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## Homework Statement

A curce has equation:

x² + 2xy - 3y² + 16 = 0

find the co-ordinates of the points on the curve where dy/dx = 0

## Homework Equations

## The Attempt at a Solution

I dont have a clue. Do I convert them into parametric equations?

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- Thread starter thomas49th
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- #1

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A curce has equation:

x² + 2xy - 3y² + 16 = 0

find the co-ordinates of the points on the curve where dy/dx = 0

I dont have a clue. Do I convert them into parametric equations?

- #2

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Differentiate Implicitly.

You should get [itex]\mathrm{d}y/ \mathrm{d}x[/itex] within the differentiated equation twice. Factorise [itex]\mathrm{d}y/ \mathrm{d}x[/itex] (by taking it to one side) and rearrange it so that it is only on one side. Then solve for co-ordinates.

You should get [itex]\mathrm{d}y/ \mathrm{d}x[/itex] within the differentiated equation twice. Factorise [itex]\mathrm{d}y/ \mathrm{d}x[/itex] (by taking it to one side) and rearrange it so that it is only on one side. Then solve for co-ordinates.

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- #3

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i tried looking at implicit differentiation here http://www.maths.abdn.ac.uk/~igc/tch/ma1002/diff/node49.html

and my book but i dont understand it.

dy/dx (x² + 2xy - 3y² + 16)

= 2x + 2y - 3y² + 16

Apparently i should use the chain rule somewhere?

Thanks

and my book but i dont understand it.

dy/dx (x² + 2xy - 3y² + 16)

= 2x + 2y - 3y² + 16

Apparently i should use the chain rule somewhere?

Thanks

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- #4

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i tried looking at implicit differentiation here http://www.maths.abdn.ac.uk/~igc/tch/ma1002/diff/node49.html

and my book but i dont understand it.

dy/dx (x² + 2xy - 3y² + 16)

= 2x + 2y - 3y² + 16

Apparently i should use the chain rule somewhere?

Thanks

We should write that first line as

d/dx (x² + 2xy - 3y² + 16) = d/dx (0)

i.e., it's "d/dx" and not "dy/dx" in front, since we want to take the derivative with respect to x.

Then use the chain rule, and when needed the product rule. For example:

[tex]

\frac{d}{dx}(x^2 y^4)

= 2x \cdot y^4 + x^2 \cdot 4 y^3 \frac{dy}{dx}

[/tex]

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- #5

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d/dx (x² + 2xy - 3y² + 16)

i get

2x + 2y - 3y²

how am i ment to use the chain rule? I dont see.

dy/dx = dy/du x du/dx is the chain rule. What is u and y

Thanks :)

- #6

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d/dx (y^4)

= d/dx (u)

= du/dx

and by the chain rule

= (du/dy) * (dy/dx)

= 4 y^3 * (dy/dx)

Moreover:

d/dx (2xy) is not 2y. You need the product rule here, applied to the functions "x" and "y".

- #7

tiny-tim

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Simple question:

dy/dx = dy/du x du/dx is the chain rule. What is u and y

Simple answer:

y is 3y², and u is y.

d(3y²)/dx = d(3y²)/dy dy/dx = 6y dy/dx.

- #8

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Thanks :)

- #9

tiny-tim

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Thanks :)

Hi thomas49th!

Your original problem was implicit differentiation of x² + 2xy - 3y² + 16:

that is, d/dx of x² + 2xy - 3y² + 16.

Obviously , d/dx of x² + 16 is really easy, and d/dx of 2xy is fairly easy … I assume you can do that …

So that leaves d/dx of 3y².

You know how to do d/dy of 3y².

So you use the chain rule: d(3y²)/dx = d(3y²)/dy dy/dx = 6y dy/dx.

Does that make sense?

- #10

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= 2(x+y)

= 2x + 2y ???

i don't think it is. I've used the product rule (dy/dx = v du/dx + u dv/dx).

I know about the chain rule, quotient and product rule. I just dont know how to differeriate parametric equations and equations with both x and y in (i can do y = ... but not y² + yx + x²).

d/dx of x² + 16 is really easy, it's 2x. easy peasy. This implicit differentiation is really confusing me. I cant find it in my book!!

Thanks :)

- #11

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if you have y^2 then different as if it was x , you should have 2y. simple right?

Now all you have to do is multiply the 2y by dy/dx. Just write 2y . dy/dx and that's the answer.

So whenever you different with the letter y, do it as if it was x but just multiply it by dy/dx. This is because of the chain rule, since y is a function of x.

- #12

tiny-tim

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is d/dx of 2xy

= 2(x+y)

= 2x + 2y ???

i don't think it is. I've used the product rule (dy/dx = v du/dx + u dv/dx).

d/dx of 2xy

= d(2xy)/dx = yd(2x)/dx + 2x dy/dx = 2y + 2x dy/dx.

- #13

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2x +

NOTE is it 2y + 2x . (dy/dx) or 2y . (dy/dx) + 2x

So now i've differntiated it i need to find these co-ordinates

Well ill set 2x +2y + 2x . (dy/dx -6y . dy/dx = 0

2x + (2x . dy/dx) = (6y . dy/dx) - 2y

But now im stuck with how to treat these dy/dx...

Thanks :)

- #14

tiny-tim

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NOTE is it 2y + 2x . (dy/dx) or 2y . (dy/dx) + 2x

You're right … it's definitely the first one.

(2y . (dy/dx) would be the d/dx of y²)

Then, as you said in your first post:

find the co-ordinates of the points on the curve where dy/dx = 0

So just put dy/dx = 0 in 2x + 2y + 2x . (dy/dx) - 6y.dy/dx

that gives you … ?

(going to bed now … :zzz:)

- #15

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2x + 2y = 0

y = x

please say that is right...

Thanks :)

Night Night.

- #16

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2x + 2y = 0

y = x

please say that is right...

Thanks :)

Night Night.

Not quite but you are really close.

2x + 2y = 0

Divide both sides of equation by 2:

x + y = 0

And then

x = ???

- #17

HallsofIvy

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No. It is NOT right.

2x+ 2y= 0 is right. But you solved that equation wrong!

2x+ 2y= 0 is right. But you solved that equation wrong!

- #18

tiny-tim

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2x + 2y = 0

y = x

please say that is right...

Thanks :)

Night Night.

… yawn … stretch … rubs eyes …

It's right!!

well … apart from the minus sign!!

Now put y = -x into the original curve equation x² + 2xy - 3y² + 16 = 0, to find the points where dy/dx = 0, as required.

- #19

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silly mistake about the minus

right anyhow

x² + 2xy - 3y² + 16 = 0

= x² - 2(x)² - 3x² = -16

-4x² = -16

x = +-2

SOOOOOOOO ill sub x = - y

y² - 2y² - 3y² = 0

well that's easy y = +-2

so (2,-2),(-2,2)....how do i not know it's (-2,-2) (2,2)?

- #20

tiny-tim

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x = +-2

SOOOOOOOO ill sub x = - y

y² - 2y² - 3y² = 0

erm … you're missing the obvious … you have x = ±2, and x = -y.

So you don't need to put y back into the equation, do you?

- #21

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yeh just sub x = 2 go get one y co-ord then x = -2

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