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# Alevel maths question help: Watch

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2. (Original post by creativebuzz)
Where's the question?
3. (Original post by Mary562)
Where's the question?
There's an image attached
4. (Original post by creativebuzz)
There's an image attached
Oops sorry just seen it now
What have you done so far?
5. (Original post by Mary562)
Oops sorry just seen it now
What have you done so far?
Sorry about the quality, but this is what I've done so far:

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6. (Original post by creativebuzz)
Sorry about the quality, but this is what I've done so far:

this is right - you can cancel the three out
7. (Original post by Mary562)
this is right - you can cancel the three out
But then I would get x = y^1/2 and I wasn't sure if that was right :/
8. (Original post by creativebuzz)
But then I would get x = y^1/2 and I wasn't sure if that was right :/
I think it is but then to find coordinates you have to substitute the value of either x or y back into the original equation
9. (Original post by Mary562)
I think it is but then to find coordinates you have to substitute the value of either x or y back into the original equation
I substituted x = y^1/2 back into the original equation to get

y^3/2 - y^3 - 2y^3/2 = 48

-2y^3/2 - y^3 = 48

y^2(2y^-3/2 + 1) = 48

And then it comes out really badly..

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10. (Original post by creativebuzz)
I substituted x = y^1/2 back into the original equation to get

y^3/2 - y^3 - 2y^3/2 = 48

-2y^3/2 - y^3 = 48

y^2(2y^-3/2 + 1) = 48

And then it comes out really badly..

Have you got there yet? Or need a hand?
11. (Original post by creativebuzz)

x3 + y3 - 3xy = 48

*differentiate both sides using implicit differentiation, remember the product rule for 3xy

3x2 + 3y2(dy/dx) - [3y + 3x(dy/dx)] = 0

*get dy/dx on its own

3x2 + 3y2(dy/dx) - 3y - 3x(dy/dx) = 0
3y2(dy/dx) - 3x(dy/dx) = 3y - 3x2
(dy/dx)(3y2 -3x) = 3y - 3x2
(dy/dx) = [(3y - 3x2)/(3y2 -3x)]
(dy/dx) = [(y - x2)/(y2 -x)]

*make (dy/dx) = 0 for stationary points

0 = [(y - x2)/(y2 -x)]
0 = y - x2
y = x2

*sub y back into the original equation as x2, to find the value for x

x3 + (x2)3 - 3x(x2) = 48
x3 + x6 - 3x3 - 48 = 0
x6 - 2x3 - 48 = 0

Spoiler:
Show

* use quadratic formula except make x3 = (-b +/- sqroot[b2 - 4ac])/2a

a=1 b=-2 c=-48
x3 = (+2 +/- sqroot[(-2)2 - 4x1x-48])/2x1
x3 = (2 +/- 14)/2
x3 = 1 +/- 7
x3 = -6 and x3 = 8
x = -6(1/3) and x = 2

*sub x back into y = x2 to find corresponding y values

when x = -6(1/3), y = 6(2/3) (-6(1/3),6(2/3))
when x = 2, y = 4 (2,4)

*to find the nature you find the second derivitave of original equation using quotient rule

(d2y/dx2) = u'v -v'u/v2
u = (y - x2) v = (y2 - x)
u'= ((dy/dx) -2x) v'= (2y(dy/dx) - 1)

(d2y/dx2) = [((dy/dx) -2x)(y2 - x)-(2y(dy/dx) - 1)(y - x2)]/(y2 - x)2
(d2y/dx2) = [a load of shiz i cant be bothered to type]
(d2y/dx2) = [((dy/dx)(2yx2 - y2 - x) -2xy + x2 + y]/(y2 - x)2

*sub (dy/dx) into equation

(d2y/dx2) = [((y - x2)/(y2 -x))(2yx2 - y2 - x) -2xy + x2 + y]/(y2 - x)2

*sub x and y of the stat pts into(d2y/dx2), if (d2y/dx2) > 0 then its a minimum tp, if (d2y/dx2) < 0 then its a maximum tp

subbing (-6(1/3),6(2/3))
(d2y/dx2) > 0 so (-6(1/3),6(2/3)) is a min tp

subbing (2,4)
(d2y/dx2) < 0 so (2,4) is a max tp

12. (Original post by ElSpencerano)
x3 + y3 - 3xy = 48

*differentiate both sides using implicit differentiation, remember the product rule for 3xy

3x2 + 3y2(dy/dx) - [3y + 3x(dy/dx)] = 0

*get dy/dx on its own

3x2 + 3y2(dy/dx) - 3y - 3x(dy/dx) = 0
3y2(dy/dx) - 3x(dy/dx) = 3y - 3x2
(dy/dx)(3y2 -3x) = 3y - 3x2
(dy/dx) = [(3y - 3x2)/(3y2 -3x)]
(dy/dx) = [(y - x2)/(y2 -x)]

*make (dy/dx) = 0 for stationary points

0 = [(y - x2)/(y2 -x)]
0 = y - x2
y = x2

*sub y back into the original equation as x2, to find the value for x

x3 + (x2)3 - 3x(x2) = 48
x3 + x6 - 3x3 - 48 = 0
x6 - 2x3 - 48 = 0

Spoiler:
Show

* use quadratic formula except make x3 = (-b +/- sqroot[b2 - 4ac])/2a

a=1 b=-2 c=-48
x3 = (+2 +/- sqroot[(-2)2 - 4x1x-48])/2x1
x3 = (2 +/- 14)/2
x3 = 1 +/- 7
x3 = -6 and x3 = 8
x = -6(1/3) and x = 2

*sub x back into y = x2 to find corresponding y values

when x = -6(1/3), y = 6(2/3) (-6(1/3),6(2/3))
when x = 2, y = 4 (2,4)

*to find the nature you find the second derivitave of original equation using quotient rule

(d2y/dx2) = u'v -v'u/v2
u = (y - x2) v = (y2 - x)
u'= ((dy/dx) -2x) v'= (2y(dy/dx) - 1)

(d2y/dx2) = [((dy/dx) -2x)(y2 - x)-(2y(dy/dx) - 1)(y - x2)]/(y2 - x)2
(d2y/dx2) = [a load of shiz i cant be bothered to type]
(d2y/dx2) = [((dy/dx)(2yx2 - y2 - x) -2xy + x2 + y]/(y2 - x)2

*sub (dy/dx) into equation

(d2y/dx2) = [((y - x2)/(y2 -x))(2yx2 - y2 - x) -2xy + x2 + y]/(y2 - x)2

*sub x and y of the stat pts into(d2y/dx2), if (d2y/dx2) > 0 then its a minimum tp, if (d2y/dx2) < 0 then its a maximum tp

subbing (-6(1/3),6(2/3))
(d2y/dx2) > 0 so (-6(1/3),6(2/3)) is a min tp

subbing (2,4)
(d2y/dx2) < 0 so (2,4) is a max tp

Awesome, thank you! By the way, when it came to substituting in your values into the second derivative did you just spend ages substituting the values into your calculator or did you use some sort of shortcut/trick?
13. (Original post by creativebuzz)
Awesome, thank you! By the way, when it came to substituting in your values into the second derivative did you just spend ages substituting the values into your calculator or did you use some sort of shortcut/trick?
Just subbed the values into my calculator because I could not be bothered simplifying (d2y/dx2​).

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