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anonymous
 one year ago
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 commaseparated list of ordered pairs.)
x = 3 cos θ, y = sin 2θ
anonymous
 one year ago
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 commaseparated list of ordered pairs.) x = 3 cos θ, y = sin 2θ

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freckles
 one year ago
Best ResponseYou've already chosen the best response.4so \[\text{ horizontal tangent for parametric equations is given by } \frac{dy}{d \theta}=0 \text{ provided } \\ \frac{d x}{d \theta} \neq 0 \\ \text{ and } \\ \text{ vertical tangent for parametric equations is given by } \frac{dx}{d \theta}=0 \text{ provided } \\ \frac{dy}{d \theta} \neq 0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so first do you know how to find dy/dtheta and dx/dtheta ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ dx }{ d \theta} = 3 \sin \theta \] \[\frac{ dy }{ d \theta } = 2 \cos 2\theta \] There we go @freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.4ok cool now we need to set both of those things equal to 0 and solve for theta

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[3 \sin(\theta)=0 \text{ and solve also } 2 \cos(2 \theta)=0 \\ \text{ or } \sin(\theta)=0 \text{ and also solve } \cos(2 \theta)=0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4can you solve either one or both or none?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was able to get \[\sin \theta = 0 \] \[\theta = \pi, 2\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4also 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That was where I was having trouble, since I am really rusty with my trig.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4well hey does the problem say anything about restrictions on theta?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4like is theta between 0 and 2pi or does it not say anything about that?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4ok so there is an infinite amount of solutions

freckles
 one year ago
Best ResponseYou've already chosen the best response.4sin(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
 one year ago
Best ResponseYou've already chosen the best response.4that just means n can be ...,5,4,3,2,1,0,1,2,3,4,5,... so on in either direction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm, sounds like things could get complicated...

freckles
 one year ago
Best ResponseYou've already chosen the best response.4do you know the period of sine is 2pi?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4in 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That was what I got for that part. :)

phi
 one year ago
Best ResponseYou've already chosen the best response.0it sounds like you should be looking for dy/dx =0 (not dy/theta or dx/dtheta)

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so 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
 one year ago
Best ResponseYou've already chosen the best response.4but of course we can go the other way around to include the negatives

freckles
 one year ago
Best ResponseYou've already chosen the best response.4http://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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Gotcha. 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
 one year ago
Best ResponseYou've already chosen the best response.4anyways I'm probably going to be lazy in continue using t instead of theta

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so now we have cos(2t)=0

freckles
 one year ago
Best ResponseYou've already chosen the best response.4solve this first where t is in [0,2pi)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's fine. Use t. cos (2t) = 0 is what is give me issues.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4ok do you know how to solve cos(u)=0 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wouldn't you take arccos of both sides?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you 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
 one year ago
Best ResponseYou've already chosen the best response.4\[\cos(u)=0 \text{ on } (0,4\pi) \\ \text{ gives } u=\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, so just keep adding or subtracting 2 pi each time...

freckles
 one year ago
Best ResponseYou've already chosen the best response.4but we wanted to solve cos(2t)=0 on (0,2pi)

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so if u=2t we have: the equations: \[2t=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}, \frac{7\pi}{2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so to find t divide both sides by 2 (this is a lazy way of writing 4 equations :p)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lol. So, \[t = \frac{ \pi }{ 4 }, \frac{ 3\pi }{ 4}, ...\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4yeah 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
 one year ago
Best ResponseYou've already chosen the best response.4that double back Z thing is just an integer

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so 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
 one year ago
Best ResponseYou've already chosen the best response.4so 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)*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right. I was actually able to find the vertical tangent lines. Finding the horizontal ones though is where I am getting stuck

freckles
 one year ago
Best ResponseYou've already chosen the best response.4are you still getting stuck on that? or is it clearer? or no?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just 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
 one year ago
Best ResponseYou've already chosen the best response.4\[\cos(2t)=0 \text{ where } 0 \le t < 2\pi \\ \text{ let } u=2t \\ \text{ solving this for } t \text{ we get } t=\frac{u}{2} \\ \text{ so we have } \cos(2t)=0 \text{ where } 0 \le t <2\pi \text{ becomes } \\ \cos(u)=0 \text{ where } 0 \le \frac{u}{2} <2\pi \\ \text{ rewriting with the inequality solved for } u \\ \text{ gives } \\ \cos(u)=0 \text{ where } 0 \le u <4\pi \\ \text{ so we first wanted to solve } \cos(u)=0 \text{ on } [0,4\pi) \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I just wanted to show why I solved cos(u)=0 on [0,4pi) instead of [0,2pi)

freckles
 one year ago
Best ResponseYou've already chosen the best response.4"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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For the problem, we have to find the points where there are vertical and horizontal tangent points to the curvegiven by the parametric equations.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4find the points (x,y) so we already found the t's

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For example, I find that the points (3, 0) and (3,0) are vertical tangents. ;)

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[(3 \cos(t),\sin(2t)) \\ \] 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
 one year ago
Best ResponseYou've already chosen the best response.4you know replace t with 3pi/4 and evaluate then replace t with 5pi/4 and evaluate then replace t with 7pi/4 and evaluate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahh, gotcha! ;) So, many tangents. I will figure those out, then, and will message you if I have any more questions. ;) Thank you, @freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.4also 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
 one year ago
Best ResponseYou've already chosen the best response.0Well done! http://prntscr.com/8ddra6

freckles
 one year ago
Best ResponseYou've already chosen the best response.4@MoonlitFate do you see from @mathmate 's visual we do indeed get 4 horizontals and 2 verticals ?
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