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Parametric curve with tangent

mark

(10 pts)

In this you’re going to use either CalcPlot3D or GeoGebra 3D to plot a groovy parametric curve together with a tangent line. In some special, degenerate cases, you might explain why there is no well defined tangent line at the point where t=0.

To get your parameterized path, choose your name from this list:

audrey

My function is

p(t) = (\cos(6t), \sin(3t), \cos(2t).

Note that p(0) = (1,0,1) and that

p'(t) = (-6\sin(6t), 3\cos(3t), -2\sin(2t)

so p'(0) = (0,3,0).
Therefore, my line is:

\vec{\ell}(t) = \langle 1,3t,1 \rangle.

Here’s my picture:

And here’s a link.

jsanfeli

My function is

p(t)=(sin(3t),cos(2t),sin(3t)).

Note that p(0)=(0,1,0) and that

p'(t)=(3cos(3t),-2sin(2t),3cos(3t))

so p'(0)=(3,0,3).
Therefore, my line is:

\vec{\ell}(t)=\langle 3t,1,3t \rangle.

Here’s my picture:

And here’s the a link

dfleming

My function was:

p(t) = (cos(4t), sin(4t), cos(2t)

When t = 0,

p(0) = (1,0,1)

Additionally,

p'(t) = (-4 sin(4t), 4 cos(4t), -2 sin(2t))

And

p'(0) = (0,4,0)

Therefore, my tangent line is:

l(t) = (1,4t,1)

My picture is:

skhadka

My Function is

p(t) = (sin(4t),cos(6t),cos(5t))

p(0)= (0,1,1). p’(t)= (4,0,0)
so the line is l(t) = <4t,1,1>

Picture is

link CalcPlot3D

kduckett

My function is

p(t) = (sin(3t),cos(5t),sin(6t))

Note that: p(0) = (0,1,0) and that

p'(t) = (3cos(3t),-5sin(5t),6cos(6t))

so p’(0) = (3,0,6)

Therefore my line is:

\vec{\ell}(t) = \langle 3t,1,6t \rangle

Here is my picture:

And here’s a link.

chughes2

My function is p(t)=(\cos(6t),\sin(6t),\cos(4t)), and p(0)=(1,0,1)
therefore p'(t)=-6\sin(6t),6cos(6t),-4\sin(4t), and p'(0)=(0,6,0)

by following the equation \vec{p}(0)+\vec{p}(0)t, we get the equation \langle1,0,1\rangle+\langle0,6,0\rangle t, therefore our tangent equation is \langle1,6t,1\rangle

here is a photo showing the result:

smarsha1

My function is \vec{p}(t) = \langle \sin(3t),\cos(4t),\cos(2t)\rangle

Note that \vec{p}(0) = \langle 0,1,1 \rangle and that

\vec{p'}(t) = \langle 3\cos(3t),-4\sin(4t),-2\sin(2t) \rangle so \vec{p'}(0) = \langle 3,0,0 \rangle.

Therefore, my line is:

\vec{\ell}(t) = \langle 3t,1,1 \rangle.

And lastly, here is my picture with a Link to the graph:

whardin3

My function is:
cos(2t),sin(2t),cos(5t)
p(0)= (1,0,1)
and p'(t)= -2sin(2t), 2cos(2t), -5sin(5t)
so p'(0)= (0, 2, 0)
Thus my line is: l(t)= <1, 2t, 1>
My picture is below: