MoonlitFate
  • MoonlitFate
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • 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
  • freckles
so first do you know how to find dy/dtheta and dx/dtheta ?
MoonlitFate
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.