September 19th, 2018, 08:01 PM  #1 
Senior Member Joined: Apr 2017 From: New York Posts: 119 Thanks: 6  Tangent Line
Let me explain my steps for the attached question and please someone confirm me. Compute a unit vector that lies tangent to the curve at the given point. Hint:If you are given a point, you need to first find the t value that produces that point. (a)r(t) =(3 cos(t),âˆ’sin(2t, t), when t=pi (b) r(t)=(sin(2âˆ’t), e^(âˆ’t2+4), t^3+ 1) at (0,1,9) (a) step 1: find r'(t) step2: plug in t=pi at the r'(t) function so you find the point of tangency let's say it is point (a,b,c) step 3: point(a,b,c)/magnitude of point (a,b,c) will give me the unit vector that lies tangent to the curve (b) step1: first match the x,y,z components of the given point (1,0,9) and x,y,z components of the given line r(t) so you can calculate the t value step 2: find r'(t) step3: insert t value you found on step 2 into r'(t) function so you will be finding the point of tangency step 4: point(a,b,c)/ magnitude (a,b,c) will give you the unit vector 
September 20th, 2018, 09:47 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,320 Thanks: 1232  

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