>>Not sure if the above is right

It looks like you have calculated the unit vector

Solved

Posted on 2011-09-19

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.

The answer I got was:

<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.

2 Comments

Comment Utility

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

>>Not sure if the above is right

It looks like you have calculated the unit vector

Comment Utility

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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