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OCR C3 (not MEI) Official Thread - Tuesday 21st June 2016

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Reply 60
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SI 1
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Original post by AlfieH
It's not the point at which the gradient is the maximum - so you may be getting confused there.

Nor is it the point where the gradient is the lowest.

Instead the gradient is always 0 at the turning points/ min or max points. It is the maximum/ minimum as in the highest/lowest POINT of the curve not gradient.

If that makes sense?


I understand that but the question asked for the point at which the gradient takes its maximum value. I assumed this was asking for the value of x at which the curve was the steepest.
Reply 61
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AlfieH
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Original post by SI 1
I understand that but the question asked for the point at which the gradient takes its maximum value. I assumed this was asking for the value of x at which the curve was the steepest.


No - that means when the curve has a minimum point so to find x you must equate it to 0.
Original post by Yazmin123
can someone help me with this question from jan 20063sin6Bcosec2B=4 where beta lies between zero and 90 degreesany help would be appreciated


image.jpg
Reply 63
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Original post by AlfieH
No - that means when the curve has a minimum point so to find x you must equate it to 0.


It can't be when the curve has a minimum point because the graph with the question showed P was not at the minimum point.
Basically when it asks for the maximum (or minimum) gradient, I make the second derivative equal to 0, solve for x and sub that value of x back into the first derivative. is that right?
Original post by SI 1
I understand that but the question asked for the point at which the gradient takes its maximum value. I assumed this was asking for the value of x at which the curve was the steepest.


If you want to find the point at which the gradient is maximum you'll have to differentiate again and set f''(x)=0, then verify if this point is a maximum or minimum from looking at the graph of dy/dx or finding the third derivative and checking if its positive or negative (yeesh).

Post the question if you are still unsure.
Reply 65
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Original post by 16characterlimit
If you want to find the point at which the gradient is maximum you'll have to differentiate again and set f''(x)=0, then verify if this point is a maximum or minimum from looking at the graph of dy/dx or finding the third derivative and checking if its positive or negative (yeesh).

Post the question if you are still unsure.


its question 8 ii on this paper
Specimen QP - C3 OCR.pdf43.6KB
Original post by SI 1
its question 8 ii on this paper


dy/dx = (2lnx)/x found using the chain rule, u=lnx

d^2y/dx^2 = (2-2lnx)/x^2 found using product/quotient rule, I will refer to it as f''(x) from now cause its less messy

when f''(x) = 0 f'(x) is maximum, which is the gradient, hence 2-2lnx = 0 (times both sides by x^2) and rearrange to get lnx = 1 . x = e therefore.

Subbing back into f'(x) we find the maximum gradient is 2/e

Because we know the gradient, we can now can find out the tangent. We know at point P x=e therefore y=1, so that gives us enough information to write the tangent as y=2x/e - 1

Which does indeed go through (0,-1)
Reply 67
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Original post by 16characterlimit
dy/dx = (2lnx)/x found using the chain rule, u=lnx

d^2y/dx^2 = (2-2lnx)/x^2 found using product/quotient rule, I will refer to it as f''(x) from now cause its less messy

when f''(x) = 0 f'(x) is maximum, which is the gradient, hence 2-2lnx = 0 (times both sides by x^2) and rearrange to get lnx = 1 . x = e therefore.

Subbing back into f'(x) we find the maximum gradient is 2/e

Because we know the gradient, we can now can find out the tangent. We know at point P x=e therefore y=1, so that gives us enough information to write the tangent as y=2x/e - 1

Which does indeed go through (0,-1)


Thank you so much!! One question, is the fact that f''(x)=0 is when f'(x) is maximum a rule or something cos I don't remember ever learning that?
Original post by SI 1
Thank you so much!! One question, is the fact that f''(x)=0 is when f'(x) is maximum a rule or something cos I don't remember ever learning that?


Try imagining f'(x) as a different function g(x), if you wanted to find the maximum of g(x) what would you do?
Reply 69
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Original post by 16characterlimit
Try imagining f'(x) as a different function g(x), if you wanted to find the maximum of g(x) what would you do?


Do you mean the maximum gradient or the maximum point?
Original post by 16characterlimit
Try imagining f'(x) as a different function g(x), if you wanted to find the maximum of g(x) what would you do?


From what I recall the second derivative being 0 at a point isn't informative of whether it's a minimum or maximum point?


Posted from TSR Mobile
Original post by SI 1
Do you mean the maximum gradient or the maximum point?


Point
Original post by drandy76
From what I recall the second derivative being 0 at a point isn't informative of whether it's a minimum or maximum point?


Posted from TSR Mobile


No, you should always check, but in the context of that question there is only one stationary point and it more or less told you you have to find a maximum.

edit: nvm I didnt read your post properly.

I think if its d^2y/dx^2 = 0 its an inflection point and you have to look at surrounding values of the graph, this very very rarely comes up in C1 and even less so in C3 though
(edited 7 years ago)
Reply 73
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Original post by 16characterlimit
Point

wouldn't you just find the stationary points and the put into second derivative to see which one is a maximum
Original post by 16characterlimit
No, you should always check, but in the context of that question there is only one stationary point and it more or less told you you have to find a maximum.

edit: nvm I didnt read your post properly.

I think if its d^2y/dx^2 = 0 its an inflection point and you have to look at surrounding values of the graph, this very very rarely comes up in C1 and even less so in C3 though


Yeah without context it could be a local minima/maxima for all we know, although I've never seen a question really use this idea, granted my sample size is rather small


Posted from TSR Mobile
Original post by SI 1
wouldn't you just find the stationary points and the put into second derivative to see which one is a maximum


Yes, so basically the stationary points of a function are the values of x of the functions integral are maxima or minima.

e.g

f'(x)=0 gives min/max of f(x)

f''(x)=0 gives min/man of f'(x)

f^n(x)=0 gives min/max of f^n-1(x) where n is the order of the derivative
Original post by 16characterlimit
Yes, so basically the stationary points of a function are the values of x of the functions integral are maxima or minima.

e.g

f'(x)=0 gives min/max of f(x)

f''(x)=0 gives min/man of f'(x)

f^n(x)=0 gives min/max of f^n-1(x) where n is the order of the derivative


Don't forget points of inflection


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Reply 77
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Original post by 16characterlimit
Yes, so basically the stationary points of a function are the values of x of the functions integral are maxima or minima.

e.g

f'(x)=0 gives min/max of f(x)

f''(x)=0 gives min/man of f'(x)

f^n(x)=0 gives min/max of f^n-1(x) where n is the order of the derivative


Thanks. hope a question like this doesn't come up tomorrow.
Original post by mrbeady9
image.jpg

How did you change it into the first form, like what formula did you use?
Original post by underestimate
How did you change it into the first form, like what formula did you use?


It's from an earlier part of the question.

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