# C3 Maths help- Trig

Watch
Announcements
Thread starter 3 years ago
#1
Hi everyone,

I was doing the C3 2013 withdrawn paper and was confused on part D of this question.
Attachment 736510

Couple of reasons,
1) If it is asking for the time taken for two revolutions, why can't I times my answer to C by 4. Because C is calculating the time taken for the first maximum to be reached, which would be half a revolution, right?
2) So the other method I did was to draw a cos graph, and from there work out that two complete revolutions would be when x= 4pi

Using the harmonic identity I'd already worked out,

Pi t/5 + 0.2187 = 4pi
But in the MS you just do
pi t/5 = 4pi

Help would be appreciated!
0
3 years ago
#2
(Original post by Mazza2000)
Hi everyone,

I was doing the C3 2013 withdrawn paper and was confused on part D of this question.
Attachment 736510

Couple of reasons,
1) If it is asking for the time taken for two revolutions, why can't I times my answer to C by 4. Because C is calculating the time taken for the first maximum to be reached, which would be half a revolution, right?
2) So the other method I did was to draw a cos graph, and from there work out that two complete revolutions would be when x= 4pi

Using the harmonic identity I'd already worked out,

Pi t/5 + 0.2187 = 4pi
But in the MS you just do
pi t/5 = 4pi

Help would be appreciated!
I remember this Ferris wheel question, but because TSR is silly the attachment isn't working for some reason.

But for 1), it depends on where 'it' starts - imagine a circle, the time to reach the top (0 degrees)would differ depending on whether you start at 180 or 270.

2) tricky as I can't see the question, but it seems like you did an extra step by using the harmonic identity; the question would be, why? Again, guesswork as I can't see the question, but the difference is between time taken vs at what time is the second revolution complete

(This paper is horrible)
0
Thread starter 3 years ago
#3
(Original post by Kevin De Bruyne)
I remember this Ferris wheel question, but because TSR is silly the attachment isn't working for some reason.

But for 1), it depends on where 'it' starts - imagine a circle, the time to reach the top (0 degrees)would differ depending on whether you start at 180 or 270.

2) tricky as I can't see the question, but it seems like you did an extra step by using the harmonic identity; the question would be, why? Again, guesswork as I can't see the question, but the difference is between time taken vs at what time is the second revolution complete

(This paper is horrible)
THanks for your response, but I'm still confused on one bit- even if I started at say 270 degrees, time for 2 complete revolutions would still be the same as if I started at 180 degrees, right?
0
3 years ago
#4
(Original post by Mazza2000)
THanks for your response, but I'm still confused on one bit- even if I started at say 270 degrees, time for 2 complete revolutions would still be the same as if I started at 180 degrees, right?
Yes. I was trying to guess what C refers to as I can't see the question. If it is indeed the time for half a revolution then it wouldn't make sense if 4*C is not the time for 2 revolutions. So we have to look at whether C is really the time for half a revolution or whether by definition it is something else.
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

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

### Poll

Join the discussion

#### What support do you need with your UCAS application?

I need help researching unis (3)
10%
I need help researching courses (2)
6.67%
I need help with filling out the application form (3)
10%
I need help with my personal statement (14)
46.67%
I need help with understanding how to make my application stand out (5)
16.67%
I need help with something else (let us know in the thread!) (1)
3.33%
I'm feeling confident about my application and don't need any help at the moment (2)
6.67%