What's the unit tangent vector T(t) for curve given by the vector-valued function r(t)= t^2 i + t j + t^2 k

Question

these are the possible answers from the back of the book....

a. (1 + 8t^2)^-1/2(2t i + j + 2t k)
b. (1 + 2t^2)^-1/2( i + t j + t k)
c. (1 + 5t^2)^-1/2(t i + j + 2t k)
d. (1 + 32t^2)^-1/2(4t i + j + 4t k)
e. (1 + 68t^2)^-1/2(8t i + j + 2t k)
f. (4 + 2t^2)^-1/2(t i + 2 j + t k)

 or is it none of these?

Helper · Accepted Answer

tangent vector is T= r'/|r'| r= < t^2, t, t^2> r'= < 2t, 1, 2t> |r'|= sqrt (4t^2+1+4t^2)= sqrt (1+8t^2) T= <2t, 1, 2t>/sqrt(1+8t^2) T = <2t,1,2t>*(1+8t^2)^( -1/2) answer is A

firat c · 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.

Erika · Answer

Take the by-made from the defining formulation, which provides: d(t) = 2t i + j + 2t ok. Take the dimensions of this vector, that's l(t) = sqrt(4t^2+a million+4t^2) = sqrt( a million+8t^2). Any non-0 vector divided by using its length is the unit vector interior the direction of the unique vector, so which you get d(t)/l(t) = (a million+8t^2)^(-a million/2)(2t i+ j + 2t ok) through fact the vector you're searching for. it is the alternative a.

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What's the unit tangent vector T(t) for curve given by the vector-valued function r(t)= t^2 i + t j + t^2 k

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