# Thread: normal vector to surface

1. Is there a theorem that says whenever we have an implicity defined continuously differentiable function generating a surface in , the vector: is normal to the surface? where is the basis.  2.

3. Yes, for suppose v is a vector which is tangent to the surface at a point p. Then by definition, v = r'(0) where r(t) is a curve on the surface which passes through p at t=0.

Since r(t) lies on the surface, f(r(t)) = 0 for all t. Taking the derivative at zero we get that <grad f, v> = 0, where <,> denotes the dot product. But this just means that

grad f is perpendicular to the surface.  4. Originally Posted by salsaonline
Yes, for suppose v is a vector which is tangent to the surface at a point p. Then by definition, v = r'(0) where r(t) is a curve on the surface which passes through p at t=0.

Since r(t) lies on the surface, f(r(t)) = 0 for all t. Taking the derivative at zero we get that <grad f, v> = 0, where <,> denotes the dot product. But this just means that

grad f is perpendicular to the surface.
The only useful note that I could possibly add to this is that for those who are not used to vector calculus, the operation of taking the gradient is acutally the operation of taking the derivative of a vector valued function and it follows the chain rule with the dot product being the applicable multiplication operation. That is precisely how you conclude that <grad f, v> = 0.  5. Originally Posted by salsaonline
Yes, for suppose v is a vector which is tangent to the surface at a point p. Then by definition, v = r'(0) where r(t) is a curve on the surface which passes through p at t=0.

Since r(t) lies on the surface, f(r(t)) = 0 for all t. Taking the derivative at zero we get that <grad f, v> = 0, where <,> denotes the dot product. But this just means that

grad f is perpendicular to the surface.
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