Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

woah i have no idea what this is xD

Answer 7

me neither, but one is a circle and the other is a line

Answer 8

Hmm...

Answer 9

oh oops

Answer 10

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

Answer 11

i must be tired "center at the origin"

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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.

Answer 18

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

Answer 19

oh i can understand what you sent

Answer 20

want to walk through it?

Answer 21

perpendicular vectors

Answer 22

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

Answer 23

She's just learned basic derivatives haha

Answer 24

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

Answer 25

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

Answer 26

oh..... nvm, negative slope.

Answer 27

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

Answer 28

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

Answer 29

it doesnt have to positive

Answer 30

all slopes are fine

Answer 31

y=-a/b x

Answer 32

got logged out!

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

thats the slope of a line y/x

Answer 37

for any line going through the center

Answer 38

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

Answer 39

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

Answer 40

brb

Answer 41

=]

Answer 42

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