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

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To calculate the binormal vector B(t) for the curve r(t) = <t, 4-t^2, 0>, T(t) is derived as <1/sqrt(1+4t^2), -2t/sqrt(1+4t^2), 0> after differentiating r(t) and normalizing. N(t) is found by differentiating T(t) and normalizing, resulting in <-2t/sqrt(1+4t^2), -1/sqrt(1+4t^2), 0>. The cross product of T(t) and N(t) should yield a non-zero binormal vector, contrary to the initial claim of <0,0,0>. The discussion highlights the importance of correctly applying the cross product and suggests a potential error in the calculations. Accurate computation of these vectors is crucial for understanding the geometry of the curve.
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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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