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Alevel maths question help:

About stationary points etc..

intergration.png5.2KB
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Avatar for Mary562
Mary562
2
Original post by creativebuzz
About stationary points etc..


Where's the question?
Original post by Mary562
Where's the question?


There's an image attached :smile:
Reply 3
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Mary562
2
Original post by creativebuzz
There's an image attached :smile:


Oops sorry just seen it now
What have you done so far?
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:

WIN_20150121_223641.JPG49.7KB
Reply 5
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Mary562
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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
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 :/
Reply 7
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Mary562
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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
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..

My answer should be
intergration.png2.5KB
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..

My answer should be

Have you got there yet? Or need a hand? :biggrin:
Original post by creativebuzz
About stationary points etc..


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


:smile:
(edited 9 years ago)
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


:smile:


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?
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​). :smile:

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