Leonardox

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

Leonardox

This is the work I have done.

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romsek

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

$\ell(u) =uT + p$

$\ell(u) =\left( \dfrac{u}{\sqrt{1+16 \pi ^2}}+2,~\dfrac{4 \pi u}{\sqrt{1+16 \pi ^2}},~1\right)$