Consider the helix r(t) =<cos (5t),sin(5t),-1t>. Compute, at t=pi/6 :A. The unit tangent vector T=<-5/squareroot 101, -5 squareroot3/squareroot 101, -2/squareroot 101> B. The unit normal vector N=<squareroot 3/2, -1/2, 0>C. The unit binormal vector B=<___,___,___>
Question
Consider the helix r(t) =<cos (5t),sin(5t),-1t>. Compute, at t=pi/6 :A. The unit tangent vector T=<-5/squareroot 101, -5 squareroot3/squareroot 101, -2/squareroot 101> B. The unit normal vector N=<squareroot 3/2, -1/2, 0>C. The unit binormal vector B=<,,___>
Solution
To find the unit binormal vector B, we use the formula B = T x N, where T is the unit tangent vector and N is the unit normal vector.
Given T = <-5/sqrt(101), -5sqrt(3)/sqrt(101), -2/sqrt(101)> and N = <sqrt(3)/2, -1/2, 0>, we can compute the cross product as follows:
B = T x N = det[ i j k -5/sqrt(101) -5sqrt(3)/sqrt(101) -2/sqrt(101) sqrt(3)/2 -1/2 0 ]
B = i(-5sqrt(3)/sqrt(101)0 - (-2/sqrt(101)(-1/2))) - j(-5/sqrt(101)*0 - (-2/sqrt(101)sqrt(3)/2)) + k(-5/sqrt(101)(-1/2) - (-5sqrt(3)/sqrt(101)*sqrt(3)/2))
Simplify to get:
B = i(1/sqrt(101)) - j(sqrt(3)/sqrt(101)) + k(-5/2sqrt(101) + 15/2sqrt(101))
So, the unit binormal vector B at t = pi/6 is B = <1/sqrt(101), -sqrt(3)/sqrt(101), 10/sqrt(101)>.
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