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 August 29th, 2018, 10:51 PM #1 Newbie   Joined: Aug 2018 From: Suisun City, CA Posts: 3 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 "AH-HA!" 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 comma-separated 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 x-values 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 29th, 2018, 11:45 PM   #2
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 Originally Posted by cvbody I am at my wits end with this question and hopefully someone can push me in the right direction or offer me some "AH-HA!" 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 comma-separated 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 x-values 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?
It is ugly, isn't it? I'd use a slightly different (non-standard) version:
$\displaystyle y = \left ( 4 \sqrt{3} \right ) x + \left ( \frac{4 \sqrt{3} - 2 \pi}{\sqrt{3}} \right )$

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 29th, 2018, 11:52 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,100 Thanks: 1093 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 $-u$ Thanks from topsquark and cvbody
 August 30th, 2018, 10:47 AM #4 Newbie   Joined: Aug 2018 From: Suisun City, CA Posts: 3 Thanks: 0 Perfect! Thanks for the confirmation. I really appreciate your help.

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