Find the unit tangent vector T(t) for the curve given by the vector-valued function r(t)= 2t^2 i + t j +2t^2 k?

Question

Find the unit tangent vector T(t) for the curve given by the vector-valued function r(t)= 2t^2 i + t j +2t^2 k?
one of these could possibly be a correct response...

(1 + 8t^2)^-1/2(2t i + j + 2t k)
(1 + 2t^2)^-1/2( i + t j + t k)
(1 + 5t^2)^-1/2(t i + j + 2t k)
(1 + 32t^2)^-1/2(4t i + j + 4t k)
(1 + 68t^2)^-1/2(8t i + j + 2t k)
(4 + 2t^2)^-1/2(t i + 2 j + t k)
or is it none of these?

Anonymous · Accepted Answer

42.

? · Answer

Take the derivative of the defining formula, which gives you: d(t) = 2t i + j + 2t k. Take the length of this vector, it is l(t) = sqrt(4t^2+1+4t^2) = sqrt( 1+8t^2). Any non-zero vector divided by its length is the unit vector in the direction of the original vector, so you get d(t)/l(t) = (1+8t^2)^(-1/2)(2t i+ j + 2t k) as the vector you are looking for. It is the choice a.

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Find the unit tangent vector T(t) for the curve given by the vector-valued function r(t)= 2t^2 i + t j +2t^2 k?

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