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

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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

Mathematics
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so \[\text{ horizontal tangent for parametric equations is given by } \frac{dy}{d \theta}=0 \text{ provided } \\ \frac{d x}{d \theta} \neq 0 \\ \text{ and } \\ \text{ vertical tangent for parametric equations is given by } \frac{dx}{d \theta}=0 \text{ provided } \\ \frac{dy}{d \theta} \neq 0\]
so first do you know how to find dy/dtheta and dx/dtheta ?
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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\[\frac{ dx }{ d \theta} = -3 \sin \theta \] \[\frac{ dy }{ d \theta } = 2 \cos 2\theta \] There we go @freckles
ok cool now we need to set both of those things equal to 0 and solve for theta
\[-3 \sin(\theta)=0 \text{ and solve also } 2 \cos(2 \theta)=0 \\ \text{ or } \sin(\theta)=0 \text{ and also solve } \cos(2 \theta)=0\]
can you solve either one or both or none?
I was able to get \[\sin \theta = 0 \] \[\theta = \pi, 2\]
2 pi
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?
are there *
That was where I was having trouble, since I am really rusty with my trig.
well hey does the problem say anything about restrictions on theta?
like is theta between 0 and 2pi or does it not say anything about that?
It doesn't. :\
ok so there is an infinite amount of solutions
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
that just means n can be ...,-5,-4,-3,-2,-1,0,1,2,3,4,5,... so on in either direction
Hmm, sounds like things could get complicated...
do you know the period of sine is 2pi?
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?
Yes.
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)
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,...
but of course we can go the other way around to include the negatives
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
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.
anyways I'm probably going to be lazy in continue using t instead of theta
so now we have cos(2t)=0
solve this first where t is in [0,2pi)
That's fine. Use t. cos (2t) = 0 is what is give me issues.
ok do you know how to solve cos(u)=0 ?
Wouldn't you take arccos of both sides?
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
\[\cos(u)=0 \text{ on } (0,4\pi) \\ \text{ gives } u=\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}\]
Oh, so just keep adding or subtracting 2 pi each time...
but we wanted to solve cos(2t)=0 on (0,2pi)
so if u=2t we have: the equations: \[2t=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}, \frac{7\pi}{2}\]
so to find t divide both sides by 2 (this is a lazy way of writing 4 equations :p)
Lol. So, \[t = \frac{ \pi }{ 4 }, \frac{ 3\pi }{ 4}, ...\]
yeah we have: \[t=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{5},... \\ \text{ we can generalize the solution } \\ t=\frac{(2n+1) \pi}{4} , n \in \mathbb{Z}\]
that double back Z thing is just an integer
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
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)*
Right. I was actually able to find the vertical tangent lines. Finding the horizontal ones though is where I am getting stuck
are you still getting stuck on that? or is it clearer? or no?
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.
\[\cos(2t)=0 \text{ where } 0 \le t < 2\pi \\ \text{ let } u=2t \\ \text{ solving this for } t \text{ we get } t=\frac{u}{2} \\ \text{ so we have } \cos(2t)=0 \text{ where } 0 \le t <2\pi \text{ becomes } \\ \cos(u)=0 \text{ where } 0 \le \frac{u}{2} <2\pi \\ \text{ rewriting with the inequality solved for } u \\ \text{ gives } \\ \cos(u)=0 \text{ where } 0 \le u <4\pi \\ \text{ so we first wanted to solve } \cos(u)=0 \text{ on } [0,4\pi) \]
I just wanted to show why I solved cos(u)=0 on [0,4pi) instead of [0,2pi)
"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?
For the problem, we have to find the points where there are vertical and horizontal tangent points to the curvegiven by the parametric equations.
find the points (x,y) so we already found the t's
For example, I find that the points (-3, 0) and (3,0) are vertical tangents. ;)
\[(3 \cos(t),\sin(2t)) \\ \] 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
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
Ahh, gotcha! ;) So, many tangents. I will figure those out, then, and will message you if I have any more questions. ;) Thank you, @freckles
k!
also a note on what @phi said earlier \[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \\ \text{ so we do have horizontal when } \frac{dy}{dt}=0 \\ \text{ and verticals when } \frac{dx}{dt}=0 \\ \text{ if there is no intersetions in the above solutions }\]
Well done! http://prntscr.com/8ddra6
nice visual
@MoonlitFate do you see from @mathmate 's visual we do indeed get 4 horizontals and 2 verticals ?

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