Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to check your work. (Enter your answers as a comma-separated list of ordered pairs.)
x = 3 cos θ, y = sin 2θ

- MoonlitFate

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- schrodinger

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- freckles

so
\[\text{ horizontal tangent for parametric equations is given by } \frac{dy}{d \theta}=0 \text{ provided } \\ \frac{d x}{d \theta} \neq 0 \\ \text{ and } \\ \text{ vertical tangent for parametric equations is given by } \frac{dx}{d \theta}=0 \text{ provided } \\ \frac{dy}{d \theta} \neq 0\]

- freckles

so first do you know how to find dy/dtheta and dx/dtheta ?

- MoonlitFate

Yes, I was able to find them.
For dx/dtheta I got -3 sin (theta)
And for dy/dtheta I got 2 cos (2 theta)
Sorry if the formatting is weird.

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## More answers

- MoonlitFate

\[\frac{ dx }{ d \theta} = -3 \sin \theta \]
\[\frac{ dy }{ d \theta } = 2 \cos 2\theta \]
There we go @freckles

- freckles

ok cool
now we need to set both of those things equal to 0 and solve for theta

- freckles

\[-3 \sin(\theta)=0 \text{ and solve also } 2 \cos(2 \theta)=0 \\ \text{ or } \sin(\theta)=0 \text{ and also solve } \cos(2 \theta)=0\]

- freckles

can you solve either one or both or none?

- MoonlitFate

I was able to get \[\sin \theta = 0 \]
\[\theta = \pi, 2\]

- MoonlitFate

2 pi

- freckles

also we are going to have an infinite amount of solutions without any restrictions
and I notice it says enter answers separated by comma
or there any restrictions on theta?

- freckles

are there *

- MoonlitFate

That was where I was having trouble, since I am really rusty with my trig.

- freckles

well hey does the problem say anything about restrictions on theta?

- freckles

like is theta between 0 and 2pi
or does it not say anything about that?

- MoonlitFate

It doesn't. :\

- freckles

ok so there is an infinite amount of solutions

- freckles

sin(t)=0
when t=0,pi,2pi,3pi,4pi,5pi,... and so on...
you can also go the negative way around
t also includes=..,-5pi,-4pi,-3pi,-2pi,-pi
in general you can say
sin(t)=0 when t=n pi
where n is an integer

- freckles

that just means n can be ...,-5,-4,-3,-2,-1,0,1,2,3,4,5,... so on in either direction

- MoonlitFate

Hmm, sounds like things could get complicated...

- freckles

do you know the period of sine is 2pi?

- freckles

in the first period of sine
that is I'm talking about in the interval [0,2pi)
we see the solutions to sin(t)=0
is t=0 or t=pi
do you agree with this?

- MoonlitFate

Yes.

- MoonlitFate

That was what I got for that part. :)

- phi

it sounds like you should be looking for dy/dx =0 (not dy/theta or dx/dtheta)

- freckles

so the period of sine being 2pi
means t=0+2pi or t=pi+2pi is a solution
simplifying gives: t=2pi or t=3pi
but we can keep going around the circle over and over again
adding another 2pi
t=2pi+2pi or t=3pi+2pi
simplifying t=4pi or t=5pi
we see a pattern here we are going to get 0,pi,2pi,3pi,4pi,5pi,6pi,...

- freckles

but of course we can go the other way around to include the negatives

- freckles

http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent.aspx
here is another source @MoonlitFate
this is why I'm saying to find dy/dtheta or dx/dtheta

- MoonlitFate

Gotcha. I have so much trig to brush up on. Okay, and I know that your values of theta you have to plug back ito your original euqations to find your coordinates.

- freckles

anyways I'm probably going to be lazy in continue using t instead of theta

- freckles

so now we have cos(2t)=0

- freckles

solve this first where t is in [0,2pi)

- MoonlitFate

That's fine. Use t. cos (2t) = 0 is what is give me issues.

- freckles

ok do you know how to solve cos(u)=0 ?

- MoonlitFate

Wouldn't you take arccos of both sides?

- freckles

you should see the first value at u=pi/2
cos(pi/2)=0
second value occurs at u=3pi/2
cos(3pi/2)=0
we could also go around again
next value would be at u=pi/2+2pi=5pi/2
we know
cos(5pi/2)=0
and the next next value would be at u=3pi/2+2pi=7pi/2
and we know also
cos(7pi/2)=0

- freckles

\[\cos(u)=0 \text{ on } (0,4\pi) \\ \text{ gives } u=\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}\]

- MoonlitFate

Oh, so just keep adding or subtracting 2 pi each time...

- freckles

but we wanted to solve cos(2t)=0 on (0,2pi)

- freckles

so if u=2t we have:
the equations:
\[2t=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}, \frac{7\pi}{2}\]

- freckles

so to find t divide both sides by 2
(this is a lazy way of writing 4 equations :p)

- MoonlitFate

Lol. So, \[t = \frac{ \pi }{ 4 }, \frac{ 3\pi }{ 4}, ...\]

- freckles

yeah
we have:
\[t=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{5},... \\ \text{ we can generalize the solution } \\ t=\frac{(2n+1) \pi}{4} , n \in \mathbb{Z}\]

- freckles

that double back Z thing is just an integer

- freckles

so let's go back to the first thing I said:
\[\text{ horizontal tangent for parametric equations is given by } \frac{dy}{d \theta}=0 \text{ provided } \\ \frac{d x}{d \theta} \neq 0 \\ \text{ and } \\ \text{ vertical tangent for parametric equations is given by } \frac{dx}{d \theta}=0 \text{ provided } \\ \frac{dy}{d \theta} \neq 0\]
so we definitely don't have any intersection in the solutions of dy/dt and dx/dt
(remember I'm being lazy; using t instead of theta)
so we have horizontal tangent is at t=(2n+1)pi/4
and vertical tangents is at t=n pi
n is an integer

- freckles

so we definitely don't have any intersection in the solutions of dy/dt=0 and dx/dt=0
(remember I'm being lazy; using t instead of theta)*

- MoonlitFate

Right. I was actually able to find the vertical tangent lines. Finding the horizontal ones though is where I am getting stuck

- freckles

are you still getting stuck on that? or is it clearer? or no?

- MoonlitFate

Just none of my solutions are correct. I plug in the values where t that make dy/dtheta = 0 to find the points, but they aren't right.

- freckles

\[\cos(2t)=0 \text{ where } 0 \le t < 2\pi \\ \text{ let } u=2t \\ \text{ solving this for } t \text{ we get } t=\frac{u}{2} \\ \text{ so we have } \cos(2t)=0 \text{ where } 0 \le t <2\pi \text{ becomes } \\ \cos(u)=0 \text{ where } 0 \le \frac{u}{2} <2\pi \\ \text{ rewriting with the inequality solved for } u \\ \text{ gives } \\ \cos(u)=0 \text{ where } 0 \le u <4\pi \\ \text{ so we first wanted to solve } \cos(u)=0 \text{ on } [0,4\pi) \]

- freckles

I just wanted to show why I solved cos(u)=0 on [0,4pi) instead of [0,2pi)

- freckles

"Just none of my solutions are correct. I plug in the values where t that make dy/dtheta = 0 to find the points, but they aren't right.
"
what are you talking about?

- MoonlitFate

For the problem, we have to find the points where there are vertical and horizontal tangent points to the curvegiven by the parametric equations.

- freckles

find the points (x,y)
so we already found the t's

- MoonlitFate

For example, I find that the points (-3, 0) and (3,0) are vertical tangents. ;)

- freckles

\[(3 \cos(t),\sin(2t)) \\ \]
let's just use the t's from the first period then since we don't actually have to find the t's :p
we had t=0,pi for vertical
we had t=pi/4,3pi/4,5pi/4,7pi/4 for horizontal
vertical:
\[(3 \cos(0),\sin(2 \cdot 0))=(3 (1),\sin(0))=(3,0) \\ (3 \cos(\pi),\sin(2 \cdot \pi))=(3(-1),0)=(-3,0)\]
so good job there
horizontal:
\[(3 \cos(\frac{\pi}{4}),\sin(2 \cdot \frac{\pi}{4}))=(3 \frac{\sqrt{2}}{2}, \sin(\frac{\pi}{2}))=(\frac{3 \sqrt{2}}{2},1)\]
still need to plug in the other three values

- freckles

you know replace t with 3pi/4
and evaluate
then replace t with 5pi/4
and evaluate
then replace t with 7pi/4
and evaluate

- MoonlitFate

Ahh, gotcha! ;) So, many tangents. I will figure those out, then, and will message you if I have any more questions. ;) Thank you, @freckles

- freckles

k!

- freckles

also a note on what @phi said earlier
\[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \\ \text{ so we do have horizontal when } \frac{dy}{dt}=0 \\ \text{ and verticals when } \frac{dx}{dt}=0 \\ \text{ if there is no intersetions in the above solutions }\]

- mathmate

Well done!
http://prntscr.com/8ddra6

- freckles

nice visual

- freckles

@MoonlitFate do you see from @mathmate 's visual we do indeed get 4 horizontals and 2 verticals ?

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