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Differentiating x² + 2xy - 3y² + 16 = 0

  • Thread starter thomas49th
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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?
 

Answers and Replies

  • #2
Air
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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.
 
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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
 
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  • #4
Redbelly98
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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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so if i do:

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
Redbelly98
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Let's suppose u = y^4, and we want to differentiate that with respect to x.

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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Hi thomas49th! :smile:

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. :biggrin:

oh yes they are …

d(3y²)/dx = d(3y²)/dy dy/dx = 6y dy/dx. :smile:
 
  • #8
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im still lost completely. I know the chain rule. I dont see how y is 3y² nor why u is y. Can you show me why. Why do you set u equal to y?
Thanks :)
 
  • #9
tiny-tim
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im still lost completely. I know the chain rule. I dont see how y is 3y² nor why u is y. Can you show me why. Why do you set u equal to y?
Thanks :)
Hi thomas49th! :smile:

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? :smile:
 
  • #10
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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).

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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thomas the idea is really simple.

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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so the final answer is

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

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 … ? :smile:

(going to bed now … :zzz:)
 
  • #15
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ahhh YOU ACUTLALY SUBST FOR dy/dx = 0
2x + 2y = 0
y = x

please say that is right...

Thanks :)
Night Night.
 
  • #16
Redbelly98
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ahhh YOU ACUTLALY SUBST FOR dy/dx = 0
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!
 
  • #18
tiny-tim
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ahhh YOU ACUTLALY SUBST FOR dy/dx = 0
2x + 2y = 0
y = x

please say that is right...

Thanks :)
Night Night.
… yawn … stretch … rubs eyes …

It's right!! :biggrin:

well … apart from the minus sign!! :rolleyes:

Now put y = -x into the original curve equation x² + 2xy - 3y² + 16 = 0, to find the points where dy/dx = 0, as required. :smile:
 
  • #19
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y = -x

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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Hi thomas49th! :smile:
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? :wink:
 
  • #21
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yeh just sub x = 2 go get one y co-ord then x = -2
 

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