Question

PLEASE HELP, DUE TODAY: Find the cosine of the angle between the curves
cos t, sin(t)/4, sin tand cos t, sin t, sin(2t)where they intersect.
The possible answers are : 7√5√17/85 or
−9√5√17/85. Please show all of your work.
Thanks

Answer 1

I won't do it for you, but try taking the derivative in the intersection point, and remember that it is the tangent of the angle between the horizontal and the tangent line

Answer 2

Okay, how do I find the intersection point? I was told to do this:
cos(t)=cos(u)
sin(t)/4=sin(u)
sin(t)=sin(2u)
So I solve for u in one of the equations, and substitute it into one of the other equations and that is supposed to get me a point of intersection...but it just doesnt seem right to say u=arccos(cos(t)).. I'm really confused... can you explain a little bit more?

Answer 3

You are right, that is really strange and I believe it is wrong. Can you put more clearly the functions of the heading, because I'm confused with what you wrote.

Answer 4

Sure. \[<\cos(t), \frac{ -\sin(t) }{ 4 }, \sin(t)> \] and \[<\cos(t), \sin(t), \sin(2t)>\]. Those are the two curves. So the point of intersection comes from setting x_1=x_2, y_1=y_2, and z_1=z_2, but like I said before, it doesnt look right...

Answer 5

Aaah ok, now I understand. That is right, you need to solve those equations now. Was that your doubt, or you need help with this part?

Answer 6

Yeah, that's the part I need help in :/

Answer 7

Well cos(t) is always the same, what about the third equation, when is sin(t)=sin(2t)? What can you do here?

Answer 8

\[\frac{ \arcsin(\sin(t)) }{ 2 }=u\]. Is that any better?

Answer 9

try logic, think about the trigonometric circle, this way it gets more complicated

Answer 10

t=pi?

Answer 11

Yes, that is a particular solution, and maybe the only one you need, but te general solution is t=n pi, where n is any integer, so 2pi is also an important solution.
Ok now go to the second equation. Remember that for both curves intersect, all 3 equations must be satisfied, so the right solution for the second one must also be a solution for the third, but try to do the general solution first.

Answer 12

Im confused again :/ what do i do with t=n*pi now? I'm sorry, I've just been stuck on this problem for about 3 hours now

Answer 13

Dont worry about n pi, solve the second one, then you come back to n pi.
In mathematics never try to think of how are you going to use that latter if you know that you did what you had to do. You need to solve those equations, that is obvious and right, so don't worry about later. Only do that if you think you already screwed up.

Answer 14

Well, another solution would be pi, wouldnt it?

Answer 15

how would you guys propose to when y is the same?

Answer 16

to find*

Answer 17

I don't understand the question, can you rephrase it?

Answer 18

actually nevermind I was being slow :/ it's easy

Answer 19

Can you help me out?

Answer 20

I think so, yeah
no interval is given for t?

Answer 21

Nope

Answer 22

well I guess maybe the shape is periodic so the angles repeat
you guys seem to be on the right track
cos t=cos t -> t is all reals
sin t=1/4sin t ->t=0
sin t=sin(2t) ->t=n*pi, n=0,1,2...

Answer 23

only one value of t fits all three solutions, so...?

Answer 24

0?

Answer 25

yeah, looks like it

Answer 26

then do what @ivanmlerner said and get the derivatives (i.e. tangent vectors) at that point

Answer 27

How do i get the derivatives if i only have t=0?

Answer 28

take the derivatives first

Answer 29

...before plugging in the numbers
that is the tangent vector function

Answer 30

Wait, for the second solution t is not only 0, it can be any n*pi also but since the angles repeat that does not make any difference

Answer 31

ah, you are right, but the choice of 0 or n*pi should give the same answer, no?

Answer 32

So i could really use either, because there are two possible answers?

Answer 33

0 or n*pi are the only solutions for the y and z equations
the sines and cosines will have the same values, and since we only have sint and cost in both our position function and tangent function I argue that it makes no difference whether we use t=0 or t=n*pi

Answer 34

though I have been known to be wrong in my day, my theory is that this shape is periodic, and we will get the same answer either way
let's try and see if we can get one of your choices...

Answer 35

Okay, so, let's try 0. I plug 0 in into \[<\cos(t), \sin(t), \sin(2t)\] and get \[<1, 0, 0>\] and that would be hwere the curves meet, right?

Answer 36

And now, the derivatives are \[<-\sin(t), -\frac{ \cos(t) }{ 4 }, \cos(t)> and <-\sin(t), \cos(t), 2\cos(2t)>\] yes?

Answer 37

Oh my god that is so cool, I graphed it and maybe it would be better to check 0 and pi.

Answer 38

@kandsy yes

Answer 39

the y component in the first function is negative? I missed that

Answer 40

yes, because -cos is the deriv of -sin

Answer 41

right, but I didn't see the -sin in your post, I think there is a typo

Answer 42

Right, i see that now, my original post was the typo, it does in fact have a negative.

Answer 43

My answer is close to your options but I'm ot sure where the 85 is coming from
anyway let's continue, you're doing fine...

Answer 44

oh I see it's just rationalized
yes I have one answer...

Answer 45

I have one issue which is that I am not getting the negative sign

Answer 46

I hate missing negative signs :S

Answer 47

oh wait, now I got it
okay, we're good :)

Answer 48

so now i plug 0 into the first curve and get \[<-\sin(0),\frac{ −\cos(0)}{ 4 },\cos(0)>\] right? And so I get: <0, -1/4, 1> and that is one of the points of intersection. Now i plug pi into the other curve, right?

Answer 49

the best way to do it in my opinion is to do the dot product of the two vectors without pluggin in the numbers until the end

Answer 50

so let\[\vec r_1'=\langle-\sin t,-\frac14\cos t,\cos t\rangle\]\[\vec r_2=\langle-\sin t,\cos t,2\cos(2t)\cos t\rangle\]

Answer 51

now apply the formula\[\vec r_1\cdot\vec r_2=\|\vec r_1\|\|\vec r_2\|\cos\theta_{12}\]

Answer 52

stupid LaTeX :/

Answer 53

I mean\[\vec r_1'\cdot\vec r_2'\]then plug in t=0
excuse me, I am sorry to confuse you

Answer 54

that should make more sense though, as those are the tangent vectors

Answer 55

Thank you VERY much :) It's made a lot of sense. You're fine. But, what youre telling me is, that i only have to plug 0 in, in order to get one of the answers? Because, I only need one.

Answer 56

then do the same for\[\|\vec r_1'\|\]and\[\|\vec r_2'\|\]

Answer 57

yes you only need to plug in t=0, I already checked

Answer 58

then once you got all that business, finally apply\[\vec r_1'\cdot\vec r_2'=\|\vec r_1'\|\|\vec r_2'\|\cos\theta_{12}\]solve for \(\cos\theta_{12}\) and you're done :)

Answer 59

Okay, great, thanks. My problem has been solved :)

Answer 60

you are welcome :D
happy to help!