Results 1 to 3 of 3

Math Help - Relative Rates

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    19

    Relative Rates

    A spotlight is on the ground 20m away from a wall and a 2m tall person is walking towards the wall at a rate of 1.5m/s . How fast is the height of the shadow changing when the person is 8m from the wall? Is the shadow increasing or decreasing in height at this time?

    i know i need to find dh/dt, which equals dx/dt * dh/x.

    Is it possible to find dh/dx? (ie the relation between height of shadow and distance from wall)

    Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Quote Originally Posted by anonymous_maths View Post
    Is it possible to find dh/dx? (ie the relation between height of shadow and distance from wall)

    Thanks in advance.
    Yes. Use the similar triangles in your sketch (?!) to say...

    \frac{h}{20} = \frac{2}{20 - x}

    Just in case a picture helps with the chain rule...



    ... where



    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to t, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    Spoiler:

    Spoiler:





    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; May 7th 2010 at 04:38 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    625
    Hello, anonymous_maths!

    A spotlight is on the ground 20m away from a wall.
    A 2m tall person is walking towards the wall at a rate of 1.5m/s.

    How fast is the height of the shadow changing when the person is 8m from the wall?
    Is the shadow increasing or decreasing in height at this time?
    Code:
          |
        S o
        : |  *
        : |     *
        : |        *  P
        h |           o
        : |           |  *
        : |           |2    *
        : |           |        *
        T o-----------o-----------o L
          :     x     Q    20-x   :
          : - - - -  20 - - - - - :

    The light is at L\!:\;\;TL = 20

    The person is: . PQ = 2
    His distance from the wall is: . x \:=\:TQ \quad\Rightarrow\quad QL \:=\:20-x
    This distance is decreasing at: . \frac{dx}{dt} \:=\:\text{-}\frac{3}{2} m/sec.

    The height of his shadow is: . h = ST.


    From the similar right triangles, we have:

    . . \frac{h}{20} \:=\:\frac{2}{20-x} \quad\Rightarrow\quad h \:=\:40(20-x)^{-1}


    Differentiate with respect to time: . \frac{dh}{dt} \;=\;\text{-}40(2-x)^{-2}(\text{-}1)\frac{dx}{dt} \;=\;\frac{40}{(20-x)^2}\frac{dx}{dt}


    When x = 8\!:\;\;\frac{dh}{dt} \;=\;\frac{40}{12^2}\left(\text{-}\frac{3}{2}\right) \;=\;-\frac{5}{12}


    The height of the shadow is decreasing at \tfrac{5}{12} meters per second.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Relative Rates
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 2nd 2011, 09:29 PM
  2. Derivatives/Relative Rates
    Posted in the Calculus Forum
    Replies: 7
    Last Post: May 11th 2010, 05:19 PM
  3. Relative Rates of Growth
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 18th 2010, 03:50 PM
  4. relative rates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 2nd 2010, 04:49 PM
  5. Relative rates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 23rd 2008, 12:26 AM

Search Tags


/mathhelpforum @mathhelpforum