Question

Find parametric equations for the tangent line at the point (cos(-4pi/6), sin(-4pi/6), -4pi/6) on the curve x = const , y = sin(t), z = t

Answer 1

i got the z part....that's it

Answer 2

x= constant?

Answer 3

yes

Answer 4

differentiate

x=0
y= cos(t)
z=1

plug in points

Answer 5

he made it soud a lot easier than it actually is

Answer 6

\[f(x,y,z)=(K,\sin t, t)\]\[f'(x,y,z)=(0,\cos t, 1)\]plugging in the points gives\[(0,-1/2,1)\] I feel like we're not done yet, but I'm thinking how to proceed...

Answer 7

I'm going to ask how to represent this parametrically, one sec...

Answer 8

After plugging in the points should have been\[(0,\cos(\sin -2\pi/3),1)=(0,0.648,1)\]
To get the parametric form of this line I think we have to say\[v=(x_0,y_0,z_0)+t(f'(x,y,z)\]\[=(1/2,-\sqrt{3}/2,-2 \pi/3)+t(0,0.647,1)\]\[=(1/2,-\sqrt{3}/2+0.674t,-2 \pi/3+t)\] These are vectors above, but <> symbols look strange on this equation program. I'm not sure if that is the right form though, but it seems right to me. You may want a second opinion.

Answer 9

well...i can try and plug them in and see what's what

Answer 10

i know for a fact...b/c it's been check already taht z = t - 4pi/6

Answer 11

yeah I have that too, is the rest right though?

Answer 12

i don't see where you have that

Answer 13

I wrote z as -2pi/3+t
same thing...

Answer 14

yea...says the x and y are wrong

Answer 15

I have
x=1/2
y=0.674t-sqrt(3)/2
z=t-2pi/3

Answer 16

yea...says z is the only one that's correct

Answer 17

hmmm....

Answer 18

only thing i can think....take derivative of each?

Answer 19

then just use the points given?

Answer 20

like the z component...the derivative is just -4pi/6 so z = t....just t - 4pi/6

Answer 21

i mean that's how we do it right?

Answer 22

like y...should be sin(t) + derivative of y component

Answer 23

I reposted the question and asked James to answer it. He's a pro so he can tell us where we messed up.

Answer 24

k

Answer 25

yeah, but then it should be
x=-1/2
y=sqrt(3)/2+tcos(t)
z=1-2pi/3
tell me at least the x is right now, I forgot the "-" sign earlier.

Answer 26

sorry went to the bathroom

Answer 27

no the x is not correct

Answer 28

Ok, the solution is on my post of this question.

Answer 29

yea i'm lookin for it

Answer 30

P=(-1/2,-sqrt(3)/2,-2pi/3)
aD=(0,a*sqrt(3)/2,a)
\[P+aD=(-1/2,-\sqrt{3}/2+a \sqrt{3}/2,a-2 \pi/3)\]

Answer 31

P=(-1/2,-sqrt(3)/2,-2pi/3)
aD=(0,a*sqrt(3)/2,a)
P+aD=(−1/2,−√3/2-a√3/2,a−2π/3)
x=-1/2 (this IS right)
y=[-sqrt(3)/2](a+1)
z=a-2pi/3