Page 1 of 3 123 LastLast
Results 1 to 15 of 39

Math Help - Related Rate-Ladder against the wall

  1. #1
    Member
    Joined
    Mar 2010
    Posts
    150

    Related Rate-Ladder against the wall

    Here is the problem I'm working on:

    A ladder 20ft long leans against a house. If the foot of the ladder is moving away from the house at the rate of 2ft/s, find how fast the slope of the ladder is decreasing when the foot of the ladder is 12ft from the house.

    Here is what I have so far:













    Given:
    • \frac{dx}{dt}=2
    • x=12
    • z=20
    Find:
    • \frac{d\theta}{dt} when x = 12
    My Work:
    • By the theorem of Pythagoras:
      • y=16
    • sin\theta = \frac{y}{z} = \frac{16}{20} = \frac{4}{5}
    • cos\theta=\frac{x}{20}
    • x=20cos\theta
    • \frac{dx}{dt} = 20(-sin\theta)*\frac{d\theta}{dt}
    • 2 = 20*\frac{-4}{5}*\frac{d\theta}{dt}
    • \frac{d\theta}{dt} = \frac{-1}{8} rad/s

    • The book, however, came up with \frac{25}{72} per second
    Any tips on where I went wrong?

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Well, the most important thing I can see is that I don't think you're solving for the right variable. It's asking for how fast the slope is changing at that point. You attempted to calculate (I haven't checked the accuracy) how fast the angle the ladder makes with the floor is changing. Those are not the same.

    What is the slope in general? How could you find its derivative in time?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2010
    Posts
    150
    Well, the general definition of slope is \frac{\Delta y}{\Delta x}

    @ 6 seconds \frac{\Delta y}{\Delta x} = \frac{9}{12} = \frac{3}{4} ??

    I'm not quite sure how I would go about finding its derivative...another hint?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2010
    Posts
    150
    What about tan \theta = \frac{y}{x}

    Taking the derivative, I get:

    sec^2 \theta*\frac{d\theta}{dt} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

    ...am I on the right track?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    What you're calculating (incorrectly there) is the actual slope at the time when you're given data. However, that's not what the problem is asking for, either. You're asked to find how fast the slope is changing. Let's say you define the slope as

    m(t)=\dfrac{y(t)-0}{0-x(t)}=-\dfrac{y(t)}{x(t)}.

    You're asked to find

    \dfrac{dm}{dt}.

    I would agree that this is a bit confusing: you're trying to take the derivative of a slope! No, it's not really a second derivative, because here you're taking the time derivative, whereas the slope is a sort of spatial derivative in this case.

    How can you find this time derivative?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Mar 2010
    Posts
    150
    I'm still stuck...can't seem to wrap my head around this one
    Follow Math Help Forum on Facebook and Google+

  7. #7
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Do you understand how I got m(t)?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Mar 2010
    Posts
    150
    Not really...I was thinking \displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Exactly. But now let's suppose the point (x_{1},y_{1}) is the point where the ladder touches the floor, and let's suppose the point (x_{2},y_{2}) is the point where the ladder touches the wall. What are x_{2} and y_{1}?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Mar 2010
    Posts
    150
    ok..i see that they are both equal to zero

    I can't just plug in numbers at this point bc I would get \frac{-4}{3}

    So I am a bit confused as to what y(t) and -x(t) are and how I would take the derivative of them?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Right. So, since there's only one x coordinate to keep track of, and only one y coordinate to keep track of, we'll drop the subscripts and get the m I have in post # 5. You follow that now? (Keep in mind also that both x and y change in time).
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Mar 2010
    Posts
    150
    I'm still stuck. I don't really understand what y(t) and -x(t) are?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    y(t) is the height of the point at which the ladder touches the wall, and x(t) is the x-coordinate of the point at which the ladder touches the floor. You have to have the negative sign in the slope there, because the y and x do not correspond to the same point. Make sense?

    Or think of it this way: y is positive up (along the wall), and x is positive to the right (along the floor). The slope of that ladder has to be negative, because it's going down as x increases. But y and x are both positive. Hence, m = - y / x. Make sense?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Mar 2010
    Posts
    150
    Yeah, that part makes sense, I just dont understand how to differentiate -y/x
    Follow Math Help Forum on Facebook and Google+

  15. #15
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Well, if you have a quotient there, what rule do you suppose you're supposed to use?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 3 123 LastLast

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 20th 2010, 04:49 AM
  2. Sliding Ladder (Related Rates)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 5th 2008, 11:09 AM
  3. Possible Related Rates Problem w/ Ladder
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 19th 2008, 06:41 AM
  4. Ladder and box (and wall)
    Posted in the Math Challenge Problems Forum
    Replies: 4
    Last Post: August 1st 2008, 11:43 AM
  5. application ladder on wall
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 11th 2007, 11:40 AM

Search Tags


/mathhelpforum @mathhelpforum