Question

Calculus Question!! Medal!!!

Answer 1

Looks like we can solve this by separation of variables. Have you tried that?

Answer 2

no i have not

Answer 3

Do you need help with that method?

Answer 4

yes please :)

Answer 5

We have \(\dfrac{dy}{dx} = -\dfrac{x}{y}\). What do you get when you separate the variables?

Answer 6

(We want all the \(y\) terms on one side and all the \(x\) terms on the other.)

Answer 7

uh ∫-x ∫y ??

Answer 8

Did you mean to say \(\displaystyle\int -x \, dx = \int y\, dy\)?

Answer 9

yes that what I meant sorry

Answer 10

Okay. What do you get when you integrate?

Answer 11

well the first one is -x^2/2

Answer 12

Yes

Answer 13

the second is y^2/2

Answer 14

Yes. So what's the new equation?

Answer 15

-x^2/2 = y^2/2 ??

Answer 16

Don't forget the constant of integration.

Answer 17

you mean + C

Answer 18

Yes.

Answer 19

-x^2/2+C = y^2/2 +C

Answer 20

You only need the constant on one side. The sum or difference of any number of arbitrary constants is just another constant.

Answer 21

-x^2/2+C = y^2/2

Answer 22

Also, you might as well remove the 2s in the denominators :).

Answer 23

-x^2+C = y^2

Answer 24

How can we determine the value of the constant?

Answer 25

we solve for C

Answer 26

Yes. How do we do that?

Answer 27

subtracting?

Answer 28

Not sure what you mean.

Answer 29

subtracting C from both side? (i don't know)

Answer 30

Well the problem is we have \(x\) and \(y\). Can we make them be specific numbers?

Answer 31

yes?

Answer 32

Yes. Use the initial condition given.

Answer 33

so y(2) = 2? Where do we use it

Answer 34

as C?

Answer 35

We plug it into our equation. What does \(y(2)=2\) mean?

Answer 36

i'm not sure :(

Answer 37

It means that \(y=2\) when \(x=2\).

Answer 38

What do you get when you plug that in?

Answer 39

-2^2+C = 2^2
4+C = 4

Answer 40

No. Careful. What's \(-2^2\)?

Answer 41

opps sorry -4

Answer 42

So what is \(C\)?

Answer 43

C= 8

Answer 44

Right. What is the equation of our solution then, in terms of \(x\) and \(y\)?

Answer 45

-4+8 = 4?

Answer 46

sorry I suck at math

Answer 47

We found \(C\), so you can plug it in into the earlier equation that had \(x\) and \(y\).

Answer 48

i sorry which equation from earlier?

Answer 49

\(-x^2+C=y^2\).

Answer 50

-2^2+8 = 2^2

Answer 51

Don't plug in \(x=y=2\); that's a specific case. Just plug in \(C\) since we found it.

Answer 52

you mean just -x^2+ 8 = y^2 ?

Answer 53

Yes.

Answer 54

What is the graph of this equation?

Answer 55

a circle

Answer 56

Good.

Answer 57

Where is the center of the circle?

Answer 58

(0,0)

Answer 59

Yeah. And what is the circle's radius?

Answer 60

hmm like -2.828

Answer 61

Negative radius???????

Answer 62

sorry didn't mean to put the negative sign

Answer 63

What is the radius in exact form?

Answer 64

2.828

Answer 65

No. That's an approximation.

Answer 66

2.8284

Answer 67

I suspect you're using graphing software. Just do the algebra. What's the equation of a circle with radius \(r\) centered at \((0,0)\)?

Answer 68

the equation?

Answer 69

It's \(x^2+y^2 = r^2\).

Answer 70

ohh thatt yeah

Answer 71

So what is \(r\) in our case?

Answer 72

8^2

Answer 73

No.

Answer 74

but isn't x an y equal to 2

Answer 75

\(x\) and \(y\) are variables. They can take on any value as long as they satisfy the equation we obtained.

Answer 76

Our equation is \(x^2 + y^2 = 8\).

Answer 77

This means that \(r^2 = 8\).

Answer 78

What is \(r\)?

Answer 79

r= -2sqrt 2

Answer 80

sorry again not negative sign

Answer 81

Do you see what the domain is now?

Answer 82

[-2sqrt 2, 2sqrt 2]

Answer 83

Yes.

Answer 84

Thank you for all the help!

Answer 85

You're welcome. From our discussion, I think you have a lot of gaps in your algebra. I strongly recommend reviewing that before tackling differential equations. :)