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MoonlitFate

  • 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 comma-separated list of ordered pairs.) x = 3 cos θ, y = sin 2θ

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  1. freckles
    • one year ago
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    so \[\text{ horizontal tangent for parametric equations is given by } \frac{dy}{d \theta}=0 \text{ provided } \\ \frac{d x}{d \theta} \neq 0 \\ \text{ and } \\ \text{ vertical tangent for parametric equations is given by } \frac{dx}{d \theta}=0 \text{ provided } \\ \frac{dy}{d \theta} \neq 0\]

  2. freckles
    • one year ago
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    so first do you know how to find dy/dtheta and dx/dtheta ?

  3. moonlitfate
    • one year ago
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    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.

  4. moonlitfate
    • one year ago
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    \[\frac{ dx }{ d \theta} = -3 \sin \theta \] \[\frac{ dy }{ d \theta } = 2 \cos 2\theta \] There we go @freckles

  5. freckles
    • one year ago
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    ok cool now we need to set both of those things equal to 0 and solve for theta

  6. freckles
    • one year ago
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    \[-3 \sin(\theta)=0 \text{ and solve also } 2 \cos(2 \theta)=0 \\ \text{ or } \sin(\theta)=0 \text{ and also solve } \cos(2 \theta)=0\]

  7. freckles
    • one year ago
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    can you solve either one or both or none?

  8. moonlitfate
    • one year ago
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    I was able to get \[\sin \theta = 0 \] \[\theta = \pi, 2\]

  9. moonlitfate
    • one year ago
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    2 pi

  10. freckles
    • one year ago
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    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?

  11. freckles
    • one year ago
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    are there *

  12. moonlitfate
    • one year ago
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    That was where I was having trouble, since I am really rusty with my trig.

  13. freckles
    • one year ago
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    well hey does the problem say anything about restrictions on theta?

  14. freckles
    • one year ago
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    like is theta between 0 and 2pi or does it not say anything about that?

  15. moonlitfate
    • one year ago
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    It doesn't. :\

  16. freckles
    • one year ago
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    ok so there is an infinite amount of solutions

  17. freckles
    • one year ago
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    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

  18. freckles
    • one year ago
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    that just means n can be ...,-5,-4,-3,-2,-1,0,1,2,3,4,5,... so on in either direction

  19. moonlitfate
    • one year ago
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    Hmm, sounds like things could get complicated...

  20. freckles
    • one year ago
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    do you know the period of sine is 2pi?

  21. freckles
    • one year ago
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    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?

  22. moonlitfate
    • one year ago
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    Yes.

  23. moonlitfate
    • one year ago
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    That was what I got for that part. :)

  24. phi
    • one year ago
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    it sounds like you should be looking for dy/dx =0 (not dy/theta or dx/dtheta)

  25. freckles
    • one year ago
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    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,...

  26. freckles
    • one year ago
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    but of course we can go the other way around to include the negatives

  27. freckles
    • one year ago
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    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

  28. moonlitfate
    • one year ago
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    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.

  29. freckles
    • one year ago
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    anyways I'm probably going to be lazy in continue using t instead of theta

  30. freckles
    • one year ago
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    so now we have cos(2t)=0

  31. freckles
    • one year ago
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    solve this first where t is in [0,2pi)

  32. moonlitfate
    • one year ago
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    That's fine. Use t. cos (2t) = 0 is what is give me issues.

  33. freckles
    • one year ago
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    ok do you know how to solve cos(u)=0 ?

  34. moonlitfate
    • one year ago
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    Wouldn't you take arccos of both sides?

  35. freckles
    • one year ago
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    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

  36. freckles
    • one year ago
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    \[\cos(u)=0 \text{ on } (0,4\pi) \\ \text{ gives } u=\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}\]

  37. moonlitfate
    • one year ago
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    Oh, so just keep adding or subtracting 2 pi each time...

  38. freckles
    • one year ago
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    but we wanted to solve cos(2t)=0 on (0,2pi)

  39. freckles
    • one year ago
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    so if u=2t we have: the equations: \[2t=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}, \frac{7\pi}{2}\]

  40. freckles
    • one year ago
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    so to find t divide both sides by 2 (this is a lazy way of writing 4 equations :p)

  41. moonlitfate
    • one year ago
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    Lol. So, \[t = \frac{ \pi }{ 4 }, \frac{ 3\pi }{ 4}, ...\]

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

  43. freckles
    • one year ago
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    that double back Z thing is just an integer

  44. freckles
    • one year ago
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    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

  45. freckles
    • one year ago
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    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)*

  46. moonlitfate
    • one year ago
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    Right. I was actually able to find the vertical tangent lines. Finding the horizontal ones though is where I am getting stuck

  47. freckles
    • one year ago
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    are you still getting stuck on that? or is it clearer? or no?

  48. moonlitfate
    • one year ago
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    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.

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

  50. freckles
    • one year ago
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    I just wanted to show why I solved cos(u)=0 on [0,4pi) instead of [0,2pi)

  51. freckles
    • one year ago
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    "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?

  52. moonlitfate
    • one year ago
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    For the problem, we have to find the points where there are vertical and horizontal tangent points to the curvegiven by the parametric equations.

  53. freckles
    • one year ago
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    find the points (x,y) so we already found the t's

  54. moonlitfate
    • one year ago
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    For example, I find that the points (-3, 0) and (3,0) are vertical tangents. ;)

  55. freckles
    • one year ago
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    \[(3 \cos(t),\sin(2t)) \\ \] 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

  56. freckles
    • one year ago
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    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

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

  58. freckles
    • one year ago
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    k!

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

  60. mathmate
    • one year ago
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    Well done! http://prntscr.com/8ddra6

  61. freckles
    • one year ago
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    nice visual

  62. freckles
    • one year ago
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    @MoonlitFate do you see from @mathmate 's visual we do indeed get 4 horizontals and 2 verticals ?

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