Results 1 to 2 of 2

Math Help - Application of Calculus problem

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    3

    Application of Calculus problem

    A street light is at the top of a 14 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 5 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 30 ft from the base of the pole?


    This is a problem where they give you what I believe is a dv/dt of 5 ft/s, but I don't know what I have to do to find the speed of the tip of her shadow. I tried drawing triangles, but I don't know exactly what I need to do, and I can't think of any equation that would help me find this, y = mx+b since it's a straight line???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,663
    Thanks
    603
    Hello, kycon90!

    Did you make a sketch?


    A streetlight is at the top of a 14 ft tall pole.
    A woman 6 ft tall walks away from the pole with a speed of 5 ft/sec.
    How fast is the tip of her shadow moving when she is 30 ft from the base of the pole?
    Code:
        A o
          |   * - C
          |       o
          |       |   *
        14|       |       *
          |      6|           *
          |       |               *
          |       |                   *
        B o-------o-----------------------o E
          : - x - D - - -  s-x  - - - - - :
          : - - - - - - s - - - - - - - - :

    The lightpole is: AB = 14\text{ ft}

    The woman is: CD = 6\text{ ft}
    Her distance from the pole is: BD = x\text{ ft},\;\;\frac{dx}{dt} = 5 \text{ ft/sec}

    The tip of her shadow (E) is s ft from the pole: . s \,=\,BE
    . . Hence: . DE \,=\,s-x

    Since \Delta ABE \sim \Delta CDE, we have: . \frac{14}{s} \:=\:\frac{6}{s-x} \quad\Rightarrow\quad s \:=\:\frac{7}{4}x

    Differentiate with respect to time: . \frac{ds}{dt} \:=\:\frac{7}{4}\,\frac{dx}{dt}


    Therefore: . \frac{ds}{dt} \:=\:\frac{7}{4}(5) \:=\:8.75\text{ ft/sec}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: September 8th 2011, 03:25 AM
  2. Application of calculus
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 23rd 2011, 01:11 AM
  3. Application of calculus
    Posted in the Calculus Forum
    Replies: 15
    Last Post: January 30th 2011, 10:47 AM
  4. Application of calculus
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 12th 2009, 03:56 AM
  5. Application of calculus?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 29th 2008, 07:10 AM

Search Tags


/mathhelpforum @mathhelpforum