Calculus Calculus Math Forum

 September 19th, 2018, 08:08 PM #1 Senior Member   Joined: Apr 2017 From: New York Posts: 165 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? y-y1=m(x-x1) in 2D September 19th, 2018, 10:17 PM   #2
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This is the work I have done.
Attached Images 85747C8B-181E-4A29-A969-8C0D4D50EDAE.jpg (20.6 KB, 4 views) D9CAFF95-10F3-4F07-A02B-D8847043948E.jpg (19.7 KB, 3 views) September 20th, 2018, 09:33 PM #3 Senior Member   Joined: Sep 2015 From: USA Posts: 2,645 Thanks: 1476 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 $\ell(u) =uT + p$ $\ell(u) =\left( \dfrac{u}{\sqrt{1+16 \pi ^2}}+2,~\dfrac{4 \pi u}{\sqrt{1+16 \pi ^2}},~1\right)$ Tags equation, line, tangent Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post frazza999 Pre-Calculus 6 June 25th, 2014 03:58 PM unwisetome3 Calculus 2 October 28th, 2012 07:52 PM kevpb Calculus 3 May 25th, 2012 11:32 PM arron1990 Calculus 5 February 9th, 2012 02:29 AM RMG46 Calculus 28 September 28th, 2011 10:21 AM

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