Results 1 to 8 of 8
Like Tree1Thanks
  • 1 Post By topsquark

Math Help - related rates word problem

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    11

    related rates word problem

    A street light is at the top of a 17 foot tall pole. A woman 6 ft tall walks away from the pole with a speed of 8 ft/s along a straight path. How fast is the tip of her shadow moving when she is 45 feet from the base of the pole?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193
    I found a very similar problem on the internet that I know will help.
    It's here and it's Example 6.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by keemariee View Post
    A street light is at the top of a 17 foot tall pole. A woman 6 ft tall walks away from the pole with a speed of 8 ft/s along a straight path. How fast is the tip of her shadow moving when she is 45 feet from the base of the pole?
    Always draw a diagram!!

    related rates word problem-capture.jpg

    Now we can use similar triangles to get

    \frac{x+y}{17}=\frac{y}{6} Now lets multiply th equation by 17 to get

    x+y=\frac{17}{6}y \iff x=\frac{11}{6}y \iff y=\frac{6}{11}x

    Now we know that the tip of the shawdow will be moving at the rate

    \frac{dx}{dt}+\frac{dy}{dt}

    Take the derivative of y=\frac{6}{11}x with respect to time gives

    \frac{dy}{dt}=\frac{6}{11}\frac{dx}{dt}

    You should be able to finish from here.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    627
    Hello, keemariee!

    With these "streetlight" problems, we must be very careful.
    Sometimes they ask how fast the tip of the shadow is moving.
    Sometimes they ask how fast how fast the shadow is growing.


    A street light is at the top of a 17 foot tall pole.
    A woman 6 ft tall walks away from the pole with a speed of 8 ft/s along a straight path.
    How fast is the tip of her shadow moving when she is 45 feet from the base of the pole?
    Code:
        A *
          |   *
          |       *
          |           *   C
       17 |               *
          |               |   *
          |              6|       *
          |               |           *
        B *---------------*---------------* E
          :       y       D      x-y      :
          : - - - - - - x - - - - - - - - :

    The streetlight is: AB = 17
    The woman is:  CD = 6

    Let y = BD.\quad\frac{dy}{dt} \,= \,8\text{ ft/sec}
    Let x \:=\:BE\qquad\text{ then: }DE \:=\: x-y

    We have: . \Delta CDE \sim \Delta ABE\quad\Rightarrow\quad \frac{x-y}{6} \:=\:\frac{x}{17}\quad\Rightarrow\quad x \:=\:\frac{17}{11}y

    Differentiate with respect to time: . \frac{dx}{dt} \:=\:\frac{17}{11}\!\cdot\!\frac{dy}{dt}

    Since \frac{dy}{dt} \,=\,8\!:\;\;\frac{dx}{dt}\:=\:\frac{17}{11}(8) \:=\:\frac{136}{11} \:=\:12\frac{4}{11}\text{ ft/sec}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member dokrbb's Avatar
    Joined
    Feb 2013
    From
    ...from out there ;)
    Posts
    236
    Thanks
    21

    Re: related rates word problem

    Quote Originally Posted by Soroban View Post
    Hello, keemariee!

    With these "streetlight" problems, we must be very careful.
    Sometimes they ask how fast the tip of the shadow is moving.
    Sometimes they ask how fast how fast the shadow is growing.


    Code:
        A *
          |   *
          |       *
          |           *   C
       17 |               *
          |               |   *
          |              6|       *
          |               |           *
        B *---------------*---------------* E
          :       y       D      x-y      :
          : - - - - - - x - - - - - - - - :

    The streetlight is: AB = 17
    The woman is:  CD = 6

    Let y = BD.\quad\frac{dy}{dt} \,= \,8\text{ ft/sec}
    Let x \:=\:BE\qquad\text{ then: }DE \:=\: x-y

    We have: . \Delta CDE \sim \Delta ABE\quad\Rightarrow\quad \frac{x-y}{6} \:=\:\frac{x}{17}\quad\Rightarrow\quad x \:=\:\frac{17}{11}y

    Differentiate with respect to time: . \frac{dx}{dt} \:=\:\frac{17}{11}\!\cdot\!\frac{dy}{dt}

    Since \frac{dy}{dt} \,=\,8\!:\;\;\frac{dx}{dt}\:=\:\frac{17}{11}(8) \:=\:\frac{136}{11} \:=\:12\frac{4}{11}\text{ ft/sec}

    Hi all, I know it's an old problem (since 2008...) but can someone show me how did you get this

    \Delta CDE \sim \Delta ABE\quad\Rightarrow\quad \frac{x-y}{6} \:=\:\frac{x}{17}\quad\Rightarrow\quad x \:=\:\frac{17}{11}y

    I know it should be something really simple but I can't see it, please
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,888
    Thanks
    326
    Awards
    1

    Re: related rates word problem

    Quote Originally Posted by dokrbb View Post
    Hi all, I know it's an old problem (since 2008...) but can someone show me how did you get this

    \Delta CDE \sim \Delta ABE\quad\Rightarrow\quad \frac{x-y}{6} \:=\:\frac{x}{17}\quad\Rightarrow\quad x \:=\:\frac{17}{11}y

    I know it should be something really simple but I can't see it, please
    You don't know how to prove they are similar triangles, or how to get the equation?

    I'll assume the equation. The two similar triangles are CDE and ABE. So if we take the ratio of any two sides of CDE you can write a "similar" expression for ABE. For example, look at the sides x - y and 6 in CDE. These are the horizontal "leg" and vertical leg of CDE. So we know that the ratio we want from ABE will involve the horizontal (x) and vertical legs.

    Thus
    \frac{\text{horizontal leg}}{\text{vertical leg}} = \frac{x - y}{6} = \frac{x}{17}

    -Dan
    Last edited by topsquark; April 10th 2013 at 10:39 AM.
    Thanks from dokrbb
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member dokrbb's Avatar
    Joined
    Feb 2013
    From
    ...from out there ;)
    Posts
    236
    Thanks
    21

    Re: related rates word problem

    Quote Originally Posted by topsquark View Post
    You don't know how to prove they are similar triangles, or how to get the equation?

    I'll assume the equation. The two similar triangles are CDE and ABE. So if we take the ratio of any two sides of CDE you can write a "similar" expression for ABE. For example, look at the sides x - y and 6 in CDE These are the horizontal "leg" and vertical leg of CDE. So we know that the ratio we want from ABE will involve the horizontal (x) and vertical legs.

    Thus
    \frac{\text{horizontal leg}}{\text{vertical leg}} = \frac{x - y}{6} = \frac{x}{17}

    -Dan
    No, I actually got it, the ratio we use thanks to the fact that we have two similar triangles,

    my question is at the 12th grade algebra level , how we actually get x = \frac{17}{11}y from \frac{x - y}{6} = \frac{x}{17}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member dokrbb's Avatar
    Joined
    Feb 2013
    From
    ...from out there ;)
    Posts
    236
    Thanks
    21

    Re: related rates word problem

    Quote Originally Posted by dokrbb View Post
    No, I actually got it, the ratio we use thanks to the fact that we have two similar triangles,

    my question is at the 12th grade algebra level , how we actually get x = \frac{17}{11}y from \frac{x - y}{6} = \frac{x}{17}
    sorry for the stupid question - I figured it out
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Related Rates Word Problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 27th 2011, 06:08 PM
  2. related rates word problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 28th 2010, 07:44 PM
  3. Related Rates Word Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 4th 2009, 12:04 PM
  4. Word Problem-Related Rates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 9th 2007, 02:39 PM
  5. related rates word problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 5th 2007, 01:29 PM

Search Tags


/mathhelpforum @mathhelpforum