How Do You Calculate the Binormal Vector B(t) for a Given Curve?

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The discussion focuses on calculating the binormal vector B(t) for the curve defined by r(t) = . The user correctly derived the tangent vector T(t) as <1/sqrt(1+4t²), -2t/sqrt(1+4t²), 0> and the normal vector N(t) as <-2t/sqrt(1+4t²), -1/sqrt(1+4t²), 0>. However, the user incorrectly concluded that the binormal vector B(t) is <0, 0, 0>, which is a mistake as the cross product of T(t) and N(t) should yield a non-zero vector.

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S-C-3-1-3
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Hi! :)

The question is for r (t) =<t, 4-t2, 0>, find N(t),T(t), and B(t).

First I took the derivative of r (t), and divided it by its length to calculate T(t), which is <1/sqrt(1+4t2),-2t/sqrt(1+4t2,0)>. Then I took the derivative of this using the product rule, and divided it by its length to calculate N(t), which is <-2t/sqrt(1+4t2),-1/sqrt(1+4t2),0)>.

Taking the cross product of these two gives the binormalvector, which is <0,0,0>.

Is this done correctly, or does anyone get a different answer?


Thanks in advance! I didn't do all of the work on here, but I hope that's okay, as I have already done the work to find the answer...?
 
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Show the steps for finding N(t). You have an error in this.

The cross-product is not zero. Did you take the scalar product?
 
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