# polar graph help

Watch
Announcements

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

https://imgur.com/a/JCQguIC

0

reply

Report

#4

(Original post by

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

**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

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

Report

#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

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

Report

#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

(Original post by

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?

**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

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

**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

0

reply

Report

#8

(Original post by

is the 9/2 there because 3 squared and then they half it because of the equation with integrating polar curves?

**Gent2324**)is the 9/2 there because 3 squared and then they half it because of the equation with integrating polar curves?

0

reply

(Original post by

Yes - you need to find 1/2 of the integral of r^2 wrt theta.

**Pangol**)Yes - you need to find 1/2 of the integral of r^2 wrt theta.

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

Report

#10

(Original post by

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?

**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?

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

(Original post by

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?

**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?

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

Report

#12

(Original post by

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?

**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?

0

reply

(Original post by

Are you sure you should have that power 2 in your second term?

**Pangol**)Are you sure you should have that power 2 in your second term?

0

reply

Report

#14

(Original post by

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.

**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.

0

reply

(Original post by

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.

**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.

0

reply

Report

#16

(Original post by

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

**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

- multiply by (cos 2 theta)^1/2

- divide by 3

- move the negative piece to the other side of the equality sign

0

reply

(Original post by

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

**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

0

reply

Report

#18

(Original post by

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 ?

**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 ?

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

(Original post by

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.

**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.

0

reply

Report

#20

(Original post by

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

**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

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

Back

to top

to top