Jhannybean
  • Jhannybean
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Jhannybean
  • Jhannybean
trying to help out a friend in understanding how to solve this but Im not sure how to explain this proof...
Jhannybean
  • 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.
Jhannybean
  • Jhannybean
So differentiating the first one, I get \[2x+2yy'=0\]\[y'=-\frac{x}{y}\]

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Jhannybean
  • Jhannybean
What do I do with the second equation? treat \(a\) and \(b\) as constants?
Jhannybean
  • Jhannybean
\[d(ax+by=0) \implies a+by' =0\] I think.........
anonymous
  • anonymous
woah i have no idea what this is xD
anonymous
  • anonymous
me neither, but one is a circle and the other is a line|dw:1444184657069:dw|
Jhannybean
  • Jhannybean
Hmm...
anonymous
  • anonymous
oh oops
anonymous
  • anonymous
one is a circle through the origin, one is a line through the origin
anonymous
  • anonymous
i must be tired "center at the origin"
anonymous
  • anonymous
|dw:1444184757878:dw|
Jhannybean
  • Jhannybean
Well that would mean that..... \[ax+by=0 \implies by = -ax \implies y=-\frac{a}{b}x\]
anonymous
  • anonymous
maybe this would help it does seem kind of obvious though http://www.sosmath.com/diffeq/first/orthogonal/orthogonal.html
Jhannybean
  • Jhannybean
So now I've got \[(1) : y' = -\frac{x}{y} \qquad (2) : y=-\frac{a}{b}x\]
anonymous
  • anonymous
'cept you did say calc 1 right? so maybe it i s something else
anonymous
  • anonymous
i should shut up because i really don't know what you are supposed to do to show this
Jhannybean
  • 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.
anonymous
  • anonymous
try this http://www.emathzone.com/tutorials/geometry/normal-line-of-circle-passes-through-origin.html
anonymous
  • anonymous
oh i can understand what you sent
anonymous
  • anonymous
want to walk through it?
dan815
  • dan815
perpendicular vectors
Jhannybean
  • Jhannybean
Well everything minus the integration part since integration hasn't been covered. That's where I'm having the problem..
Jhannybean
  • Jhannybean
She's just learned basic derivatives haha
dan815
  • dan815
|dw:1444185390633:dw|
dan815
  • dan815
v= ---> 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
Jhannybean
  • Jhannybean
How did get the `v=` and not `v=<-x,y>`?
Jhannybean
  • Jhannybean
oh..... nvm, negative slope.
Jhannybean
  • Jhannybean
Can I just take ax + by = 0 in standard form, and state that the slope for that line is positive? lol....
dan815
  • dan815
v=<-y,x> is fine too if u have clockwise direction tangents v= is ccw tangents directions
dan815
  • dan815
it doesnt have to positive
dan815
  • dan815
all slopes are fine
dan815
  • dan815
y=-a/b x
anonymous
  • anonymous
got logged out!
anonymous
  • 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
anonymous
  • anonymous
i guess more presicely you get \[m=\frac{y_0}{x_0}\] for the line since it is a number
Jhannybean
  • Jhannybean
how did you get y' = y/x for the line? O_o
dan815
  • dan815
thats the slope of a line y/x
dan815
  • dan815
for any line going through the center
Jhannybean
  • 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
Jhannybean
  • 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
Jhannybean
  • Jhannybean
brb
dan815
  • dan815
=]
dan815
  • dan815
dont think about it too much you know this property of circle
dan815
  • dan815
|dw:1444186328326:dw|
Jhannybean
  • Jhannybean
Ya
dan815
  • dan815
thats all they want u to show
dan815
  • dan815
The Tangent vector at x,y on a circle is perpendicular to line from center to (x,y)
dan815
  • dan815
that is what they want u to show basically
Jhannybean
  • Jhannybean
I see I see
dan815
  • dan815
yeah
carlyleukhardt
  • carlyleukhardt
Fine dan block me whatever...
carlyleukhardt
  • carlyleukhardt
ill just find someone else ig.....
carlyleukhardt
  • 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....
carlyleukhardt
  • carlyleukhardt
ima go sit in the shower for 5 hours......crying......bye dan (ಥ_ಥ)
carlyleukhardt
  • carlyleukhardt
oh yeah one more thing......never cheat on me ever again....and never fake to like me please.

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