Results 1 to 5 of 5

Math Help - Related Rate Shadow & Spotlight Question

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    7

    Related Rate Shadow & Spotlight Question

    I have a question that I am stuck on for the Related Rates problems. I did it and got a negative answer when it's suppose to be positive I did it again a little differently and still got a negative answer!! Here is the problem:

    Joe is 1.8 meters tall is walking directly towards a spotlight located on the ground 20 meters away from the theatre. If Joe walks at a constant rate of 1/5 m/s find the rate at which the length of his shadow on the wall is changing when he is 6 meters from the lamp.


    So the 2nd time I went through it I set it up as:

    Given: dx/dt = 1/5 m/s
    Find: dy/dt when x = 6

    y/20 = 1.8/20-x
    y = 36/(20-x)
    dy/dt = [-36/(20-x)^2](dx/dt)
    dy/dt = [-36/(20-6)^2](1/5)
    dy/dt = -9/245

    Worked from there. First time I got an answer of -1/5, second time an answer of -9/245. But I know if he is moving towards the spotlight the answer will have to be a positive rate. I think there is something fundamentally wrong with my procedure, if anyone could show me their through process on this I would greatly appreciate it!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,806
    Thanks
    116
    Quote Originally Posted by Sajiko View Post
    I have a question that I am stuck on for the Related Rates problems. I did it and got a negative answer when it's suppose to be positive I did it again a little differently and still got a negative answer!! Here is the problem:

    Joe is 1.8 meters tall is walking directly towards a spotlight located on the ground 20 meters away from the theatre. If Joe walks at a constant rate of 1/5 m/s find the rate at which the length of his shadow on the wall is changing when he is 6 meters from the lamp.


    So the 2nd time I went through it I set it up as:

    Given: dx/dt = 1/5 m/s
    Find: dy/dt when x = 6

    y/20 = 1.8/20-x
    y = 36/(20-x)
    dy/dt = [-36/(20-x)^2](-1)(dx/dt) You have to use the chain rule and you forgot about the derivation of the inner function
    dy/dt = [-36/(20-6)^2](1/5)
    dy/dt = -9/245

    Worked from there. First time I got an answer of -1/5, second time an answer of -9/245. But I know if he is moving towards the spotlight the answer will have to be a positive rate. I think there is something fundamentally wrong with my procedure, if anyone could show me their through process on this I would greatly appreciate it!
    ...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    7
    Quote Originally Posted by earboth View Post
    ...
    AHhhhhhhhhhh!!!! I always make such stupid mistakes! Thank you so much!

    But otherwise then does 9/245 look like the right answer?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,806
    Thanks
    116
    Quote Originally Posted by Sajiko View Post
    AHhhhhhhhhhh!!!! I always make such stupid mistakes! Thank you so much!

    But otherwise then does 9/245 look like the right answer?
    Correct

    Don't forget that a speed of \dfrac15\ \dfrac ms describes a more snail like movement.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2009
    Posts
    7
    Thank you and yeah such a slow rate lol, but that is what the teacher gave us.

    If you wouldn't mind could you take a look at the other topic I have in this forum? It's about 2 trough questions I have no approach to tackle them; I know how to solve a normal trough with isosceles triangles at each end when they give you a certain volume but not with time.

    If you could that would be great!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: April 26th 2010, 12:13 AM
  2. Related Rates, shadow and streetlight problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 5th 2010, 11:00 AM
  3. Stuck on Related Rate question
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 4th 2010, 03:31 PM
  4. Related Rates Question /Shadow of a Man
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 18th 2010, 11:18 AM
  5. The Spotlight: a related rates problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 27th 2007, 02:07 PM

Search Tags


/mathhelpforum @mathhelpforum