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Tangent to a curve and normal to the surface

  1. Nov 13, 2008 #1

    i have a very basic question in calculus. i know that the tangent to the curve y=f(x) is denoted by dy/dx which makes sense to me geometrically. But normal to a surface k(x1,x2,x3) is Grad k. how is this? it doesnt make sense to me geometrically. can anyone throw more light upon this please
  2. jcsd
  3. Nov 14, 2008 #2


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    Unfortunately you know wrong! Or at least you are not being precise enough. The slope of the tangent line to the curve y= f(x) is equal to dy/dx (evaluated at the point of tangency), not the tangent itself.

    There is no "tangent line" to a surface- there is a tangent plane. And while you can write a line in terms of a vector in its direction, a plane is determined by its normal vector.

    If a problem says "here is a curve in 3 dimensions, find it tangent line at this point" you can use the derivative to find that tangent line. If the problem says "here is a plane in 3 dimensions, find its tangent plane at this point" you can again use the derivative precisely because that derivative (gradient) is normal to the surface.

    A line, in 3 dimensions, is itself one-dimensional and so has a 3-1= 2 dimensional normal plane. A plane, conversely is two dimensional and has a 3-2= 1 dimensional normal line. Your geometric intuition is not making the jump from 2 to 3 dimensions.
  4. Nov 14, 2008 #3
    Connecting the two ideas is not straight forward, because you are dealing with different objects. OP seems to perceive this as a problem of dimensions but that will not be fruitful. Let's take a circle of radius [tex]r=1[/tex]. One way to express this is [tex]y_1(x)=\pm \sqrt{1-x^2}[/tex]

    another way starts with a bucket:

    The height of the function (respectively the z-coordinate) is the square of the distance of a point given by x,y from the origin. So [tex]z(x,y)=r^2=x^2+y^2[/tex]. This is like your [tex]k(x,y,z)[/tex] but we just use [tex]k(x,y)[/tex].

    If you set z(x,y) equal to a constant, then you cut the bucket at a certain height, and get a circle. Actually if we set it equal to 1 we get the circle given by [tex]y_1(x)[/tex]:
    [tex]x^2 + y_2^2=1[/tex]
    [tex]y_2=\pm \sqrt{1-x^2} = y_1[/tex]

    The derivative [tex]y_1'(x)[/tex] gives you the growth rate of [tex]y_1[/tex] when proceeding along the x axis, which also happens to be the slope of the tangent line.

    The gradient of [tex]z(x,y)[/tex] points outwards in the x,y plane in the direction where the bucket grows the fastest.
    So it has two components. The reason why it is normal to the tangent line on the circle which we get when we cut the bucket at a certain height is as follows:
    All the points of the cut are at the same height. So the growth along this line is 0. This means that this direction is perpendicular to the direction of strongest growth or strongest decline. In a way it is the midpoint between the two. (The exact reasons lie in the linearity of tangents but I wont elaborate on this.)

    To summarize: The gradient will give the normal to a constant surface of a function that lives one dimension higher. And the derivative gives the growth rate of a function with respect to a certain variable. It doesn't point anywhere, and to make it a tangent you have to discuss the equation of the tangent line.

    Sorry this might be a bit dense, if you have not seen theses expressions for circles before.
    Last edited: Nov 14, 2008
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