Edexcel FP2 June 2015 - Official Thread Watch

simonli2575
Badges: 2
Rep:
?
#161
Report 3 years ago
#161
(Original post by Anshul6974)
Out of interest would we be expected to sketch something like arg((z-z1)(z-z2)=pi/3 and if so what would it look like?
It would be part of a circle going from z1 to z2.
0
reply
Maths degree
Badges: 11
Rep:
?
#162
Report 3 years ago
#162
for q7a in 3e. surely you would just take over the 3/4. so then you would be applying de moivre's theorem with 4/3 ?
0
reply
Maths degree
Badges: 11
Rep:
?
#163
Report 3 years ago
#163
(Original post by Mjmuk)
I just have a couple questions:1. Do we need to know how to prove things like the integrating factor etc. or is it just De Moivres that we'll be expected to prove in the exam?2. In Exercise 3E Q7a When they are solving it they add the 2k(pi) AFTER they multiply the equation by 4 using De Moivres, and I tried to solve it by adding 2kpi before the multiplication but got different answers. So when we get a question like this do we only add 2kpi after multiplying and before ''dividing (by 3 in this case)'' and why does it make a difference?3.In Example 31 in Complex Numbers Pg 48 , For finding Min/Max values does the line ALWAYS have to pass through the centre of the circle regardless of where it starts ie max from z-1 4. And finally how do you draw arg(1-w)=pi/2 ?Thanks in advance

for q7a in 3e. surely you would just take over the 3/4. so then you would be applying de moivre's theorem with 4/3 ?
0
reply
crashMATHS
Badges: 16
Rep:
?
#164
Report 3 years ago
#164
help!

So I'm doing this polar coordinate question for the area of S. On the diagram I attached, I worked out the area of A with limits pi/6 to pi/2, doubled it and then worked out the area of that semi-circle with limits from pi to 2pi.

The bit in bold is apparently wrong - why is this? :/
Attached files
0
reply
username1259045
Badges: 17
Rep:
?
#165
Report 3 years ago
#165
(Original post by kingaaran)
help!

So I'm doing this polar coordinate question for the area of S. On the diagram I attached, I worked out the area of A with limits pi/6 to pi/2, doubled it and then worked out the area of that semi-circle with limits from pi to 2pi.

The bit in bold is apparently wrong - why is this? :/
b/c you're only including the bottom half of the circle, you also have to include the part under A above the x-axis
0
reply
crashMATHS
Badges: 16
Rep:
?
#166
Report 3 years ago
#166
(Original post by mmms95)
b/c you're only including the bottom half of the circle, you also have to include the part under A above the x-axis
But by doubling my answer for the area between pi/6 to pi/2 don't I already have that shaded part above the x-axis?
0
reply
username1259045
Badges: 17
Rep:
?
#167
Report 3 years ago
#167
(Original post by kingaaran)
But by doubling my answer for the area between pi/6 to pi/2 don't I already have that shaded part above the x-axis?
if you divide the shaded part above the axis in two sections, forming a sector and a smaller section you'll see that you've calculated the area of both of the smaller parts but not the area of the sector
0
reply
crashMATHS
Badges: 16
Rep:
?
#168
Report 3 years ago
#168
(Original post by mmms95)
if you divide the shaded part above the axis in two sections, forming a sector and a smaller section you'll see that you've calculated the area of both of the smaller parts but not the area of the sector
Could you draw what you're talking about on paint or something for me?

I still don't get it
0
reply
chemlover12
Badges: 3
Rep:
?
#169
Report 3 years ago
#169
(Original post by kingaaran)
help!

So I'm doing this polar coordinate question for the area of S. On the diagram I attached, I worked out the area of A with limits pi/6 to pi/2, doubled it and then worked out the area of that semi-circle with limits from pi to 2pi.

The bit in bold is apparently wrong - why is this? :/
To work out the top part. you need to integrate the circle from 0 to pi/6. then yu have to integrate the other equation from pi/6 to pi/2.
0
reply
username1259045
Badges: 17
Rep:
?
#170
Report 3 years ago
#170
not sure if this will help but i tried
Attached files
0
reply
crashMATHS
Badges: 16
Rep:
?
#171
Report 3 years ago
#171
(Original post by chemlover12)
To work out the top part. you need to integrate the circle from 0 to pi/6. then yu have to integrate the other equation from pi/6 to pi/2.
(Original post by mmms95)
if you divide the shaded part above the axis in two sections, forming a sector and a smaller section you'll see that you've calculated the area of both of the smaller parts but not the area of the sector
Basically, I worked out that area shaded in green as did the mark scheme. Then, like the mark scheme, I doubled it.

By doubling it, I think I have worked out both of the parts shaded in green, hence the only bit left is the bottom bit.

I don't understand what I am missing...?
Attached files
0
reply
crashMATHS
Badges: 16
Rep:
?
#172
Report 3 years ago
#172
(Original post by mmms95)
not sure if this will help but i tried
Got it. Thanks
0
reply
chemlover12
Badges: 3
Rep:
?
#173
Report 3 years ago
#173
One thing guys, does anyone know a question from a past paper where they asked to find out the locus given the arg(z/z1)
0
reply
Anshul6974
Badges: 2
Rep:
?
#174
Report 3 years ago
#174
(Original post by chemlover12)
One thing guys, does anyone know a question from a past paper where they asked to find out the locus given the arg(z/z1)
Don't think so which is why I think it's going to come up tomorrow!
0
reply
crashMATHS
Badges: 16
Rep:
?
#175
Report 3 years ago
#175
(Original post by Anshul6974)
Don't think so which is why I think it's going to come up tomorrow!
Great... I'm screwed then.
0
reply
bobabob
Badges: 2
Rep:
?
#176
Report 3 years ago
#176
http://www.thestudentroom.co.uk/show....php?t=3156193
0
reply
chemlover12
Badges: 3
Rep:
?
#177
Report 3 years ago
#177
(Original post by Anshul6974)
Don't think so which is why I think it's going to come up tomorrow!
Yea, so do I. Also, there is this topic on finding the maximum and minimum values of arg(z), which I came across recently. It's in the textbook page 51, question 16. That might come up as well.
0
reply
Maths degree
Badges: 11
Rep:
?
#178
Report 3 years ago
#178
can anyone help me out. how do you draw polar curves such as r= 2sec(theta-pie/3) ? basically all cosec and sec graphs in polar curves?
0
reply
math42
Badges: 19
Rep:
?
#179
Report 3 years ago
#179
(Original post by Maths degree)
can anyone help me out. how do you draw polar curves such as r= 2sec(theta-pie/3) ? basically all cosec and sec graphs in polar curves?
All cosec and sec graphs are straight lines.
You know at y = 0, theta = 0 and at x = 0, theta = pi/2

So sub in theta = 0 and theta = pi/2 to find where the line crosses the axes
Then just draw the line that goes through those two points
0
reply
Maths degree
Badges: 11
Rep:
?
#180
Report 3 years ago
#180
(Original post by 13 1 20 8 42)
All cosec and sec graphs are straight lines.
You know at y = 0, theta = 0 and at x = 0, theta = pi/2

So sub in theta = 0 and theta = pi/2 to find where the line crosses the axes
Then just draw the line that goes through those two points
oh okay. thank alot !
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
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

University open days

  • Cranfield University
    Cranfield Forensic MSc Programme Open Day Postgraduate
    Thu, 25 Apr '19
  • University of the Arts London
    Open day: MA Footwear and MA Fashion Artefact Postgraduate
    Thu, 25 Apr '19
  • Cardiff Metropolitan University
    Undergraduate Open Day - Llandaff Campus Undergraduate
    Sat, 27 Apr '19

Have you registered to vote?

Yes! (356)
37.63%
No - but I will (74)
7.82%
No - I don't want to (66)
6.98%
No - I can't vote (<18, not in UK, etc) (450)
47.57%

Watched Threads

View All