C3 Differentiation Urgent Help needed!!!

OP

I'm still unsure about what I need to do with 4/3pi

OP

Anyone?

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.

(edited 8 years ago)

Reply 63

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.

(edited 8 years ago)

Reply 64

OP

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

Reply 65

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.

Reply 66

OP

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?

Reply 67

Original post by jordanwu

So cos(x/2)>-1/2 when 0<x<4pi/3?

Correct. Do you understand why?

Reply 68

OP

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?

Reply 69

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. :smile:

Reply 70

OP

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. :smile:

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 :smile:

Reply 71