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September 19th, 2018, 07:01 PM   #1
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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),−sin(2t, t), when t=pi
(b) r(t)=(sin(2−t), e^(−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
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September 20th, 2018, 08:47 PM   #2
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