Results 1 to 7 of 7

Math Help - Tough vectors question...

  1. #1
    Member
    Joined
    Mar 2008
    From
    ???
    Posts
    171

    Tough vectors question...

    Let r be a postition vector of a viariable point in cartesian plane OXY such that
    r.(10j - 8i - r/|r| ) = 40 and p1 = max{(|r + 2i - 3j|)^2} , p2 = min{(|r + 2i - 3j|)^2}
    . A tanget line is drawn to the curve y = 8/(x^2) at the point A with abcissa 2. The drawn line cuts x-axis at a point B.
    Now answer the following questions based on above paragraph

    Q1 p2 is equal to ?

    Q2 p1+p2 = ?

    Q3 vector AB . vector OB = ?

    please explain exactly what is vector r ??
    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
    Let's say that \mathbf{r}=\langle r_{1},r_{2}\rangle. If you plug this into the condition

    \mathbf{r}\cdot\left(\langle -8,10\rangle-\dfrac{\mathbf{r}}{\|\mathbf{r}\|}\right)=40,

    you'll end up with the equation for an ellipse in the coordinates \langle r_{1},r_{2}\rangle.

    You can verify that the equation for the ellipse is the following:

    63r_{1}^{2}+99r_{2}^{2}-160r_{1}r_{2}+640r_{1}-800r_{2}+1600=0.

    So the answer to your question of "what is \mathbf{r}?" is this: \mathbf{r} is the coordinate for a point on the ellipse described by the above equation.

    Does that help?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2008
    From
    ???
    Posts
    171
    how do you know that it is an ellipse ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2008
    From
    ???
    Posts
    171
    I mean... how did you come to know that it is an ellipse at first sight..
    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
    Quote Originally Posted by ice_syncer View Post
    I mean... how did you come to know that it is an ellipse at first sight..
    I didn't. I recognized it as an ellipse after I worked out the equation in post # 2.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Mar 2008
    From
    ???
    Posts
    171
    Ah ok.. Now I get it... so to find p2 or p1 we jsut substitute r as <x,y> ... dot product it with itself, then d/dx the expression and equate d/dx = 0, find the maxima or minima depending on p1 or p2's demand .... the n find y from the ellipse equation... nice.. if you have an easier way please suggest. Thanks for making me understand what vector r is.
    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
    I would highly recommend drawing a picture. Drawing a picture enabled me to conclude that, after all, it's not an ellipse. If you look here, you will see that you need the discriminant to be negative, which is not true in this case. The graph looks something like an hyperbola, but I don't think it's an hyperbola. The graph to which I've linked is centered about the point (-2,3), which figures largely in your expressions for p1 and p2. I do think that p2 is well-defined. The point (-2,3) lies just above the upper branch of the relation in question, and finding the minimum distance from that point to the graph should be doable, at least theoretically. However, considering that the graph appears to go on to infinity in multiple directions, I don't think your p1 is well-defined.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tough question
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 2nd 2010, 11:36 AM
  2. tough question
    Posted in the Algebra Forum
    Replies: 15
    Last Post: September 4th 2009, 08:02 AM
  3. Tough problem... Vectors, points, and a cube
    Posted in the Pre-Calculus Forum
    Replies: 11
    Last Post: August 16th 2009, 01:43 AM
  4. Very tough polynomial question.
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: December 1st 2008, 07:57 AM
  5. Needs help on a tough question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 25th 2006, 10:44 PM

Search Tags


/mathhelpforum @mathhelpforum