Understanding the Unit Normal Vector in Multivariable Differential Calculus

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The discussion focuses on proving the relationship between the cross products of the derivatives of a space curve and the unit normal vector in multivariable differential calculus. The curve is represented by r, mapping from R to R3, where r'(t) denotes the tangent vector and r''(t) indicates the change in the tangent vector. The cross product of r'(t) and r''(t) lies in the osculating plane, and taking the cross product with r'(t) yields a vector normal to the curve. The expression provided relates to the normalization of this vector to define the unit normal vector N(t). Understanding the geometric behavior of cross products is crucial for comprehending this relationship.
issisoccer10
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This question comes from my multivariable differential calculus course. I cannot figure how to prove that the following is true...

How does...

__r'(t) x r''(t) x r'(t)_ = N(t) ?
||r'(t) x r''(t) x r'(t)||

any help would be appreciated...thanks
 
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What are r and N?
 
issisoccer10 said:
This question comes from my multivariable differential calculus course. I cannot figure how to prove that the following is true...

How does...

__r'(t) x r''(t) x r'(t)_ = N(t) ?
||r'(t) x r''(t) x r'(t)||

any help would be appreciated...thanks

Just think about how the cross product behaves geometrically. I'm assuming r is a map from R into R3 defining a space curve and that N is a normal vector to the curve, defined by being normal to the tangent vector at each point. r'(t) would then define a tangent vector at the point r(t). r''(t) describes the change in the tangent vector at that point, so for an arbitrarily small window, it should lie in the osculating plane of the curve. What direction does the cross product of these two vectors go in with respect to these two vectors? What then happens when you take the cross product of this vector with the tangent vector?
 

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