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# Differentiation - past paper question watch

1. ok the question is basically:

Find the two set of co-ordinates of the turning point on the curve:

y^3 + 3xy^2 - x^3 = 3

I know that inorder to find turning points, I must get dy/dx, and then substitute the equation dy/dx to equal 0

Differentiating the above equation, led me to get:

dy/dx = (3x^2 - 3y^2)/(3y^2 + 6xy)

......Im not sure where to go from here, coz if I change that differentiated equation into 0, I'll have a bit of controversy between x and ys.......
2. y³+3xy²-x³ = 3

Differentiates to

3y²(y')+3(2xy(y')+y²)-3x² = 0
3y²(y')+6xy(y')+3y² = 3x²
y'(3y²+6xy) = 3x²-3y²
y' = (3x²-3y²)/(3y²+6xy)

Equate to 0:

0 = (3x²-3y²)/(3y²+6xy)

The denominator must equal 0. Hence,

0 = 3x²-3y²
3x² = 3y²
x² = y²

Tricky. I'll plug this in to my graphical calculator and see what happens.
3. (Original post by ~Revol~)
ok the question is basically:

Find the two set of co-ordinates of the turning point on the curve:

y^3 + 3xy^2 - x^3 = 3

I know that inorder to find turning points, I must get dy/dx, and then substitute the equation dy/dx to equal 0

Differentiating the above equation, led me to get:

dy/dx = (3x^2 - 3y^2)/(3y^2 + 6xy)

......Im not sure where to go from here, coz if I change that differentiated equation into 0, I'll have a bit of controversy between x and ys.......
Whoops
4. Fudge, I can't figure out how to make y the subject, I've got both y² and y³ to deal with.
5. At a turning point, the top of the dy/dx fraction has to equal 0. So x = y or x = -y.

Substitute x = y into y^3 + 3xy^2 - x^3 = 3 and solve, then do the same for x = -y. Scroll down if you get stuck.

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Case 1: y = x.
y^3 + 3xy^2 - x^3 = 3
=> x^3 + 3x^3 - x^3 = 3
=> 3x^3 = 3
=> x^3 = 1
=> x = 1
=> y = 1.

Case 2: y = -x.
y^3 + 3xy^2 - x^3 = 3
=> -x^3 + 3x^3 - x^3 = 3
=> x^3 = 3
=> x = 3^(1/3)
=> y = -3^(1/3).
6. (Original post by Jonny W)
At a turning point, the top of the dy/dx fraction has to equal 0. So x = y or x = -y.
Ok, I see how dy/dx=0 when x=y, but i dont understand how it can also be x=-y???
7. (Original post by ~Revol~)
Ok, I see how dy/dx=0 when x=y, but i dont understand how it can also be x=-y???
Because x^2 = y^2 = (-y)^2
8. (Original post by ~Revol~)
Ok, I see how dy/dx=0 when x=y, but i dont understand how it can also be x=-y???
x² = y²
=>
+/-x = +/-y (taking square roots of both sides)
So x = y or x = -y.
9. I attach a graph.
Attached Images

10. (Original post by Jonny W)
I attach a graph.
Doesn't that show three turning points, whereas your answer only has two?
11. (Original post by ZJuwelH)
Doesn't that show three turning points, whereas your answer only has two?
No, the thing on the left isn't a turning point.

Ben
12. (Original post by Ben.S.)
No, the thing on the left isn't a turning point.

Ben
Oh yeah, just thought it through.
13. (Original post by ZJuwelH)
Doesn't that show three turning points, whereas your answer only has two?
A turning point is where the tangent is parallel to the x-axis (so that dy/dx = 0), not where the tangent is parallel to the y-axes (so that dy/dx is undefined).
14. I'm offering rep for the first person who finds, numerically, the equation of each of the graph's three asymptotic lines.
15. Asymptotes at x=-1.5 x = 1 y = 1

Is that right?
16. (Original post by Jonny W)
I'm offering rep for the first person who finds, numerically, the equation of each of the graph's three asymptotic lines.
Roughly ...

y = -2.89x
y = -0.65x
y = 0.53x
17. (Original post by shiny)
Roughly ...

y = -2.89x
y = -0.65x
y = 0.53x
Correct!

Dividing y^3 + 3xy^2 - x^3 = 3 by x^3 gives

(y/x)^3 + 3(y/x)^2 - 1 = 3/x^3,

which is approximately

(y/x)^3 + 3(y/x)^2 - 1 = 0

when |x| is large. Numerically solving the cubic k^3 + 3k^2 - 1 = 0 gives k = 0.532089, -0.652704 and -2.87939.
Attached Images

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