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April 14th, 2012, 07:27 PM   #1
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Unit outward normal vectors for spherical surface

Can anybody please solve this,
For the spherical surface x^2+y^2+z^2=1, what is the unit outward normal vector at the point (1/sqrt 2. 1/sqrt 2, 0)

bharathsf is offline  
April 15th, 2012, 06:47 AM   #2
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Re: Unit outward normal vectors for spherical surface

Analyzing the problem, you discover that you are asked to calculate the normal vector of a circle lying in the xy-plane. This circle has equation x^2+y^2=1, and hence has radius 1.

You know that the graph of a function f(x) is a graph y=f(x) in the plane. More important, it is also a level curve to a function g(x,y).

Now, imagine a function g(x,y) = f(x) - y. The function y=f(x) is then the level surface g(x,y)=0. For the next step you should recall that the gradient of a function is perpendicular to the level curves of a function (assuming they exist).

Now, y=f(x)=sqrt(1-x^2) (This is the function of the part of the circle above, and on, the x-axis)

Calculating the gradient:

grad(g(x,y))=(-x/sqrt(1-x^2))i - j

The gradient will in this case be a vector pointing inward, but all you have to do is multiply it by (-1) for it to point in the direction wanted. Also, to express it as a unit vector, you only have to divide it by its length.

Furthermore, this is one way to mathematically calculate the direction of the normal vector, but because of the circles symmetri it is possible to just state that the vector must make an angle pi/4 with the positive x-axis.
fredde_fisk is offline  
April 15th, 2012, 06:56 AM   #3
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Re: Unit outward normal vectors for spherical surface

Or you could just calculate the derivative of the cirlce and use the fact that two perpendicular lines' slopes satisfy (slope1)*(slope2)=-1. This method is much faster and not as messy, but the other method is applicable to more dimensions, which is why I chose it.
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