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
x=-1/2
y=sqrt(3)/2+t(cost) ?????????
z=t-2pi/3

Answer 2

Ok, so the displacement vector as a function of time is

s(t) = (-1/2, sin t, t)
and thus the velocity vector which is everywhere tangent is

d/dt s(t) = v(t) = (0, cos t, 1)

At the point P = (cos(-4pi/6), sin(-4pi/6), -4pi/6) , t = -4pi/6 = -2pi/3

Hence v(t) = v(-2pi/3) = (0, sqrt(3)/2, 1), call this D for direction.

Answer 3

The line which is thus tangent at P has parametric form

P + aD,
for real numbers a

Answer 4

or lambda, or whatever you like.

Answer 5

I got
x=-1/2
y=-sqrt(3)/2+a(cost) ?????????
z=a-2pi/3

Answer 6

Nooo. Note that we have calculated D explicitly for the t which corresponds to the point P.

Answer 7

There should be absolutely no t in the expression.

D = (0, sqrt(3)/2, 1)

Answer 8

Hence the line
P + aD = ....what?

Answer 9

P = s(-4pi/6), that's where we got t from.

Answer 10

and we use that value of t to evaluate the tangent direction v(t)

Answer 11

D = v(-4pi/6)

Answer 12

P=(-1/2,-sqrt(3)/2,-2pi/3)
aD=(0,a*sqrt(3)/2,a)
and then just add them?

Answer 13

Yes.

The parametric form of a line is a point on the line, P, plus a scalar times the direction of the line.

E.g., (1,a,0) is a straight lin in the xy-plane that cross the x-axis at (1,0,0) and parallel to the y-axis because the direction of the line is (0,1,0)

Answer 14

Ok, got it now, thanks a bunch. I need to take the Multivariable Calculus class on OCW apparently...

Answer 15

good.

Answer 16

how do you plug this in for each individual variable?

Answer 17

such that x = y = z = t-4pi/6

Answer 18

P=(-1/2,-sqrt(3)/2,-2pi/3)
aD=(0,a*sqrt(3)/2,a)
and then just add them:
\[x=-1/2\]\[y=a \sqrt{3}/2-\sqrt{3}/2\]\[z=a-2 \pi/3\]

Answer 19

Follow James' explanation to understand.
But you said x=-1/2 was wrong, which James doesn't seem to agree with, so...?

Answer 20

So D = v(t = -2pi/3) = (0, sin(t), 1) = (0, -sqrt(3)/2, 1)

Answer 21

And P = (cos(-4pi/6), sin(-4pi/6), -4pi/6) = (-1/2, -sqrt(3)/2, -2pi/3)

Answer 22

I thought I corrected that part already, but sonofa said the x=-1/2 part was wrong.

Answer 23

hence the line has parametric form

\[(-1/2, -\sqrt{3}/2(a + 1), a - 2\pi/3)\]

Check every step, but I think this is right; I'm not doing this on paper and the probability of errors is higher.

The x-ordinate is right because cos(-4pi/6) = -1/2 and the curve doesn't vary for x otherwise.

Answer 24

I'd bring 1/6 outside the who thing to make it neater.

Answer 25

P=(-1/2,-sqrt(3)/2,-2pi/3)
aD=(0,-a*sqrt(3)/2,a)
P+aD:
x=−1/2
y=-a√3/2−√3/2=-sqrt(3)/2(a+1)
z=a−2π/3
right, that's what I meant. I see the typo on the negative for the y, but you agree that x=-1/2, which was what sonofa said was wrong. He must have typed in x=1/2 or something. I don't know where he went though.

Answer 26

i have instant check thing...i can check my answers immediately...i've tried 1/2 and -1/2 and both get marked incorrect

Answer 27

says y is incorrect also

Answer 28

Are you sure x = constant, and not x = cos(t)?

Answer 29

James is an ex-professor of mathematics and came to the same conclusion as me, as can be seen above, so something is amiss here. Neither of us seem positive about the y part, but you can see we are both very confident that x=-1/2. I have my doubts about your program I guess.

Answer 30

right, that's what I was thinking too

Answer 31

I asked him at the beginning if he meant constant and he said yes

Answer 32

the curve x=cost, y=sint, z=t

Answer 33

that's copied and pasted

Answer 34

it's not my program...it's what we do our homework on

Answer 35

I asked you at the beginning if x=constant and you said yes!

Answer 36

it's called webwork...you submit your answers and tells you if your right or wrong

Answer 37

ah ha! s(t) = (cos t, sin t, t)
So v(t) = (-sin t, cos t, 1)

For t = -4pi/6 = -2pi/3,
v(t) = (-sqrt(3)/2, -1/2, 1) = D

Now P = (cos(-4pi/6), sin(-4pi/6), -4pi/6)) = (-1/2, -sqrt(3)/2,-2pi/3)

Thus
P + aD = (-1/2, -sqrt(3)/2,-2pi/3) + a(-sqrt(3)/2, -1/2, 1)

Answer 38

wait...what's the x ?

Answer 39

and what's the y?

Answer 40

-1/2-a sqrt(3)/2

Answer 41

Seriously Sonofa, can't you see how to pull out the x-ordinate here, or y-ordinate?

Answer 42

i can but it looks like your still saying that x is -1/2 and i'm sayin it's getting marked incorrect when i submit that

Answer 43

P+aD
gotta add them dude!

Answer 44

No, you're not reading correctly, which makes me very nervous about any verification you're doing.

Answer 45

and what does the verification program expect for the parameter: a, t, lambda, .... ?

Answer 46

t

Answer 47

it's not a verification again....it's what our homework is submitted on

Answer 48

whatever,
then
x=-1/2-t sqrt(3)/2

Answer 49

P + tD
= (-1/2, -sqrt(3)/2,-2pi/3) + t(-sqrt(3)/2, -1/2, 1)
= (-1/2 - sqrt(3)/2 . t, -sqrt(3)/2 - t/2, - 2pi/3 + t)

Answer 50

okay...now y and z are correct....x is still saying it's wrong

Answer 51

Oh, yes, my fault
-sin(-4pi/6) = sqrt(3)/2. Hence
P + tD = (-1/2 + sqrt(3)/2 . t, -sqrt(3)/2 - t/2, - 2pi/3 + t)

Answer 52

i got it!

Answer 53

yes...exactly lol....it was + sqrt(3)/2 * t

Answer 54

i could never thank you enough for helpign me with that

Answer 55

would have never ever gotten that on my own

Answer 56

both of you.....i owe you huge

Answer 57

Nah, it's all thanks to James, as usual.