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# C4 - Differentiation Questions watch

1. Hey, I have a few questions..

1. (1 + x) (2 + y) = x² + y²

(a) Work ouy dy/dx. I got dy/dx =
2x + y - 2
1 + x - 2y

And then.. (the bits I'm stuck on)
(b) Find the gradient of the curve (1 + x) (2 + y) = x² + y² at each of the 2 points where the curve meets the axis

I did:
Meets y axis at x=0, so dy/dx = y-2/1+x
Meets x axis at y=0, so dy/dx = 2x-2/1+x

But I dunno what to do next

(c) Show also that there are 2 points at which the tangents to this curve are parallel to the y axis.

Parallel to the y axis.. Does that mean dy/dx = 0? I'm not sure to do with it though..

2. A curve has equation 7x² + 48xy - 7y² + 75 = 0.
A and B are 2 distinct points on the curve and at each of the points the gradient = 2/11. Use implicit differentiation to show that x + 2y = 0 at the points A and B.

By implicit differentiation i got dy/dx =
- 14x - 48y
48x - 14y

And i substitued dy/dx = 2/11 into i dont think its right cos i started getting really big numbers... What do I do after differentiating... Or did i differentiate wrong?

2. 2. You've done the differentiation correctly. Just note that the division can be simplified, since each term contains a factor of 2, so you get:

dy/dx = (-7x-24y)/(24x-7y)

Now you'll avoid the 'big numbers' since everything will be a factor of 2 less. But you've done the right thing. Just need to keep plugging away.
3. (Original post by mockel)
2. You've done the differentiation correctly. Just note that the division can be simplified, since each term contains a factor of 2, so you get:

dy/dx = (-7x-24y)/(24x-7y)

Now you'll avoid the 'big numbers' since everything will be a factor of 2 less. But you've done the right thing. Just need to keep plugging away.
Oh yeh..! Wow, that was easy, I thought it was lots of working out!
Thanks
4. 1.
a)
2 + 2x + y +xy = x² + y²
2 + dy/dx + y + x(dy/dx) = 2x + 2y(dy/dx)

dy/dx[1+x-2y] = 2x-y-2

dy/dx = (2x-y-2)/(1+x-2y)

[Slightly different to yours]

b) At x-axis, y=0
=> (1+x)(2) = x²
x²-2x-2 = 0
x = (2 ± √12)/2 = 1 ± √3

At y-axis, x=0
=> 2+y = y²
(y-2)(y+1) = 0
y = 2 or y = -1

Hmm. This suggests that there are 4 points at which the curve meets the axis. Sure it doesn't say just 1 of the axis?
Anyway, now sub in these x and y values into dy/dx (which you've calculated in b) )

c) Parallel to y-axis means dy/dx = ∞, since it's just a vertical line.

dy/dx = (2x-y-2)/(1+x-2y)

dy/dx = ∞, when the denominator = 0

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

Erm...not too sure why that means "2 points at which...."
Meh, someone else will figure it out.
5. Ooh i see..
So does that mean gradients parallel to the x axis = 0 ?
6. (Original post by violet)
Ooh i see..
So does that mean gradients parallel to the x axis = 0 ?
Yes, common sense really, since x-axis is 0 gradient, any lines parallel will also be 0, according to the definition of a gradient!
7. Yep, i totally knew that.. I just wanted to, erm, check if you did!

8. Sorry, i've just downloaded mathtype and wanted something to test it out on

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