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Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Stationary points...Help!

RamocitoMorales

Basically, I need to find the stationary points of the equation

x^2+y^2+xy=3

This is how far I've gotten,

x^2+y^2+xy=3

Differentiating with respect to

x

2x+2y.\dfrac{dy}{dx}+x.\dfrac{dy}{dx}+y=0

Then, bringing the

\dfrac{dy}{dx}

together, I get,

\dfrac{dy}{dx}(2y+x)=-2x-y

\dfrac{dy}{dx}=\dfrac{-2x-y}{2y+x}

At stationary points,

\dfrac{dy}{dx}=0

, so,

\dfrac{-2x-y}{2y+x}=0

-2x-y=0

y=-2x

I don't see where I can go from here and how I can find the stationary points. I need help please?

Thanks.

RamocitoMorales

x^2+y^2+xy=3

y=-2x

You have two equations there, what can you do with them? :smile:

:smile:

Where (on that curve) is y equal to -2x?

RamocitoMorales

OP

LetoKynes

You have two equations there, what can you do with them? :smile:

:smile:

Simultaneously solve it?

RamocitoMorales

OP

LetoKynes

You have two equations there, what can you do with them? :smile:

:smile:

It still doesn't work, this is what happens.

x^2+y^2+xy=3

y=-2x

Substituting in the y value into the first equation, I get,

x^2+(-2x)^2+x(-2x)=3

x^2+4x^2-4x^2=3

x^2=3

x=\sqrt{3}

That's not the right answer, the right answer should be (1,-2), (-1, 2).

Since when has x(-2x) equalled -4x^2?

RamocitoMorales

It still doesn't work, this is what happens.

x^2+y^2+xy=3

y=-2x

Substituting in the y value into the first equation, I get,

x^2+(-2x)^2+x(-2x)=3

x^2+4x^2-4x^2=3

Mistake on this line

x^2=3

x=\sqrt{3}

That's not the right answer, the right answer should be (1,-2), (-1, 2).

Do you get the right answer if you correct that?

RamocitoMorales

OP

LetoKynes

Do you get the right answer if you correct that?

I can be such a plum sometimes, sorry about that.

So, continuing form the mistake, I get,

x^2+4x^2-2x^2=3

3x^2=3

x^2=\dfrac{3}{3}

x^2=1

x=\sqrt{1}

x=1

or

x=-1

Thank you, that is the correct answer for

x

. Now I will sub it back into the original equation to find the value for

y

.

x^2+y^2+xy=3

x=1

in this situation, so,

1+y^2+y=3

y^2+y=3-1

y^2+y=2

So now what do I do?

Thanks.

Solve the quadratic equation...?

Yeap, just solve the quadratic equation.

RamocitoMorales

OP

generalebriety

Solve the quadratic equation...?

So basically,

y^2+y=2

Which can be simplified to,

y^2+y-2=0

Which can be factorised as,

(y+2)(y-1)=0

So

y=1

or

y=-2

But it's still wrong, since on the answers it says that y should equal to -2 and 2, not -2 and 1.

hmmmm

I would give you reputation, but it's worthless.

You can use the x = -1 and you should get the other y value :smile:

:smile:

Although I'm not sure how you're supposed to know which value to use.. :p:

:p:

RamocitoMorales

OP

LetoKynes

You can use the x = -1 and you should get the other y value :smile:

:smile:

Although I'm not sure how you're supposed to know which value to use.. :p:

:p:

Ah of course, I'd forgotten all about the

x=-1

Now when

x=-1

in the equation

x^2+y^2+xy=3

,

(-1)^2+y^2+(-1)y=3

1+y^2-y=3

y^2-y+1-3=0

y^2-y-2=0

(y-2)(y+1)=0

So,

y=2

, or

y=-1

This is a very confusing question.

RamocitoMorales

Ah of course, I'd forgotten all about the

x=-1

Now when

x=-1

in the equation

x^2+y^2+xy=3

,

(-1)^2+y^2+(-1)y=3

1+y^2-y=3

y^2-y+1-3=0

y^2-y-2=0

(y-2)(y+1)=0

So,

y=2

, or

y=-1

This is a very confusing question.

It's not that confusing. If x = 1 then y = 1 or -2; if x = -1 then y = -1 or 2. Which values are valid (remembering that y = -2x)? They will then all satisfy the original equation, so you're done.

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