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What is a tangent line

  1. Jul 23, 2014 #1

    The tangent to a curve in a plane at a particular point has the same Gradient as the curve has at that point.

    More generally, the (n-1)-dimensional tangent hyperplane to an (n-1)-dimensional surface in n-dimensional space at a particular point has the same Gradient as the surface has at that point.

    So if [itex]A\,=\,(a_1,a_2,\cdots a_n)[/itex] is a point on a surface defined by the equation [itex]F(x_1,x_2,\cdots x_n) = 0[/itex], then the tangent hyperplane to the curve through [itex]A[/itex] is [itex]\frac{\partial F}{\partial x_1}\arrowvert_A(x_1 - a_1)\,+\,\frac{\partial F}{\partial x_2}\arrowvert_A(x_2 - a_2)\,+\,\cdots\,\frac{\partial F}{\partial x_2}\arrowvert_A(x_n - a_n)\,=\,0[/itex]

    If a curve in n dimensions is defined using a parameter t as [itex]A(t)\,=\,(a_1(t),a_2(t),\cdots a_n(t))[/itex] , then its tangent is:
    [itex](x_1 - a_1) / \frac{da_1}{dt}\,=\,(x_2 - a_2) / \frac{da_2}{dt}\,=\,\cdots\,=\,(x_n - a_n) / \frac{da_n}{dt}[/itex]


    For example, if the point [itex]A\,=\,(x_0,y_0)[/itex] lies on the circle:

    [tex]F(x,y)\,=\,(x-p)^2+(y-q)^2\,-\,r^2\,=\,0[/tex] (1)

    then [tex]\frac{\partial F}{\partial x}\arrowvert_A\,=\,2(x-p)\arrowvert_A\,=\,2(x_0-p)[/tex]
    and [tex]\frac{\partial F}{\partial y}\arrowvert_A\,=\,2(y-q)\arrowvert_A\,=\,2(y_0-q)[/tex]

    and so the equation of the tangent at [itex]A[/itex] is:

    [tex](x_0-p)(x-x_0)\,+\,(y_0-q)(y-y_0)=0[/tex] (2)

    Alternatively, the same circle can be defined by [itex]A(\theta)\,=\,(p+r\cos\theta, q+r\sin\theta)[/itex]
    and so the equation of the tangent at [itex]A(\theta)[/itex] is:


    Extended explanation

    We will transform the equation (2) into more convenient type for better way of memorizing and using the formula. Because of [itex]M(x_1,y_1) /in K[/itex]:

    [tex](x_1-p)+(y_1-q)^2=r^2[/tex] (3)

    If we sum the equations (2) and (3), we get:



    [tex](x_1-p)(x-p)+(y_1-q)(y-q)=r^2[/tex] (4)

    The equation (4) is equation of tangent of the circle in the point [itex]M(x_1,y_1) \in K[/itex].

    If the K have center (0,0), i.e [itex]K: x^2+y^2=r^2[/itex], then p=q=0, so the equation of the tangent is:


    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
  2. jcsd
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