A community for students.
Here's the question you clicked on:
 0 viewing
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.
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.

This Question is Closed

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0trying to help out a friend in understanding how to solve this but Im not sure how to explain this proof...

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0So differentiating the first one, I get \[2x+2yy'=0\]\[y'=\frac{x}{y}\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0What do I do with the second equation? treat \(a\) and \(b\) as constants?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0\[d(ax+by=0) \implies a+by' =0\] I think.........

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0woah i have no idea what this is xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0me neither, but one is a circle and the other is a linedw:1444184657069:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0one is a circle through the origin, one is a line through the origin

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i must be tired "center at the origin"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444184757878:dw

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Well that would mean that..... \[ax+by=0 \implies by = ax \implies y=\frac{a}{b}x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0maybe this would help it does seem kind of obvious though http://www.sosmath.com/diffeq/first/orthogonal/orthogonal.html

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0So now I've got \[(1) : y' = \frac{x}{y} \qquad (2) : y=\frac{a}{b}x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0'cept you did say calc 1 right? so maybe it i s something else

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i should shut up because i really don't know what you are supposed to do to show this

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Haha yeah I'm trying to figure it out. http://www.slader.com/textbook/9780538497817stewartcalculus7thedition/163/exercises/49/ Im not really understanding the solution here either.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0try this http://www.emathzone.com/tutorials/geometry/normallineofcirclepassesthroughorigin.html

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh i can understand what you sent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0want to walk through it?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Well everything minus the integration part since integration hasn't been covered. That's where I'm having the problem..

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0She's just learned basic derivatives haha

dan815
 one year ago
Best ResponseYou've already chosen the best response.1v=<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

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0How did get the `v=<y,x>` and not `v=<x,y>`?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0oh..... nvm, negative slope.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Can I just take ax + by = 0 in standard form, and state that the slope for that line is positive? lol....

dan815
 one year ago
Best ResponseYou've already chosen the best response.1v=<y,x> is fine too if u have clockwise direction tangents v=<y,x> is ccw tangents directions

dan815
 one year ago
Best ResponseYou've already chosen the best response.1it doesnt have to positive

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0looks 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
 one year ago
Best ResponseYou've already chosen the best response.0i guess more presicely you get \[m=\frac{y_0}{x_0}\] for the line since it is a number

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0how did you get y' = y/x for the line? O_o

dan815
 one year ago
Best ResponseYou've already chosen the best response.1thats the slope of a line y/x

dan815
 one year ago
Best ResponseYou've already chosen the best response.1for any line going through the center

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0oh 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
 one year ago
Best ResponseYou've already chosen the best response.0Ok 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1dont think about it too much you know this property of circle

dan815
 one year ago
Best ResponseYou've already chosen the best response.1thats all they want u to show

dan815
 one year ago
Best ResponseYou've already chosen the best response.1The Tangent vector at x,y on a circle is perpendicular to line from center to (x,y)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1that is what they want u to show basically

carlyleukhardt
 one year ago
Best ResponseYou've already chosen the best response.0Fine dan block me whatever...

carlyleukhardt
 one year ago
Best ResponseYou've already chosen the best response.0ill just find someone else ig.....

carlyleukhardt
 one year ago
Best ResponseYou've already chosen the best response.0okay 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

carlyleukhardt
 one year ago
Best ResponseYou've already chosen the best response.0ima go sit in the shower for 5 hours......crying......bye dan (ಥ_ಥ)

carlyleukhardt
 one year ago
Best ResponseYou've already chosen the best response.0oh yeah one more thing......never cheat on me ever again....and never fake to like me please.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.