Question

calculus - two variables
pic

Answer 1

If it only depends on \(r\) then the level sets have got to be circles.

Answer 2

don't circles depend on Theta though?

Answer 3

This is because if \(f(r) = k\) then \(f^{-1}(k)=r = \sqrt{x^2+y^2}\)

Answer 4

Remember that \(r\) is the distance from the origin.

Answer 5

If we solve for \(r\) in terms of \(k\), we're realizing the distance from the origin is constant.

Answer 6

Since \(f^{-1}(k)\) is constant.

Answer 7

So if the distance from the origin is constant... it is equidistant: circles!

Answer 8

you lost me with f^{-1}(k)

Answer 9

\[f^{-1}(k)\]

Answer 10

do we begin with \[f(x,y)=\]

Answer 11

Well a level curve at \(k\) is basically saying \(f(r,\theta)=k\). Do you follow with that?

Answer 12

\[f(x,y)=x^2+y^2\]

Answer 13

Okay sure...

Answer 14

for polar co-ordinates? yea i follow

Answer 15

\[
f(x,y) = g(r)=g(x^2+y^2)
\]

Answer 16

So we'll say \(g(r)=k\) to be notation consistent.

Answer 17

yep! i understnad now

Answer 18

so \[x^2+y^2=(\sqrt(k))^2\]

Answer 19

We can find roots of \(g(r)-k\). Clearly \(r\) is going to be equal to a couple constants.

Answer 20

No @chris00 , you cant assume \(g(r)=r\). It could be \(g(r) = 1/\tan (r)\) for all we know.

Answer 21

The point is regardless, our level curves are equations of the form \(g(r)=k\). Now since \(k\) is constant, then \(r\) is constant... thus circles.

Answer 22

ohhhh i see where i went wrong. mi bad. silly moment

Answer 23

is that my description for a)?

Answer 24

Yep

Answer 25

so it would still be level circles if \[g(r) = 1/\tan (r)\]

Answer 26

?

Answer 27

ah, so we are just generalising here

Answer 28

Yeah.

Answer 29

We have to be general because we don't have any other info.

Answer 30

yea that makes perfect sense

Answer 31

now with b, its obvious to see the domain cannot include 0 but i don't know how to explain this

Answer 32

\[
\sin(xy)\neq 0\\
xy\neq n\pi \quad n\in \mathbb Z
\]

Answer 33

-_- surely thats not all thats required though?

Answer 34

so \[x,y \in R \ n\]

Answer 35

oops

Answer 36

\[x,y \in R excepts cannot contain n*Pi

Answer 37

omg.
\[x,y \in R \] except xy is not equal to n*Pi ?

Answer 38

\[
\forall x,y\in \mathbb R\;\forall n\in \mathbb Z\quad xy\neq n\pi
\]

Answer 39

much better notation. haha

Answer 40

your a boss at math wio.

Answer 41

Thank you very much.

Answer 42

Do you know how to write a testimony?

Answer 43

i'll get to it!

Answer 44

Thank you. I am very close to 99.

Answer 45

i am glad i can help you get there!

Answer 46

one thing wio, i'm still not grasping why u say \[f^{-1}(k)=r = \sqrt{x^2+y^2}\]

Answer 47

why do we deal with inverses here?

Answer 48

@wio

Answer 49

Nevermind that. I was just trying to show how \(r\) was constant.

Answer 50

is it relevant to the question?

Answer 51

Not really. there are better ways of saying it.

Answer 52

how would you say it? need to understnad it a bit better. sorry for the inconvienience

Answer 53

I suppose you could say :\[
g(r)=k
\]Since \(k\) is constant, then \(r\) must be constant

Answer 54

is it appropriate to begin with \[f(x,y) = g(r)=g(x^2+y^2)\] ?

because we can have many different types of circles, thats just the easiest form?

Answer 55

Actually \(r=\sqrt{x^2+y^2}\) I did a typo back there.

Answer 56

\[f(x,y) = g(r)=g(\sqrt(x^2+y^2))\]

Answer 57

By the way \(\sqrt{x^2+y^2}=|(x,y),(0,0)|\). Literally the distance from the origin.

Answer 58

Trust me, they don't expect you to be that rigorous. It's a question not a proof.

Answer 59

so you would just say the level curves look like circles because r is constant?

Answer 60

and not dependent on theta?

Answer 61

Yeah.

Answer 62

and r is constant because g(r)=k where k is constnat?