Results 1 to 4 of 4

Math Help - Tangent to a curve

  1. #1
    Junior Member
    Joined
    Oct 2005
    Posts
    30

    More help with differentiation please!

    1. The equations of a tangent and normal to the curve xy=4 at the point where x=2 are y=4-x and y=x.
    Show that the normal does intersect the curve again and find the coordinates of the point of intersection.

    2. Find the equations of the normals to the curve xy=3 which are parallel to the line 3x-y-2=0

    Thanks so much if you can help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    " 1. The equations of a tangent and normal to the curve xy=4 at the point where x=2 are y=4-x and y=x.
    Show that the normal does intersect the curve again and find the coordinates of the point of intersection."

    If two curves intersect, then their coordinates at their point of intersection are the same.
    xy = 4 ----(1)
    y = 2 ------(2)

    y of (1) = y of (2),
    4/x = x
    4 = x^2
    x = +,-sqrt(4) = +,-2
    That means the two curves intersect at x=2 and x = -2.
    At x = -2,
    y = x = -2
    So, (-2,-2) is the other point of intersection.

    -----------
    2. Find the equations of the normals to the curve xy=3 which are parallel to the line 3x-y-2=0

    Slope of tangent lines to xy=3:
    Differentiate both sides with respect to x,
    x(dy/dx) +y(1) = 0
    dy/dx = -y/x --------slope of tangent lines to xy=3 curve.

    3x -y -2 = 0
    -y = -3x +2
    y = 3x -2 ------y=mx+b form.
    So, slope is 3.
    Any line parallel to [3x -y -2 = 0] will have a slope of 3 also.
    So if the normals to the xy=3 curve are parallel to [3x -y -2 = 0], then their slope is 3 also.

    The normals and tangent lines to xy=3, at the same points, have slopes that are negative reciprocals.
    Meaning, if the slope of the normal line is 3, then the slope of the corresponding tangent line is -1/3.
    So,
    -y/x = -1/3
    Cross multiply,
    -y*3 = x*(-1)
    -3y = -x
    3y = x
    y = x/3 -----------the normal lines to xy=3 curve.

    Doing it in another way:

    If the slope of the tangent lines to xy=3 is -y/x, then the slope of the normal lines is -1 / (-y/x) = x/y

    x/y = 3
    x = 3y
    y = x/3 -------normal lines to xy=3.
    Last edited by ticbol; November 14th 2005 at 12:10 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2005
    Posts
    30
    Thanks so much. You're method for the last question seemed really logical however the answer is supposed to be y=3x+/-8. Do you know why this may be?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    " You're method for the last question seemed really logical however the answer is supposed to be y=3x+/-8. Do you know why this may be? "

    -------
    Umm, that made me review my answer yesterday. And, yes, the correct answer is y = 3x +8 , or y = 3x -8.

    In my answer yesterday I stopped at y = x/3.
    What I did not figure out then is that at the point of normalcy (?) y = x/3 only. The equation of the normal line is not yet the y = x/3.

    So, let me finish the answer yesterday.

    At the point of "normalcy", y = x/3, so, substitute that into xy=3,
    x(x/3) = 3
    x*x = 3*3
    x^2 = 9
    x = +,-sqrt(9)
    x = +,-3 -------the x's at the points of normalcy.

    The y's:

    When x = 3,
    xy = 3
    (3)y = 3
    y = 3/3 = 1
    So, (3,1) is one point of normalcy.

    When x = -3,
    xy = 3
    (-3)y = 3
    y = 3/(-3) = -1
    So, (-3,-1) is the other point of normalcy.

    For the equations of the two normal lines:

    At (3,1), and with a slope of 3, by point-slope form,
    y-1 = 3(x-3) ----------answer.
    Or,
    y -1 = 3x -9
    y = 3x -9 +1
    y = 3x -8 -----------in y=mx+b form, answer.

    At (-3, -1), and with a slope of 3, by point-slope form,
    y -(-1) = 3(x -(-3))
    y +1 = 3(x+3) ----------answer.
    Or,
    y +1 = 3x +9
    y = 3x +9 -1
    y = 3x +8 -----------in y=mx+b form, answer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tangent on a curve
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 30th 2010, 06:32 AM
  2. Replies: 6
    Last Post: April 7th 2010, 03:34 PM
  3. tangent to curve
    Posted in the Calculus Forum
    Replies: 7
    Last Post: August 6th 2009, 06:19 PM
  4. Tangent Curve to a Curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 25th 2009, 01:03 AM
  5. Tangent to the curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 20th 2008, 05:25 AM

Search Tags


/mathhelpforum @mathhelpforum