Results 1 to 8 of 8

Thread: Walking Up a Hill

  1. #1
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591

    Walking Up a Hill

    I am good with the first part of this question. I just Cant seem to pull out the final result about the angle of the line joining the two people, This trig just got ugly for me.

    -Bobak
    Attached Thumbnails Attached Thumbnails Walking Up a Hill-picture-1.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Take coordinate axes as follows. The x- and y-axes are both in the inclined plane (that's the key thing), with the x-axis horizontal and the y-axis up the line of greatest slope. The z-axis will then be perpendicular to the plane, so it will not be vertical. In fact, a vertical line will be in the direction of the vector $\displaystyle (0,\sin\theta,\cos\theta)$.

    If Arthur is at the point A and Bertha is at B, then the line joining O to the mid-point of AB will make an angle $\displaystyle \gamma = {\textstyle\frac12}(\alpha+\beta)$ with the y-axis, and it's easy to see from this that the line AB will be in the direction $\displaystyle (\cos\gamma,\sin\gamma,0)$. Now all you have to do is to take the inner product of this vector with the one in the vertical direction.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    Opalg I don't really understand why this problem needs to be done with 3 dimensions or vectors.

    I have attached my working for the first part. I am fairly sure the last result should be be pulled out form a bit of trig using the diagram i drew. Or am i barking up the wrong tree ?
    Attached Thumbnails Attached Thumbnails Walking Up a Hill-img018.jpg  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2008
    Posts
    154
    that looks right.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    That's fine. Now notice that the line OM makes an angle $\displaystyle \textstyle\frac12(\alpha+\beta)$ with the axis Oy. Therefore AB makes the same angle with the axis Ox.

    Quote Originally Posted by bobak View Post
    Opalg I don't really understand why this problem needs to be done with 3 dimensions or vectors.
    An inclined plane is an intrinsically 3-dimensional concept, so I don't think you can avoid using 3 dimensions in the solution. The simplest way to find the angle that AB makes with the vertical is the method I suggested in my previous comment.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    I understand how the inner product of the two vectors give the results I am just having problems defining the vectors properly in the co-ordinate system. This is mainly cause I am struggling with defining the x y and z axis properly on a diagram. This may be asking for a bit much but could you possibly show me a little sketch of this.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by bobak View Post
    I understand how the inner product of the two vectors give the results I am just having problems defining the vectors properly in the co-ordinate system. This is mainly cause I am struggling with defining the x y and z axis properly on a diagram. This may be asking for a bit much but could you possibly show me a little sketch of this.
    Not sure if this helps. It's supposed to show why the vertical line is in the direction of the vector $\displaystyle (0,\sin\theta,\cos\theta)$.
    Attached Thumbnails Attached Thumbnails Walking Up a Hill-plane.png  
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    That is very clear now. Thank you very much Opalg. I really do appreciate you help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. walking speed
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Sep 2nd 2010, 01:10 PM
  2. Walking to Work
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Mar 2nd 2010, 10:31 PM
  3. Walking robots
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: Jun 22nd 2009, 10:32 PM
  4. Average speed of a man walking up a hill.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Dec 8th 2008, 11:03 AM
  5. walking the block
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Nov 2nd 2008, 01:39 PM

Search Tags


/mathhelpforum @mathhelpforum