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A vector parallel to the tangent to the curve x=3t^(4/3) y=2t^3 -1 z= 2/(t^2) at the point (3,-3,2) on the curve is:

the answer is <-2,3,2>

how do you get this??
10 years ago

Answers

Steve
the tangent vector is ∇v = (dx/dt,dy/dt,dz/dt)
∇v = <4∛t,6t,-4/t^3>
v(-1) = <3,-3,2>, so
∇v = <-4,6,4>

Note that <-2,3,2> is parallel to ∇v(-1)
10 years ago

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