August 29th, 2018, 11:51 PM  #1 
Newbie Joined: Aug 2018 From: Suisun City, CA Posts: 7 Thanks: 0  Parallel unit vectors
I am at my wits end with this question and hopefully someone can push me in the right direction or offer me some "AHHA!" advice. Find the unit vectors that are parallel to the tangent line to the curve y = 8 sin(x) at the point (π/6, 4). (Enter your answer as a commaseparated list of vectors.) I have found the equation of the tangent line at the given point to be: y=4sqrt(3)x[(2sqrt(3)pi)/3]+4 My understanding is to choose two arbitrary xvalues on the tangent line to create a vector, then divide by their magnitude to create the needed unit vector. When I do this, I start to get some really ugly looking algebra. Am I going about this correctly? If not, please correct me. If I am right, what can I do to get some better looking algebra? 
August 30th, 2018, 12:45 AM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,041 Thanks: 815 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle y = \left ( 4 \sqrt{3} \right ) x + \left ( \frac{4 \sqrt{3}  2 \pi}{\sqrt{3}} \right )$ It'll make things a little easier. As far as the unit vector I would choose x = 0 and $\displaystyle x = \frac{1}{4 \sqrt{3}}$. It's not great but not too bad. Dan  
August 30th, 2018, 12:52 AM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,311 Thanks: 1223 
The line parallel to the tangent that passes through the point of tangency does indeed require all that messy algebra. But they just ask for a vector parallel to the tangent at that point. Any vector with a slope of $4\sqrt{3}$ will suffice. $v = (1, 4\sqrt{3})$ is one such vector and the corresponding unit vector is $u = \dfrac{1}{7}(1,~4\sqrt{3})$ The other unit vector parallel to the tangent is just $u$ 
August 30th, 2018, 11:47 AM  #4 
Newbie Joined: Aug 2018 From: Suisun City, CA Posts: 7 Thanks: 0 
Perfect! Thanks for the confirmation. I really appreciate your help.


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parallel, tangent, unit, vectors 
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