September 19th, 2018, 07:08 PM  #1 
Senior Member Joined: Apr 2017 From: New York Posts: 104 Thanks: 6  equation of a tangent line
Compute an equation of the tangent line to the curve q(s) = (s,sin(πs^2),cos(3πs^2))at the point (2,0,1) where s∈R in this questions my steps will be these: step1: match x,y,z components of the point and q(s) to find parameter s step2: find q'(s) step3: insert parameter s into q'(s) to find the point of tangency step 4: equation of tangent line is what in 3D? yy1=m(xx1) in 2D 
September 19th, 2018, 09:17 PM  #2 
Senior Member Joined: Apr 2017 From: New York Posts: 104 Thanks: 6 
This is the work I have done.

September 20th, 2018, 08:33 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,120 Thanks: 1101 
you seem to be misunderstanding this a bit $q(s) = (s,~\sin(\pi s^2),~\cos(3 \pi s^2)$ $p=(2,~0,~1) \Rightarrow s=2$ We find the tangent vector at $p$ by differentiating $q(s)$ and letting $s=2$ $\dfrac{dq}{ds} = (1,~2 \pi s \cos \left(\pi s^2\right),~6 \pi s \sin \left(3 \pi s^2\right))$ $\left . \dfrac{dq}{ds}\right_{s=2} = (1,~4 \pi ,~0)$ and our unit tangent vector at $p$ is thus $T = \dfrac{1}{\sqrt{16\pi^2 + 1}}(1,~4\pi,~0)$ and the equation for our line is simply $\ell(u) =uT + p$ $\ell(u) =\left( \dfrac{u}{\sqrt{1+16 \pi ^2}}+2,~\dfrac{4 \pi u}{\sqrt{1+16 \pi ^2}},~1\right)$ 

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equation, line, tangent 
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