostrophe space when space straight h space equals space 5 space is space space space space space space space space space space space space space space space space space space space space space fraction numerator negative 20 over denominator straight pi left parenthesis 15 right parenthesis squared end fraction space equals space fraction numerator negative 20 over denominator straight pi cross times 225 end fraction space equals space fraction numerator negative 4 over denominator 45 space straight pi end fraction cm divided by sec.

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