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

- schrodinger

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- Jhannybean

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

- 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

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Jhannybean

What do I do with the second equation? treat \(a\) and \(b\) as constants?

- Jhannybean

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

- anonymous

woah i have no idea what this is xD

- anonymous

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

- Jhannybean

Hmm...

- anonymous

oh oops

- anonymous

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

- anonymous

i must be tired "center at the origin"

- anonymous

|dw:1444184757878:dw|

- Jhannybean

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

- anonymous

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

- Jhannybean

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

- anonymous

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

- anonymous

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

- 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

try this
http://www.emathzone.com/tutorials/geometry/normal-line-of-circle-passes-through-origin.html

- anonymous

oh i can understand what you sent

- anonymous

want to walk through it?

- dan815

perpendicular vectors

- Jhannybean

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

- Jhannybean

She's just learned basic derivatives haha

- dan815

|dw:1444185390633:dw|

- 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

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

- Jhannybean

oh..... nvm, negative slope.

- Jhannybean

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

- dan815

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

- dan815

it doesnt have to positive

- dan815

all slopes are fine

- dan815

y=-a/b x

- anonymous

got logged out!

- 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

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

- Jhannybean

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

- dan815

thats the slope of a line y/x

- dan815

for any line going through the center

- 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

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

brb

- dan815

=]

- dan815

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

- dan815

|dw:1444186328326:dw|

- Jhannybean

Ya

- dan815

thats all they want u to show

- dan815

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

- dan815

that is what they want u to show basically

- Jhannybean

I see I see

- dan815

yeah

- carlyleukhardt

Fine dan block me whatever...

- carlyleukhardt

ill just find someone else ig.....

- 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

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

- carlyleukhardt

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.