Application of Derivatives

  • Question 1
    CBSEENMA12034955

    Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm.

    Solution
    Let A be area of circle of radius r
    therefore space space space space space space space space space space space straight A space equals space πr squared
    Rate of change of area with respect to straight r space equals space dA over dr
                                                           equals space straight d over dr left parenthesis πr squared right parenthesis space equals space 2 πr
    when r = 5,    rate of change of area  = 2 straight pi space cross times space 5 space equals space 10 straight pi space cm squared divided by cm
    Question 2
    CBSEENMA12034957

    Find the rate of change of the area of a circle with respect to its radius r when
    (a) r = 3 cm   (b) r = 4 cm

    Solution
    Let A be area of circle of radius r
    therefore space space space space space space space space space space space space space space space straight A space equals space πr squared
    Rate of change of area with respect to straight r space equals space dA over dr
                                                             equals space straight d over dr left parenthesis πr squared right parenthesis space equals space 2 πr

    (a) When r = 3, rate of change of area = 2straight pi × 3 = 6  cm2/cm.
    (b) When r = 4, rate of change of area = 2 straight pi × 4 = 8 straight pi cm2/cm.

    Question 3
    CBSEENMA12034960

    Find the rate of change of the volume of a ball with respect to its radius r. How fast is the volume changing with respect to the radius when the radius is 2 m?

    Solution
    Let V be volume of ball of radius r
    therefore space space space space space space space space space straight V equals space 4 over 3 πr cubed
    Rate of change of volume with respect to straight r space equals space dV over dr
                                    equals space straight d over dr open parentheses 4 over 3 πr cubed close parentheses space equals space 4 over 3 straight pi straight d over dr left parenthesis straight r cubed right parenthesis space equals fraction numerator 4 straight pi over denominator 3 end fraction cross times 3 straight r squared space equals space 4 πr squared
    When r = 2 m, rate of change of volume = 4 straight pi (2)2 = 16 straight pi m3/m.
    Question 4
    CBSEENMA12034962

    How fast is the volume of a ball changing with respect to its radius when the radius is 3 m?

    Solution
    Let V be volume of ball of radius r.
     therefore space space space space space space space straight V space equals space 4 over 3 πr cubed
    Rate of change of volume with respect to straight r space equals space dV over dr
                                              equals space straight d over dr open parentheses fraction numerator 4 straight pi over denominator 3 end fraction straight r cubed close parentheses space equals space fraction numerator 4 straight pi over denominator 3 end fraction cross times 3 straight r squared space equals space 4 πr squared
    When r = 3 m, rate of change of volume  = 4 space straight pi space left parenthesis 3 right parenthesis squared space equals space 36 space straight pi space space straight m cubed divided by straight m

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