## 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.

1. Jhannybean

trying to help out a friend in understanding how to solve this but Im not sure how to explain this proof...

2. Jhannybean

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

So differentiating the first one, I get $2x+2yy'=0$$y'=-\frac{x}{y}$

4. Jhannybean

What do I do with the second equation? treat $$a$$ and $$b$$ as constants?

5. Jhannybean

$d(ax+by=0) \implies a+by' =0$ I think.........

6. anonymous

woah i have no idea what this is xD

7. anonymous

me neither, but one is a circle and the other is a line|dw:1444184657069:dw|

8. Jhannybean

Hmm...

9. anonymous

oh oops

10. anonymous

one is a circle through the origin, one is a line through the origin

11. anonymous

i must be tired "center at the origin"

12. anonymous

|dw:1444184757878:dw|

13. Jhannybean

Well that would mean that..... $ax+by=0 \implies by = -ax \implies y=-\frac{a}{b}x$

14. anonymous

maybe this would help it does seem kind of obvious though http://www.sosmath.com/diffeq/first/orthogonal/orthogonal.html

15. Jhannybean

So now I've got $(1) : y' = -\frac{x}{y} \qquad (2) : y=-\frac{a}{b}x$

16. anonymous

'cept you did say calc 1 right? so maybe it i s something else

17. anonymous

i should shut up because i really don't know what you are supposed to do to show this

18. Jhannybean

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
20. anonymous

oh i can understand what you sent

21. anonymous

want to walk through it?

22. dan815

perpendicular vectors

23. Jhannybean

Well everything minus the integration part since integration hasn't been covered. That's where I'm having the problem..

24. Jhannybean

She's just learned basic derivatives haha

25. dan815

|dw:1444185390633:dw|

26. dan815

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

How did get the v=<y,-x> and not v=<-x,y>?

28. Jhannybean

oh..... nvm, negative slope.

29. Jhannybean

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

30. dan815

v=<-y,x> is fine too if u have clockwise direction tangents v=<y,-x> is ccw tangents directions

31. dan815

it doesnt have to positive

32. dan815

all slopes are fine

33. dan815

y=-a/b x

34. anonymous

got logged out!

35. anonymous

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

i guess more presicely you get $m=\frac{y_0}{x_0}$ for the line since it is a number

37. Jhannybean

how did you get y' = y/x for the line? O_o

38. dan815

thats the slope of a line y/x

39. dan815

for any line going through the center

40. Jhannybean

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

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

brb

43. dan815

=]

44. dan815

dont think about it too much you know this property of circle

45. dan815

|dw:1444186328326:dw|

46. Jhannybean

Ya

47. dan815

thats all they want u to show

48. dan815

The Tangent vector at x,y on a circle is perpendicular to line from center to (x,y)

49. dan815

that is what they want u to show basically

50. Jhannybean

I see I see

51. dan815

yeah

52. carlyleukhardt

Fine dan block me whatever...

53. carlyleukhardt

ill just find someone else ig.....

54. carlyleukhardt

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

ima go sit in the shower for 5 hours......crying......bye dan (ಥ_ಥ)

56. carlyleukhardt

oh yeah one more thing......never cheat on me ever again....and never fake to like me please.