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C3 Differentiation Urgent Help needed!!! watch

1. I'm still unsure about what I need to do with 4/3pi
2. Anyone?
3. (Original post by jordanwu)
Well the y values are increasing as x increases (after 4pi/3) until 2pi then the y values start to decrease from 1
Yes, but we're looking at cos from 0 to pi. Do you know why?

x = 4pi/3 is what we're interested in but we're dealing with cosx/2, so we're looking at costheta when theta = 2pi/3 (where theta = x/2 but we draw the graph as any cos function).

It may be a good idea to draw a graph of cosx/2 for 0 < x < 2pi and look at x = 4pi/3 (which is cos 2pi/3) and draw conclusions from that.

If you're still stuck I will show you the graph and what you need to see.

Just to recap what we've done: looked at dy/dx and identified that dy/dx must be positive if y is increasing.

Looked at the expression for dy/dx given in the question and realised that the numerator is important.

Looked at the numerator to see that it must be positive, so 1/2 + cosx/2 > 0.

Rearranged and found the critical value, x = 4pi/3.

Drew the graph of costheta or cosx/2 and looking for the values where cosx/2 > -1/2, knowing the critical value x=4pi/3.
Since x is between 0 and 2pi we're only interested in what cos(x/2) is between 0 and pi.
4. (Original post by jordanwu)
Anyone?
Drew the graph of costheta or cosx/2 and looking for the values where cosx/2 > -1/2, knowing the critical value x=4pi/3. Since x is between 0 and 2pi we're only interested in what cos(x/2) is between 0 and pi.

It may be this step that is unclear.

Also if cos(x/2)>-1/2 then dy/dx is positive so y is increasing.
5. (Original post by SeanFM)
Drew the graph of costheta or cosx/2 and looking for the values where cosx/2 > -1/2, knowing the critical value x=4pi/3. Since x is between 0 and 2pi we're only interested in what cos(x/2) is between 0 and pi.

It may be this step that is unclear.

Also if cos(x/2)>-1/2 then dy/dx is positive so y is increasing.
On the graph of cos(x/2) the x value for y=-1/2 is 8.3775
6. (Original post by jordanwu)
On the graph of cos(x/2) the x value for y=-1/2 is 8.3775
That is correct, but for costheta, when theta = 2pi/3 costheta = -1/2, so if theta = x/2 then x = 4pi/3 is also a solution.

You got yours from theta = 4pi/3, so x = 8pi/3 = 8.3775...

In the graph of cos(x/2) we're only interested in x values between 0 and 2pi (given in the question) so we start at cos0 and end up at cos(2pi/2) = cos(pi) = -1. So we are only looking at the output between cos0 and cospi. The solution that you have, x= 8.3375 is OUTSIDE of the range of x, so we are not interested in it. It is true that cos(8.3375/2) = -1/2 but it is OUTSIDE of the range.

There is a point at which cos(x/2) becomes < - 1/2. We have already worked out that point, it is when costheta = - 1/2.
But we are looking for the points where cos(x/2) > -1/2 because then the numerator is POSITIVE in the expression dy/dx so we have answered the question.

Cos(x/2) starts at x/2= 0 , y = 1 and DECREASES as it moves towards x = 2pi/2 , y = -1. So between x/2 = 0 and x/2 = pi, y is > -1/2 so that is when the numerator is positive so dy/dx is positive so y is increasing. The point at which y = 0 (anything after is less than 0, so that's negative is x/2 = 2pi/3, which is x = 4pi/3.

So the answer you are looking for is 0<x/2 < 2pi/3, so 0<x<4pi/3.
7. (Original post by SeanFM)
That is correct, but for costheta, when theta = 2pi/3 costheta = -1/2, so if theta = x/2 then x = 4pi/3 is also a solution.

You got yours from theta = 4pi/3, so x = 8pi/3 = 8.3775...

In the graph of cos(x/2) we're only interested in x values between 0 and 2pi (given in the question) so we start at cos0 and end up at cos(2pi/2) = cos(pi) = -1. So we are only looking at the output between cos0 and cospi. The solution that you have, x= 8.3375 is OUTSIDE of the range of x, so we are not interested in it. It is true that cos(8.3375/2) = -1/2 but it is OUTSIDE of the range.

There is a point at which cos(x/2) becomes < - 1/2. We have already worked out that point, it is when costheta = - 1/2.
But we are looking for the points where cos(x/2) > -1/2 because then the numerator is POSITIVE in the expression dy/dx so we have answered the question.

Cos(x/2) starts at x/2= 0 , y = 1 and DECREASES as it moves towards x = 2pi/2 , y = -1. So between x/2 = 0 and x/2 = pi, y is > -1/2 so that is when the numerator is positive so dy/dx is positive so y is increasing. The point at which y = 0 (anything after is less than 0, so that's negative is x/2 = 2pi/3, which is x = 4pi/3.

So the answer you are looking for is 0<x/2 < 2pi/3, so 0<x<4pi/3.
So cos(x/2)>-1/2 when 0<x<4pi/3?
8. (Original post by jordanwu)
So cos(x/2)>-1/2 when 0<x<4pi/3?
Correct. Do you understand why?
9. (Original post by SeanFM)
Correct. Do you understand why?
So basically from the beginning:

I need to find the range of x values for which y is increasing.
If y is increasing then dy/dx must be greater than 0.
Set dy/dx>0.
Bring the denominator over to the RHS as numerator is more important in determining whether or not dy/dx is positive or not, denominator can be positive or negative as you have a term being squared.
Now we have 1/2 + cos(x/2)>0.
Bring 1/2 to RHS to get cos(x/2)>-1/2 so now I must consider where the graph of y=cos(x/2)>-1/2, but I first need the critical value.
To find the critical value x/2=cos^-1(-1/2) so x/2=2pi/3, therefore x=4pi/3.
Now we have the critical value we look at the graph of cos(x/2) between 0 and 2pi as in the question and we find that the x value for -1/2 is the critical value that we obtained, therefore we want all the x values between 0 and 4pi/3 as the shape of the curve indicates that the y values increase as the x values decrease towards 0.
So final solution is 0<x<4pi/3.

Is that ok?
10. (Original post by jordanwu)
So basically from the beginning:

I need to find the range of x values for which y is increasing.
If y is increasing then dy/dx must be greater than 0.
Set dy/dx>0.
Bring the denominator over to the RHS as numerator is more important in determining whether or not dy/dx is positive or not, denominator can be positive or negative as you have a term being squared.
Now we have 1/2 + cos(x/2)>0.
Bring 1/2 to RHS to get cos(x/2)>-1/2 so now I must consider where the graph of y=cos(x/2)>-1/2, but I first need the critical value.
To find the critical value x/2=cos^-1(-1/2) so x/2=2pi/3, therefore x=4pi/3.
Now we have the critical value we look at the graph of cos(x/2) between 0 and 2pi as in the question and we find that the x value for -1/2 is the critical value that we obtained, therefore we want all the x values between 0 and 4pi/3 as the shape of the curve indicates that the y values increase as the x values decrease towards 0.
So final solution is 0<x<4pi/3.

Is that ok?
Almost perfect, just the reasoning for the last bit.

For the graph cos(x/2) y = -1/2 when x=4pi/3. Since we are looking for x in the range 0 to 2pi x/2 is in the range 0 to pi, so y ranges from cos0 to cospi. When y>-1/2 dy/dx is positive etc. and looking at the graph of cos(x/2), between 0 and pi y is >1/2 between 0 and 4pi/3, so those are the values that we want. Well done.
11. (Original post by SeanFM)
Almost perfect, just the reasoning for the last bit.

For the graph cos(x/2) y = -1/2 when x=4pi/3. Since we are looking for x in the range 0 to 2pi x/2 is in the range 0 to pi, so y ranges from cos0 to cospi. When y>-1/2 dy/dx is positive etc. and looking at the graph of cos(x/2), between 0 and pi y is >1/2 between 0 and 4pi/3, so those are the values that we want. Well done.
Thanks for all the help you gave me yesterday and today I've been stuck on this same question for the whole day but you were really helpful, definitely deserving of a rep
12. (Original post by jordanwu)
Thanks for all the help you gave me yesterday and today I've been stuck on this same question for the whole day but you were really helpful, definitely deserving of a rep
No problem

If you need help with any more questions it may be worth making a new thread.
13. dy/dy = [ (2+cos(1/2x)) X (1/2cos(1/2x)) ] - [ sin(1/2x) X -1/2sin(1/2x) ] / [2+cos(1/2x)]^2
dy/dx = cos(1/2x) + 1/2cos^2(1/2x) + 1/2sin^2(1/2x) / [2+cos(1/2x)]^2
dy/dy = cos(1/2x) + 1/2 [cos^2(1/2x) + sin^2(1/2x) ] / [2+cos(1/2x)]^2
dy/dy = cos(1/2x) + 1/2 X 1 / [2+cos(1/2x)]^2
dy/dy = [1/2 + cos(1/2x)] / [2+cos(1/2x)]^2

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