Results 1 to 4 of 4

Math Help - Rate of change question

  1. #1
    Junior Member
    Joined
    Aug 2010
    Posts
    37

    Rate of change question

    Hi Guys,

    Am currently stuck with this rate of change question.

    question is as per attached file.

    Please help thanks !
    Attached Thumbnails Attached Thumbnails Rate of change question-rate-change-question.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Aug 2010
    Posts
    37
    up for help guys !

    Thanks !
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    779
    Hello, liukawa!

    \text{In the diagram, }P(p,\,p^2)\text{ is a variable point on the curve }y\,=\,x^2
    \text{where }p > 0.\:\text{ The tangent at }P\text{ intersects the }x\text{-axis and }y\text{-axis}
    \text{at }Q\text{ and }R\text{, respectively.}

    \text{(a) Express the coordinates of }Q\text{ and }R\text{ in terms of }p.
    Code:
              |
              |       *
              |         / 
              |        /
              |      */
              |      /
         *    |     o P
          *   |   */
        - - - * - o - - - - - - -
              |  /Q
              | /
              |/
              o R 
             /|

    The slope of the tangent is: . y' \,=\,2x

    At \,P, the slope is: . m \,=\,2p

    The equation of the tangent is:
    . . y = p^2 \:=\:2p(x-p) \quad\Rightarrow\quad y \:=\:2px - p^2


    When y = 0,\;2px-p^2 \:=\:0 \quad\Rightarrow\quad x \,=\,\frac{1}{2}p .[1]

    . . The coordinates of \,Q are: . \left(\frac{1}{2}p,\:0\right)


    When x = 0,\;y \:=\:2p(0)-p^2 \quad\Rightarrow\quad y \,=\,-p^2 .[2]

    . . The coordinates of \,R are: . (0,\:-p^2)




    \text{(b) When }p = 2\text{, the rate of increase of }p\text{ is 3 units/sec.}

    \text{Find the corresponding rate of change of:}

    . . \text{(i) the }x\text{-coordinate of }Q.

    We are told: when p = 2,\;\dfrac{dp}{dt} \,=\,3


    From [1]: . x \:=\:\frac{1}{2}p

    Differentiate with respect to time: . \dfrac{dx}{dt} \:=\:\dfrac{1}{2}\!\cdot\!\dfrac{dp}{dt}

    . . Therefore: . \dfrac{dx}{dt} \:=\:\frac{1}{2}(3) \:=\:\frac{3}{2}\text{ units/sec}



    \text{(ii) the area of triangle }OQR.

    The base of the triangle is \,OQ\!:\;x \,=\,\frac{1}{2}p

    The height of the triangle is |OR|\!:\;|y| \,=\,p^2

    The area is: . A \;=\;\frac{1}{2}\left(\frac{1}{2}p\right)\left(p^2  \right) \;=\;\frac{1}{4}p^3

    Differentiate with respect to time: . \dfrac{dA}{dt} \;=\;\dfrac{3}{4}p^2\left(\dfrac{dp}{dt}\right)

    Therefore: . \dfrac{dA}{dt} \;=\;\dfrac{3}{4}(2^2)(3) \;=\;9\text{ units}^2\!\text{/sec}


    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2010
    Posts
    37
    Quote Originally Posted by Soroban View Post
    Hello, liukawa!


    The slope of the tangent is: . y' \,=\,2x

    At \,P, the slope is: . m \,=\,2p

    The equation of the tangent is:
    . . y = p^2 \:=\:2p(x-p) \quad\Rightarrow\quad y \:=\:2px - p^2


    When y = 0,\;2px-p^2 \:=\:0 \quad\Rightarrow\quad x \,=\,\frac{1}{2}p .[1]

    . . The coordinates of \,Q are: . \left(\frac{1}{2}p,\:0\right)


    When x = 0,\;y \:=\:2p(0)-p^2 \quad\Rightarrow\quad y \,=\,-p^2 .[2]

    . . The coordinates of \,R are: . (0,\:-p^2)





    We are told: when p = 2,\;\dfrac{dp}{dt} \,=\,3


    From [1]: . x \:=\:\frac{1}{2}p

    Differentiate with respect to time: . \dfrac{dx}{dt} \:=\:\dfrac{1}{2}\!\cdot\!\dfrac{dp}{dt}

    . . Therefore: . \dfrac{dx}{dt} \:=\:\frac{1}{2}(3) \:=\:\frac{3}{2}\text{ units/sec}




    The base of the triangle is \,OQ\!:\;x \,=\,\frac{1}{2}p

    The height of the triangle is |OR|\!:\;|y| \,=\,p^2

    The area is: . A \;=\;\frac{1}{2}\left(\frac{1}{2}p\right)\left(p^2  \right) \;=\;\frac{1}{4}p^3

    Differentiate with respect to time: . \dfrac{dA}{dt} \;=\;\dfrac{3}{4}p^2\left(\dfrac{dp}{dt}\right)

    Therefore: . \dfrac{dA}{dt} \;=\;\dfrac{3}{4}(2^2)(3) \;=\;9\text{ units}^2\!\text{/sec}



    Hi Soroban !

    thanks for your reply !

    however, there is one step which i dont quite understand which is the equation of tangent.

    how do you find the tangent from there?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rate of change question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 22nd 2011, 11:39 PM
  2. Replies: 3
    Last Post: April 12th 2011, 10:51 AM
  3. Rate of change question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 10th 2010, 01:55 AM
  4. Rate of change question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 15th 2008, 03:24 PM
  5. Another rate of change question
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 25th 2007, 04:06 PM

Search Tags


/mathhelpforum @mathhelpforum