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Difference between Turning point and Stationary point

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Original post by Crystalclearmagic
They're used rather interchangeably in maths even at A level - no?

If they're two different things, then Edexcel has misled me for 2 years then... :/


This is frequently mis-taught at A level. Another wrong idea is that 'if d2y/dx^2 = 0 then you have a point of inflection'. It's true you may have, but on the simple case of y = x^4 the second derivative is zero at the stationary point at the origin. But it's clearly a minimum.
Reply 21
Original post by ian.slater
This is frequently mis-taught at A level. Another wrong idea is that 'if d2y/dx^2 = 0 then you have a point of inflection'. It's true you may have, but on the simple case of y = x^4 the second derivative is zero at the stationary point at the origin. But it's clearly a minimum.


If d2y/dx2=0 and d3y/dx3 not= 0 then its a point of inflection, right?
Original post by TenOfThem
Sorry


My question related to the fact that y=x^3-x has no such point at (0,0)


Unless I've made some silly mistake, it does. All cubics have rotational symmetry about their point of inflection, which in this case is at (0,0). This curve switches from curving downwards for x<0 to curving upwards for x>0. The circle of curvature swaps sides at that point. But it's not a stationary point and is therefore a good example for me to use.
Original post by raheem94
If d2y/dx2=0 and d3y/dx3 not= 0 then its a point of inflection, right?


I think so ... that way the gradient has either a local maximum or minimum, which causes the curvature to 'flip' whether or not it's a stationary point.
Reply 24
Original post by ian.slater
I think so ... that way the gradient has either a local maximum or minimum, which causes the curvature to 'flip' whether or not it's a stationary point.


Thanks


THat just proves tenofthem right

Points of inflections are stationary points.
(edited 11 years ago)
Reply 26
Original post by TheGrinningSkull
THat just proves tenofthem right

Points of inflections are stationary points.


It is written Inflection points may be stationary points

It is written 'may'?

http://answers.yahoo.com/question/index?qid=20090507015050AAaHwNp
Original post by raheem94
It is written Inflection points may be stationary points

It is written 'may'?

http://answers.yahoo.com/question/index?qid=20090507015050AAaHwNp


Thanks, I needed an example because it said may yet it offered no examples.

I still can't picture it visually though :s
Original post by TheGrinningSkull
Thanks, I needed an example because it said may yet it offered no examples.

I still can't picture it visually though :s


http://www.wolframalpha.com/input/?i=x^3+-x
Reply 29
Original post by verello12
Is there a Difference between Turning point and Stationary point or are they the same thing.
Im doing C3 OCR MEI btw.
I know how to find turning point of a curve you do dy/dx=0 to get x then substiute x in curve equation to get y

If a question asked you to find stationary point of a curve would you just do the same thing?


Yeah, but if it does equal o, and it could be a point of inflection to check, you put in the value of x in the original equation then x+0.1 and x-0.1 to check if it definitely is.

Have you tried checking the site for help.....it's bit hard trying to complain online
Original post by ian.slater
http://www.wolframalpha.com/input/?i=x^3+-x


Thanks. I think it makes sense.

Because d2y/dx2 is 6x for that, so x is 0 gives a point of inflection but not a stationary point.

That's quite interesting because usually what you get for the 2nd differential is a constant due to terms only being quadratic so it's something new. Thanks. :smile:

EDIT: Usually it's not a constant, sorry, but usually when we do find the point of inflection, it's due to finding the gradient to be 0 in the first place, I get it now :tongue:
(edited 11 years ago)
Original post by ian.slater
This is frequently mis-taught at A level. Another wrong idea is that 'if d2y/dx^2 = 0 then you have a point of inflection'. It's true you may have, but on the simple case of y = x^4 the second derivative is zero at the stationary point at the origin. But it's clearly a minimum.


True actually; at A level they stop at d2y/dx^2 - but because I did additional maths FSMQ, I know the use of d3y/dx^3

Ah, I don't believe even in maths there can be so many "lies" that they teach! thought it was only in science when they have to simplify things and so "lie" to us at GCSE.... haha
Reply 32
Original post by verello12
Is there a Difference between Turning point and Stationary point or are they the same thing.
Im doing C3 OCR MEI btw.
I know how to find turning point of a curve you do dy/dx=0 to get x then substiute x in curve equation to get y

If a question asked you to find stationary point of a curve would you just do the same thing?


All turning points are stationary points but not all stationary points are turning points.

In AS and A2 there are 3 types of stationary points;
Maximum where dy/dx is 0 , think about the top of a negative quadratic ,n shaped, where it has 'zero' gradient, this is a turning point
Minimum where dy/dx is 0, think about the bottom of a positive quadratic, u shaped, where it has 'zero' gradient, this is a turning point
The last type of stationary point is a point of infection, this is the tricky one as dy/dx (gradient/first derivative) may not be equal to zero. In this case, a point of inflection is the point at which the rate of change of the gradient changes sign, which can be stated as d^2y/dx^2 = 0 (second derivative). There are two classic examples of a point of inflection. The easy one is a graph of x^3 at negative x values the gradient is getting more negative, at positive x the gradient is getting more positive so where the change in the gradient goes from negative to positive there is a point of inflexion (in this case at the origin and this is the case at the point of inflection the gradient happens to be zero). The second example is a graph of sin(x) from -180 to 180 (degrees), at the origin the curve's gradient goes from negative to positive, which means the origin is a point of inflection however in this case the gradient (first derivative) at this point is not zero but the rate of change of the gradient (second derivative) is.

A bit long winded but that should cover all you need to know about stationary points.

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