x Turn on thread page Beta
 You are Here: Home >< Maths

# How to find Stationary points when there are 2 varaibles. watch

1. So question is find stationary points of the curve : x^3 + y^3 -3xy = 48 and determine their nature.
I am not sure. I have tried diffentiating x and then y then rearranging/simultaneuos equations but i keep on getting zeros when the the stationary point is meant to be 2,4 for max and -6^1/3, 6^2/3 for min.
2. (Original post by Nirm)
So question is find stationary points of the curve : x^3 + y^3 -3xy = 48 and determine their nature.
I am not sure. I have tried diffentiating x and then y then rearranging/simultaneuos equations but i keep on getting zeros when the the stationary point is meant to be 2,4 for max and -6^1/3, 6^2/3 for min.
You need to differentiate implicitly with respect to x.
differentiates normally
you differentiate with respect to y, then multiply by
you use product rule(-3x and y), when you differentiate y, differentiate it with respect to y then multiply by
becomes 0.

Rearrange the final equation to get on one side, then find the values of x and y for which the differential equals 0.
3. (Original post by morgan8002)
You need to differentiate implicitly with respect to x.
differentiates normally
you differentiate with respect to y, then multiply by
you use product rule(-3x and y), when you differentiate y, differentiate it with respect to y then multiply by
becomes 0.

Rearrange the final equation to get on one side, then find the values of x and y for which the differential equals 0.
Okay so for x^3, i get 3x^2
y^3 is y^3 times by dy/dx
and for -3xy i get -3y+(-3x times by dy/dx)

Just want to make sure if it is right.
4. (Original post by Nirm)
Okay so for x^3, i get 3x^2
y^3 is y^3 times by dy/dx
and for -3xy i get -3y+(-3x times by dy/dx)

Just want to make sure if it is right.
The second one is incorrect.

You should have
5. (Original post by morgan8002)
The second one is incorrect.

You should have
Ahh okay thanks!. Also i was looking at your stats just found it really weird how in c4 you got 100% but in c1 c2 you got worse ( silly mistakes i guess )
6. (Original post by Nirm)
Ahh okay thanks!. Also i was looking at your stats just found it really weird how in c4 you got 100% but in c1 c2 you got worse ( silly mistakes i guess )
C1, C2 and S1 were mostly self taught in year 11. I didn't have so much time and didn't know the most efficient ways of studying etc. yet, so didn't do so well. You might notice that my worst module last year was equal to my best module the year before.
7. (Original post by morgan8002)
C1, C2 and S1 were mostly self taught in year 11. I didn't have so much time and didn't know the most efficient ways of studying etc. yet, so didn't do so well. You might notice that my worst module last year was equal to my best module the year before.
WOW, amazing. You did it in year 11!
8. (Original post by Nirm)
So question is find stationary points of the curve : x^3 + y^3 -3xy = 48 and determine their nature.
I am not sure. I have tried diffentiating x and then y then rearranging/simultaneuos equations but i keep on getting zeros when the the stationary point is meant to be 2,4 for max and -6^1/3, 6^2/3 for min.
If you wanted further insight, this is called Implicit Differentiation: http://en.wikipedia.org/wiki/Implicit_function

The method behind differentiating some function of y w.r.t x is via the chain rule:

We imagine y as a function of x, y(x). In the case of . Let , thus

.

Chain rule: http://en.wikipedia.org/wiki/Chain_rule

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 14, 2015
Today on TSR

### How do I turn down a guy in a club?

What should I do?

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams