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Jhannybean

  • one year ago

Calculus I Show that the given family of curves are orthogonal trajectories to eachother; that every curve in one family is orthogonal to every curve in the other family \[x^2+y^2=r^2 \qquad \qquad ax+by=0\] Without using integration or differential equations.

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  1. Jhannybean
    • one year ago
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    trying to help out a friend in understanding how to solve this but Im not sure how to explain this proof...

  2. Jhannybean
    • one year ago
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    I know to find whether they're orthogonal to one another I will eventually take their product, but as for getting there i'm a little cnofused.

  3. Jhannybean
    • one year ago
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    So differentiating the first one, I get \[2x+2yy'=0\]\[y'=-\frac{x}{y}\]

  4. Jhannybean
    • one year ago
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    What do I do with the second equation? treat \(a\) and \(b\) as constants?

  5. Jhannybean
    • one year ago
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    \[d(ax+by=0) \implies a+by' =0\] I think.........

  6. anonymous
    • one year ago
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    woah i have no idea what this is xD

  7. anonymous
    • one year ago
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    me neither, but one is a circle and the other is a line|dw:1444184657069:dw|

  8. Jhannybean
    • one year ago
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    Hmm...

  9. anonymous
    • one year ago
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    oh oops

  10. anonymous
    • one year ago
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    one is a circle through the origin, one is a line through the origin

  11. anonymous
    • one year ago
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    i must be tired "center at the origin"

  12. anonymous
    • one year ago
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    |dw:1444184757878:dw|

  13. Jhannybean
    • one year ago
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    Well that would mean that..... \[ax+by=0 \implies by = -ax \implies y=-\frac{a}{b}x\]

  14. anonymous
    • one year ago
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    maybe this would help it does seem kind of obvious though http://www.sosmath.com/diffeq/first/orthogonal/orthogonal.html

  15. Jhannybean
    • one year ago
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    So now I've got \[(1) : y' = -\frac{x}{y} \qquad (2) : y=-\frac{a}{b}x\]

  16. anonymous
    • one year ago
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    'cept you did say calc 1 right? so maybe it i s something else

  17. anonymous
    • one year ago
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    i should shut up because i really don't know what you are supposed to do to show this

  18. Jhannybean
    • one year ago
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    Haha yeah I'm trying to figure it out. http://www.slader.com/textbook/9780538497817-stewart-calculus-7th-edition/163/exercises/49/ Im not really understanding the solution here either.

  19. anonymous
    • one year ago
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    try this http://www.emathzone.com/tutorials/geometry/normal-line-of-circle-passes-through-origin.html

  20. anonymous
    • one year ago
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    oh i can understand what you sent

  21. anonymous
    • one year ago
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    want to walk through it?

  22. dan815
    • one year ago
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    perpendicular vectors

  23. Jhannybean
    • one year ago
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    Well everything minus the integration part since integration hasn't been covered. That's where I'm having the problem..

  24. Jhannybean
    • one year ago
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    She's just learned basic derivatives haha

  25. dan815
    • one year ago
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    |dw:1444185390633:dw|

  26. dan815
    • one year ago
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    v=<y,-x> ---> tangent vectors to a circle and since you have ax+by=0 this is all possible slopes of lines from the origin, so whereever it intersects your circle, there is always going to be some tangent vector at that point that is orthogonal

  27. Jhannybean
    • one year ago
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    How did get the `v=<y,-x>` and not `v=<-x,y>`?

  28. Jhannybean
    • one year ago
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    oh..... nvm, negative slope.

  29. Jhannybean
    • one year ago
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    Can I just take ax + by = 0 in standard form, and state that the slope for that line is positive? lol....

  30. dan815
    • one year ago
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    v=<-y,x> is fine too if u have clockwise direction tangents v=<y,-x> is ccw tangents directions

  31. dan815
    • one year ago
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    it doesnt have to positive

  32. dan815
    • one year ago
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    all slopes are fine

  33. dan815
    • one year ago
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    y=-a/b x

  34. anonymous
    • one year ago
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    got logged out!

  35. anonymous
    • one year ago
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    looks likke @dan815 got it under control but in brief you get \[y'=-\frac{x}{y}\] for the circle and \[y'=\frac{y}{x}\] for the line

  36. anonymous
    • one year ago
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    i guess more presicely you get \[m=\frac{y_0}{x_0}\] for the line since it is a number

  37. Jhannybean
    • one year ago
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    how did you get y' = y/x for the line? O_o

  38. dan815
    • one year ago
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    thats the slope of a line y/x

  39. dan815
    • one year ago
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    for any line going through the center

  40. Jhannybean
    • one year ago
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    oh i wa thinking about it in terms for actually solving the ax+by = 0 and not thinking about the actual slope of the line xD Yeah i cnnected the two nowhaha

  41. Jhannybean
    • one year ago
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    Ok I think I can explain this now. Just understanding that ax+by = 0 is a `line` and x\(^2\) + y\(^2\) = r\(^2\) is a circle centered at (0,0) means that by taking the product of both of their slopes I get -1

  42. Jhannybean
    • one year ago
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    brb

  43. dan815
    • one year ago
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    =]

  44. dan815
    • one year ago
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    dont think about it too much you know this property of circle

  45. dan815
    • one year ago
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    |dw:1444186328326:dw|

  46. Jhannybean
    • one year ago
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    Ya

  47. dan815
    • one year ago
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    thats all they want u to show

  48. dan815
    • one year ago
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    The Tangent vector at x,y on a circle is perpendicular to line from center to (x,y)

  49. dan815
    • one year ago
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    that is what they want u to show basically

  50. Jhannybean
    • one year ago
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    I see I see

  51. dan815
    • one year ago
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    yeah

  52. carlyleukhardt
    • one year ago
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    Fine dan block me whatever...

  53. carlyleukhardt
    • one year ago
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    ill just find someone else ig.....

  54. carlyleukhardt
    • one year ago
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    okay listen, shes offline now but.... i have no idea why the heck you would block me.....i knew that u really hated me, but who cares im useless anyway...when i saw that u blocked me...sickness went to my stomach immediatly.........Im sorry that i have been annoying recently @dan815 but im just lonley.... but whatever ill just go cry now :( u broke my heart and i just fixed it.... </3

  55. carlyleukhardt
    • one year ago
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    ima go sit in the shower for 5 hours......crying......bye dan (ಥ_ಥ)

  56. carlyleukhardt
    • one year ago
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    oh yeah one more thing......never cheat on me ever again....and never fake to like me please.

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