Results 1 to 2 of 2

Math Help - Related rate of change problem. Help please

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    2

    Related rate of change problem. Help please

    Chris, who is 6 feet tall, is walking away from a street light pole 30 deet high at a rate of 2 feet per second.

    a. How fast is his shadow increasing in length when chris is 24ft from the pole? 30 feet? -- I already found that to be 1/2 ft per sec.

    b. How fast is the tip of his shadow moving?

    c. To follow the tip of his shadow, at what angular rate must Chris be lifting his eyes when his shadow is 6ft long?



    Please help with b and c!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,713
    Thanks
    632
    Hello, cedwards08!

    Chris, who is 6 feet tall, is walking away from a street light pole 30 feet high
    at a rate of 2 ft/sec. . {\color{blue}\frac{dx}{dt} = 2 \text{ ft/sec}}

    a. How fast is his shadow increasing in length when Chris is 24 ft from the pole? 30 feet?
    Code:
          *
          |   *
          |       *
       30 |       |   *
          |      6|       *
          |       |           * 
          *-------*---------------*
          : - x - : - - - s - - - :

    From the similar right triangles: . \frac{s}{6} \:=\:\frac{x+s}{30} \quad\Rightarrow\quad s \:=\:\frac{1}{4}x

    Differentiate with respect to time: . \frac{ds}{dt} \;=\;\frac{1}{4}\left(\frac{dx}{dt}\right)

    Therefore: . \frac{ds}{dt} \:=\:\frac{1}{4}(2) \:=\:\frac{1}{2}\text{ ft/sec} always.




    b. How fast is the tip of his shadow moving?
    Code:
          *
          |   *
          |       *
       30 |       |   *
          |      6|       *
          |   x   |    y-x    *
          *-------*---------------*
          : - - - - - y - - - - - :

    We have: . \frac{y}{30} \:=\:\frac{y-x}{6} \quad\Rightarrow\quad y \:=\:\frac{5}{4}x

    Differentiate with respect to time: . \frac{dy}{dt} \;=\;\frac{5}{4}\left(\frac{dx}{dt}\right)

    Therefore: . \frac{dy}{dt} \;=\;\frac{5}{4}(2) \;=\;\frac{5}{2}\text{ ft/sec} always.




    c. To follow the tip of his shadow, at what angular rate must Chris
    be lifting his eyes when his shadow is 6ft long?
    Code:
          *
          |   *
          |       * - - - - - 
       30 |       |   *   θ
          |      6|       *
          |       |       θ   *
          *-------*---------------*
          : - x - : - - - s - - - :

    We have: . \tan\theta \:=\:\frac{6}{s} \quad\Rightarrow\quad \tan\theta \:=\:6s^{-1}

    Differentiate with respect to time: . \sec^2\!\theta\left(\frac{d\theta}{dt}\right) \:=\:-6s^{-2}\left(\frac{ds}{dt}\right)

    . . and we have: . \frac{d\theta}{dt} \;=\;-\frac{6}{s^2}\cos^2\theta\left(\frac{ds}{dt}\right  ) .[1]


    We are given: . s = 6

    From (a), we have: . \frac{ds}{dt} = \frac{1}{2}

    When s = 6, the right triangle is isosceles.

    . . Hence: . \theta = \frac{\pi}{4}\;\;\text{ and }\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}


    Substitute into [1]: . \frac{d\theta}{dt} \;=\;-\frac{6}{6^2}\left(\frac{1}{\sqrt{2}}\right)^2\lef  t(\frac{1}{2}\right) \;=\;-\frac{1}{24}

    Therefore, Chris should raise his eyes at the rate of \frac{1}{24} radians/sec.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 12th 2011, 09:51 AM
  2. differential equation from related rate of change.
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: February 21st 2011, 01:37 AM
  3. Rate of Change Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 19th 2010, 08:35 AM
  4. Rate of change problem.
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 19th 2010, 11:31 AM
  5. Related Rate of Change of an Angle
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: March 21st 2009, 12:59 AM

Search Tags


/mathhelpforum @mathhelpforum