Results 1 to 3 of 3

Math Help - Shadow Problem

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    1

    Shadow Problem

    Didn't know if this was the right category to put this into but anyway.

    I got the following problem. There is a lantern that has a height of 11 meters
    a man with a height of 2.20 meters is standing at the lantern's base. The man starts moving away from the lantern with a velocity of v=2 m/s. I need to find with what velocity his shadow grows. Please this is urgent. Thx from now.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by pokopikos View Post
    Didn't know if this was the right category to put this into but anyway.

    I got the following problem. There is a lantern that has a height of 11 meters
    a man with a height of 2.20 meters is standing at the lantern's base. The man starts moving away from the lantern with a velocity of v=2 m/s. I need to find with what velocity his shadow grows. Please this is urgent. Thx from now.
    Search the forum for "shadow length". You'll get 57 examples completely done for your problem. Pick the one which is most convenient for you.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2008
    Posts
    191
    Quote Originally Posted by pokopikos View Post
    Didn't know if this was the right category to put this into but anyway.

    I got the following problem. There is a lantern that has a height of 11 meters
    a man with a height of 2.20 meters is standing at the lantern's base. The man starts moving away from the lantern with a velocity of v=2 m/s. I need to find with what velocity his shadow grows. Please this is urgent. Thx from now.
    Firstly you need to find the general equation relating the variables. You can draw a diagram to represent the general situation. At this time man's height is 2.20 meters and the lantern that has a height of 11 m. What varies in the general case? Well, man is moving away from the lantern so the distance from him to the base of latern is changing. We'll label that distance "x". Of course we are interested in the shadow, so let's label the length of shadow "y".

    Now if you see the diagram you made, notice that there are two right triangles in the picture, one with base "y" and height 2.20, the other with base "x+y" and and height 11. These two triangles are similar triangles and their angles are the same so that their ratios of their corresponding edges should be equal:

    \frac{2.20}{y} = \frac{11}{x+y}

    2.20(x+y) = 11y

     2.20x = 8.8y

     x=4y

    The rate of change of the length of the shadow is the derivative of y. So we want to find \frac{dy}{dt}. Since he is moving away from the lantern at 2 m/s, we know that \frac{dx}{dt}=2 (if he was moving towards the lantern it'd be -2)

    Now you must differentiate the general equation implicitly

    \frac{dx}{dt} = 4 \frac{dy}{dt}

    Do you see how it's done?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. moving shadow
    Posted in the Calculus Forum
    Replies: 0
    Last Post: August 18th 2011, 01:39 PM
  2. Related Rates, shadow and streetlight problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 5th 2010, 11:00 AM
  3. Sandbag Shadow Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 24th 2009, 01:54 PM
  4. A moving shadow
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 2nd 2009, 01:20 AM
  5. Replies: 0
    Last Post: February 19th 2009, 06:10 PM

Search Tags


/mathhelpforum @mathhelpforum