Results 1 to 7 of 7

Math Help - slope of the tangent line to the curve of intersection of the vertical plane &surface

  1. #1
    Junior Member
    Joined
    Apr 2014
    From
    USA
    Posts
    42

    slope of the tangent line to the curve of intersection of the vertical plane &surface

    Hello,

    vert plane: x - root(3)*y + 2*root(3) - 1 = 0
    surface: z = x^2 + y^2

    slope of the< tangent line to the curve of intersection of the vertical plane &surface at point (1,2,5)

    First, I took the gradients and plugged in (1,2,5)

    del F (1,2,5) = <1, - root 3, 0> && del G (1,2,5) = <2,4,-1>

    Then take cross product, gives < root 3, 1, 4+2*root(3)>. The cross product is tangent to the curve of intersection.

    Now in order to get slope, I think I have to do a directional derivative, but not sure.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,987
    Thanks
    1650

    Re: slope of the tangent line to the curve of intersection of the vertical plane &sur

    What, exactly, do you mean by the "slope" of a line in three dimensions?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2014
    From
    USA
    Posts
    42

    Re: slope of the tangent line to the curve of intersection of the vertical plane &sur

    I am not sure exactly. This is a practice problem my teacher posed. But I am pretty sure I have to do a directional derivative since it is defined as the "slope of the tangent line to the curve formed by the intersection of z = f (x; y) and the vertical plane through a point p parallel to a unit vector u". He gave us a hint, use Du F (1,2)

    edit:

    So the cross product gives us < root 3, 1, 4+2*root(3)> and this is the direction of the tangent line.

    So, if i make a unit vector from it and then dot it with del F, that should satisfy my above definition?
    Last edited by SNAKE; May 1st 2014 at 08:51 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2014
    From
    USA
    Posts
    42

    Re: slope of the tangent line to the curve of intersection of the vertical plane &sur

    Can nobody help the poor reptile?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    Re: slope of the tangent line to the curve of intersection of the vertical plane &sur

    It sounds like you have already found the direction vector for the line of intersection (I didn't check your calculations, though).

    HallsofIvy is right - the idea of slope in this problem makes no sense. But I'm guessing what it means is the z component of the direction vector divided by the component in the x-y plane. If so, that's easy to calculate: v_z=4+2\sqrt{3} and v_{xy}=\sqrt{v_x^2+v_y^2}=2, so \text{slope}=2+\sqrt{3}.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Apr 2014
    From
    USA
    Posts
    42

    Re: slope of the tangent line to the curve of intersection of the vertical plane &sur

    I understand the slope oddity and your suggestion makes sense to me but he told us we would have to use D_u f (1,2), I don't know for which vector though.

    unit vector u = cross product/ |cross product| = <sqrt(3), 1, 4+2*sqrt(3)> / 7.73.

    okay, but what vector am I dotting u with, what would f be?
    Last edited by SNAKE; May 1st 2014 at 07:52 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    Re: slope of the tangent line to the curve of intersection of the vertical plane &sur

    Since you have z as a function of x and y ( z=x^2+y^2), and you know the direction you're going in the x-y plane \bold{v}=<\sqrt{3},1>, you should be able to take the directional derivative and get the same answer.

    The gradient is <2x,2y> = <2,4>, so the directional derivative is <2,4>\cdot\frac{\bold{v}}{|\bold{v}|} = \frac{1}{2}<2,4>\cdot<\sqrt{3},1>=2+\sqrt{3}.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. slope of tangent line to a curve
    Posted in the Calculus Forum
    Replies: 7
    Last Post: March 11th 2011, 09:11 PM
  2. slope of tangent line to polar curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 26th 2010, 09:00 AM
  3. slope of tangent line on a curve?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 20th 2009, 10:39 PM
  4. Replies: 0
    Last Post: November 9th 2008, 04:13 PM
  5. Slope of the tangent line to the curve
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 31st 2008, 10:47 AM

Search Tags


/mathhelpforum @mathhelpforum