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Vector Calculus: Finding the vector equation of the line tangent to r(t) at point P

Posted on 2011-09-19
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Last Modified: 2012-05-12
Find a vector equation of the line tangent to the graph of r(t) at the point P on the curve:

r(t) = (2t-1)i + sqrt(3t+4)j ; P(-1,2)

The answer I got was:
T(t) =
r'(t) / || r'(t) || =

<2, 3/2(3t+4)^(-1/2)>
/
sqrt(4 + ( 9 / (12t + 16) )

Not sure if the above is right. First time my professor said it was wrong, then claimed it was right, but when asked a third time to see if it really was, never got a clear answer so I'm confused.
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Question by:Zenoture
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BigRat earned 500 total points
ID: 36567783
http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/arc/arc.html#tangent

>>Not sure if the above is right

It looks like you have calculated the unit vector

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Author Closing Comment

by:Zenoture
ID: 36571628
Yes, I calculated the unit tangent vector instead of the vector tangent to the line, i.e. r'(t), doh! Over thinking things is bad haha.
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