How come the normal vector of the normal plane is the derivative of the curve r(t)?

doesn't the normal plane defines all the tangents or directional derivatives at the point t of a 3 dimensional curve?

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1 Answer

I'm not sure what you mean about directional derivatives; a curve is a function of just one variable, so we can't really take directional derivatives, and there is only one tangent, which is the derivative.

Think about it: a normal plane has to be perpendicular to the curve. What vector could we use as the normal vector for the plane? It should be tangent to the curve, because if the vector is tangent to the curve and the plane is normal to the vector, then the plane will be normal to the curve. Well, the derivative works as a tangent vector.

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err curve r(t) you're right it's not 3d, but how come r'(t) is the normal vector of the plane? –  Math Nov 28 '12 at 3:02
@Math: Well, that's what my second paragraph says. $r'(t)$ is tangent to the curve. If we want our plane to be normal to the curve, it should be normal to a vector that is tangent to the curve. Try to picture it in your head or maybe make a drawing. That usually helps. –  Javier Badia Nov 28 '12 at 3:08
what does normal mean? –  Math Nov 28 '12 at 3:20
it's kinda confusing because it's not "normal" as you say, yet they say it is normal. –  Math Nov 28 '12 at 3:21