Gent2324
Badges: 22
Rep:
?
#1
Report Thread starter 1 year ago
#1
im stuck with part a and also dont know how to do part b. for part A i have tried integrating with 9cos2theta with limits up to pi/2, but i get 0, im not to sure what limits i should be using? thanks
https://imgur.com/a/JCQguIC
0
reply
NotNotBatman
Badges: 20
Rep:
?
#2
Report 1 year ago
#2
Wheres the question?
0
reply
Gent2324
Badges: 22
Rep:
?
#3
Report Thread starter 1 year ago
#3
(Original post by NotNotBatman)
Wheres the question?
link in op
0
reply
Pangol
Badges: 15
Rep:
?
#4
Report 1 year ago
#4
(Original post by Gent2324)
im stuck with part a and also dont know how to do part b. for part A i have tried integrating with 9cos2theta with limits up to pi/2, but i get 0, im not to sure what limits i should be using? thanks
https://imgur.com/a/JCQguIC
First, note that you should be integrating 1/2 of r^2 with respect to theta - you are missing the 1/2. Not that that is your problem, as 1/2 of 0 is still 0!

Your problem is that you are integrating over a region which includes some parts where the curve is not defined. You should be able to see that the theta values in these parts give you negative values of r, and regions enclosed by these negative r values count as negative areas (just like regions under the x-axis in conventional integration count as negative areas), so they will cancel with the positive areas to give you zero.

If you look at the diagram and the given permissible values of theta, you can see using symmetry that if you integrate from 0 to pi/4, you won't have all of the area you need, but you will have part of it. How much?
0
reply
NotNotBatman
Badges: 20
Rep:
?
#5
Report 1 year ago
#5
Why are you integrating up to pi/2 ?

If theta = 0 and you begin increasing theta you have the initial line that then traverses the upper region of one segment of the shape up to theta =pi/4.

Plug some values in and you'll see this. When theta =pi/4 , r=0 and the direction traversed is anticlockwise. Thus integrating 1/2r^2 wrt theta from 0 to pi/4 gives the area of that sort of semi circular region. Theres four of them so then multiply by 4
0
reply
Pangol
Badges: 15
Rep:
?
#6
Report 1 year ago
#6
I've only just noticed that you have included the mark scheme with the question, which is using exactly the method I've suggested. But do you see why it works and why your first go didn't?
1
reply
Gent2324
Badges: 22
Rep:
?
#7
Report Thread starter 1 year ago
#7
(Original post by Pangol)
I've only just noticed that you have included the mark scheme with the question, which is using exactly the method I've suggested. But do you see why it works and why your first go didn't?
(Original post by NotNotBatman)
Why are you integrating up to pi/2 ?

If theta = 0 and you begin increasing theta you have the initial line that then traverses the upper region of one segment of the shape up to theta =pi/4.

Plug some values in and you'll see this. When theta =pi/4 , r=0 and the direction traversed is anticlockwise. Thus integrating 1/2r^2 wrt theta from 0 to pi/4 gives the area of that sort of semi circular region. Theres four of them so then multiply by 4
is the 9/2 there because 3 squared and then they half it because of the equation with integrating polar curves?
0
reply
Pangol
Badges: 15
Rep:
?
#8
Report 1 year ago
#8
(Original post by Gent2324)
is the 9/2 there because 3 squared and then they half it because of the equation with integrating polar curves?
Yes - you need to find 1/2 of the integral of r^2 wrt theta.
0
reply
Gent2324
Badges: 22
Rep:
?
#9
Report Thread starter 1 year ago
#9
(Original post by Pangol)
Yes - you need to find 1/2 of the integral of r^2 wrt theta.
ah right that makes sense now, for part b however im still stuck. i get that PQ is 6, and i tried to calculate PS by using the fact that when theta = pi/4, the radius is 3(cospi/4)^1/2, using trig i can calculate the missing side which is sinpi/4 x 3(cospi/4)^1/2 and then multiply by 2 but for some reason that doesnt get me the answer.

doing it the differentiation way i can differentiate it fine but the simplifying bit im a bit lost, on line 4 part b not sure what they are doing?
0
reply
Pangol
Badges: 15
Rep:
?
#10
Report 1 year ago
#10
(Original post by Gent2324)
ah right that makes sense now, for part b however im still stuck. i get that PQ is 6, and i tried to calculate PS by using the fact that when theta = pi/4, the radius is 3(cospi/4)^1/2, using trig i can calculate the missing side which is sinpi/4 x 3(cospi/4)^1/2 and then multiply by 2 but for some reason that doesnt get me the answer.

doing it the differentiation way i can differentiate it fine but the simplifying bit im a bit lost, on line 4 part b not sure what they are doing?
You've forgotten to use 2 theta instead of just theta when substituting pi/4 into the original equation. If you do this with 2 theta, you'll see what that isn't going to work.

I think in the line you are referring to they are multiplying by (cos 2 theta)^1/2 and dividing by 3, as well as moving part of the equation to the other side. Is this the bit you mean?
0
reply
Gent2324
Badges: 22
Rep:
?
#11
Report Thread starter 1 year ago
#11
(Original post by Pangol)
You've forgotten to use 2 theta instead of just theta when substituting pi/4 into the original equation. If you do this with 2 theta, you'll see what that isn't going to work.

I think in the line you are referring to they are multiplying by (cos 2 theta)^1/2 and dividing by 3, as well as moving part of the equation to the other side. Is this the bit you mean?
yep just tried it and got a complex number...

yes i just multiplied by cos 2 theta ^1/2 and i got -3sin2theta sintheta + 3(cos2theta)^2 costheta
how is it still cos2theta ^1/2 and not to the power of 2?
0
reply
Pangol
Badges: 15
Rep:
?
#12
Report 1 year ago
#12
(Original post by Gent2324)
yep just tried it and got a complex number...

yes i just multiplied by cos 2 theta ^1/2 and i got -3sin2theta sintheta + 3(cos2theta)^2 costheta
how is it still cos2theta ^1/2 and not to the power of 2?
Are you sure you should have that power 2 in your second term?
0
reply
Gent2324
Badges: 22
Rep:
?
#13
Report Thread starter 1 year ago
#13
(Original post by Pangol)
Are you sure you should have that power 2 in your second term?
ah yes its 1 not 2, im at where they are at now after dividing by cos2theta ^1/2. i dont see at all what they are doing on the line after though.
0
reply
Pangol
Badges: 15
Rep:
?
#14
Report 1 year ago
#14
(Original post by Gent2324)
ah yes its 1 not 2, im at where they are at now after dividing by cos2theta ^1/2. i dont see at all what they are doing on the line after though.
Can't see a line where they divide by this, only one where they multiply by it. If you can be more specific about the line you're looking at, I'm sure we can help out.
0
reply
Gent2324
Badges: 22
Rep:
?
#15
Report Thread starter 1 year ago
#15
(Original post by Pangol)
Can't see a line where they divide by this, only one where they multiply by it. If you can be more specific about the line you're looking at, I'm sure we can help out.
on line 4 ive got to where they are, they just put the cos2theta on the denominator, i dont understand what they do to get sin2theta sintheta = cos2theta cos theta on the line after
0
reply
Pangol
Badges: 15
Rep:
?
#16
Report 1 year ago
#16
(Original post by Gent2324)
on line 4 ive got to where they are, they just put the cos2theta on the denominator, i dont understand what they do to get sin2theta sintheta = cos2theta cos theta on the line after
This is what we were talking about before. From the line you agree with (the one that starts "At max/min;"

- multiply by (cos 2 theta)^1/2
- divide by 3
- move the negative piece to the other side of the equality sign
0
reply
Gent2324
Badges: 22
Rep:
?
#17
Report Thread starter 1 year ago
#17
(Original post by Pangol)
This is what we were talking about before. From the line you agree with (the one that starts "At max/min;"

- multiply by (cos 2 theta)^1/2
- divide by 3
- move the negative piece to the other side of the equality sign
yep im there now thanks, which double angle rule are they using on the lhs where they get sin squared? and also how do they know that costheta(1-4sin^2theta) = 0 ?
0
reply
Pangol
Badges: 15
Rep:
?
#18
Report 1 year ago
#18
(Original post by Gent2324)
yep im there now thanks, which double angle rule are they using on the lhs where they get sin squared? and also how do they know that costheta(1-4sin^2theta) = 0 ?
They are using the standard identity for sin(2theta) in terms of sin(theta) and cos(theta), and combining it with the sin(theta) that is already there.

On the next line, they have brought everything onto one side and factorised out the cos(theta).

These are all bits from a regular maths A level - they shouldn't be giving you too much trouble if you're doing a question at this level.
0
reply
Gent2324
Badges: 22
Rep:
?
#19
Report Thread starter 1 year ago
#19
(Original post by Pangol)
They are using the standard identity for sin(2theta) in terms of sin(theta) and cos(theta), and combining it with the sin(theta) that is already there.

On the next line, they have brought everything onto one side and factorised out the cos(theta).

These are all bits from a regular maths A level - they shouldn't be giving you too much trouble if you're doing a question at this level.
we havent done these yet in normal maths and for some reason our mock exams for further maths is the entire syllabus so we have just had to kind of rush through it when revising. that being said i pretty much always lose marks on simplifying trig, can never figure out which identity to use
0
reply
Pangol
Badges: 15
Rep:
?
#20
Report 1 year ago
#20
(Original post by Gent2324)
we havent done these yet in normal maths and for some reason our mock exams for further maths is the entire syllabus so we have just had to kind of rush through it when revising. that being said i pretty much always lose marks on simplifying trig, can never figure out which identity to use
Ah, that makes sense if you are doing the regular course and the further course in parallel rather than in series.

It might be worth asking your teacher for a list of topics from the regular course that you are expected to know for the further course you are taking. But I would say that pretty much all of the trig stuff is essential (maybe not harmonic form, or the R-alpha thing as most people call it).
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Have you made your mind up on your five uni choices?

Yes, and I've sent off my application! (3)
100%
I've made my choices but havent sent my application yet (0)
0%
I've got a good idea about the choices I want to make (0)
0%
I'm researching but still not sure which universities I want to apply to (0)
0%
I haven't started researching yet (0)
0%
Something else (let us know in the thread!) (0)
0%

Watched Threads

View All