For the given position vectors r(t) compute the unit tangent vector T(t) for the given value of t A) let r(t)=<cos2t,sin2t> Then T(pi/4) <__,__>B) let r(t)=<t^2,t^3> Then T(1)=<_,_>c) let r(t)= e^2t i+ e^-t j+ t K Then T(2)= _i+_j+_k
Question
For the given position vectors r(t) compute the unit tangent vector T(t) for the given value of t A) let r(t)=<cos2t,sin2t> Then T(pi/4) <,>B) let r(t)=<t^2,t^3> Then T(1)=<,>c) let r(t)= e^2t i+ e^-t j+ t K Then T(2)= _i+_j+_k
Solution
A) The unit tangent vector T(t) is given by the derivative of the position vector r(t) divided by the magnitude of the derivative of r(t).
For r(t) = <cos2t, sin2t>, the derivative r'(t) = <-2sin2t, 2cos2t>.
The magnitude of r'(t) is sqrt((-2sin2t)^2 + (2cos2t)^2) = sqrt(4sin^2(2t) + 4cos^2(2t)) = sqrt(4) = 2.
So, T(t) = r'(t) / ||r'(t)|| = <-sin2t, cos2t>.
Substituting t = pi/4, we get T(pi/4) = <-sin(pi/2), cos(pi/2)> = <-1, 0>.
B) For r(t) = <t^2, t^3>, the derivative r'(t) = <2t, 3t^2>.
The magnitude of r'(t) is sqrt((2t)^2 + (3t^2)^2) = sqrt(4t^2 + 9t^4) = t*sqrt(4 + 9t^2).
So, T(t) = r'(t) / ||r'(t)|| = <2t/tsqrt(4 + 9t^2), 3t^2/tsqrt(4 + 9t^2)> = <2/sqrt(4 + 9t^2), 3t/sqrt(4 + 9t^2)>.
Substituting t = 1, we get T(1) = <2/sqrt(4 + 9), 3/sqrt(4 + 9)> = <2/sqrt(13), 3/sqrt(13)>.
C) For r(t) = e^2t i + e^-t j + t k, the derivative r'(t) = 2e^2t i - e^-t j + k.
The magnitude of r'(t) is sqrt((2e^2t)^2 + (-e^-t)^2 + 1^2) = sqrt(4e^4t + e^-2t + 1).
So, T(t) = r'(t) / ||r'(t)|| = <2e^2t/sqrt(4e^4t + e^-2t + 1), -e^-t/sqrt(4e^4t + e^-2t + 1), 1/sqrt(4e^4t + e^-2t + 1)>.
Substituting t = 2, we get T(2) = <2e^4/sqrt(4e^8 + e^-4 + 1), -e^-2/sqrt(4e^8 + e^-4 + 1), 1/sqrt(4e^8 + e^-4 + 1)>.
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