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Normal to graph of a function

  1. Jun 26, 2012 #1
    Suppose you have a single variable differentiable function r: R -> R restricted to an interval, like [0,1] for simplicity, if you want. Consider the surface of revolution obtained by rotating the graph of r around the x axis. How do I find the NORMAL VECTOR to the surface at each of its points (x,y,z)? I guess I could find the normal vector to the graph of r in the xy plane and then rotating it with the same rotation I use to generate the surface of revolution. Somehow though this process seems to become unfeasible at some point so I must be missing something or doing it the wrong way altogether.
     
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  3. Jun 26, 2012 #2

    HallsofIvy

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    To be more specific, lets assume you have y= f(x) rotated around the x-axis. That is, of course, a two dimensional surface and can be written in terms of two parameters, x and the angle, [itex]\theta[/itex], of rotation. At each x, the point (x,y) rotates around a circle of radius y= f(x) and so we have x= x, [itex]y= f(x)cos(\theta)[/itex], and [itex]z= f(x)sin(\theta)[/itex].

    We can write that as a vector function: [itex]\vec{r}(x,\theta)= x\vec{i}+ f(x)cos(\theta)\vec{j}+ f(x)sin(\theta)\vec{k}[/itex].

    The derivatives of that vector function, with respect to the two variables:
    [tex]\vec{r}_x= \vec{i}+ f'(x)cos(\theta)\vec{j}+ f'(x)sin(\theta)\vec{k}[/tex]
    [tex]\vec{r}_\theta= -f(x)sin(\theta)\vec{j}+ f(x)cos(\theta)\vec{k}[/tex]
    are in the tangent plane.

    Their cross product
    [tex]\begin{bmatrix}\vec{i} & \vec{j} & \vec{k}\\ 0 & -f sin(\theta) & f cos(\theta) \\ 1 & f' cos(\theta) & f' sin(\theta) \end{bmatrix}= -ff'\vec{i}+ f cos(\theta)\vec{j}+ f sin(\theta)\vec{k}[/tex]
    is the outward normal at each point.
     
  4. Jun 27, 2012 #3
    Hi HallsofIvy, this vector is something similar to what I got (with inverted signs though, I probably did the cross product wrong). I don't see why it should have norm 1 in general, though and i need it to have norm 1 (Sorry, I just found out I should have written UNIT normal vector in the original post). But then again I guess I could just divide by sqrt[(rr')^2 + r^2] and I'd get a vector with norm 1. So it's fine I hope.
     
    Last edited: Jun 27, 2012
  5. Jun 27, 2012 #4

    HallsofIvy

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    There are, of course, an infinite number of vectors perpendicular to a surface at a point, of differing lengths and directions. I chose the order of cross product to get the "outward pointing" normal (away from the x-axis which lies in the interior of the surface). You didn't say anything about a "unit vector" in your original post. If you want a unit vector, divide by the length of this vector.
     
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