Year 13 Maths Help Thread

Announcements Posted on
How helpful is our apprenticeship zone? Have your say with our short survey 02-12-2016
    • Thread Starter
    Offline

    2
    ReputationRep:
    Would the following problem be accessible for a C4 student?

    Find the total area between the curves (measured positively) \sin x and its inverse function from  x=0 to x=1 and from x=-1 to x=0.
    Online

    3
    ReputationRep:
    (Original post by Palette)
    Would the following problem be accessible for a C4 student?

    Find the total area enclosed by the curves (measured positively) \sin x and its inverse function from  x=0 to x=1 and from x=-1 to x=0.
    Wording is weird... can you make it clearer?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    Wording is weird... can you make it clearer?
    I could rephrase it as 'Find the area between the graphs of sin x and arcsin x between x=-1 and x=0. Then find the area between the graphs of sin x and arcsin x between x=0 and x=1. What is the sum of these two areas?'
    Online

    3
    ReputationRep:
    (Original post by Palette)
    I could rephrase it as 'Find the area between the graphs of sin x and arcsin x between x=-1 and x=0. Then find the area between the graphs of sin x and arcsin x between x=0 and x=1. What is the sum of these two areas?'
    You mean this?

    Name:  Capture.PNG
Views: 149
Size:  29.4 KB
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    You mean this?

    Name:  Capture.PNG
Views: 149
Size:  29.4 KB
    Yes.
    Online

    3
    ReputationRep:
    (Original post by Palette)
    Yes.
    Just rephrase it to \int^0_{-1} arcsin(x)\ dx + \int^1_0 sin(x)\ dx

    I'm unsure if C4 integrates arcsine, but it's in the formula booklet anyway so it's doable.

    Update: Yes it's doable. Basic inverse trig integration is covered in C3
    Online

    3
    ReputationRep:
    (Original post by RDKGames)
    I'm unsure if C4 integrates arcsine, but it's in the formula booklet anyway so it's doable.

    You're getting confused with differentiation and integration. The way to integrate these is to use IBP with u = \arcsin x and \mathrm{d}v = 1.
    Online

    3
    ReputationRep:
    (Original post by Zacken)
    You're getting confused with differentiation and integration. The way to integrate these is to use IBP with u = \arcsin x and \mathrm{d}v = 1.
    Yeah I wasn't sure if it's integration or differentiation that is given to you in the formula booklet, still there so students can figure it out I'm sure.
    • Thread Starter
    Offline

    2
    ReputationRep:
    Proving various trigonometric identities is by far my weakest point in C3.

    How can I start proving that \frac{\sec \theta -1}{\sec \theta +1}= \tan^2 \frac{\theta}{2}?
    Online

    3
    ReputationRep:
    (Original post by Palette)
    Proving various trigonometric identities is by far my weakest point in C3.

    How can I start proving that \frac{\sec \theta -1}{\sec \theta +1}= \tan^2 \frac{\theta}{2}?
    The obvious way; convert everything to sines and cosines first off: \frac{1 - \cos \theta}{1 + \cos \theta}, then well, everybody should be seeing that we want \frac{\theta}{2} so the only sensible plan of attack is to use \cos \theta = 1 - 2\sin^2 \frac{\theta}{2} and \cos \theta = 2\cos^2 \frac{\theta}{2} - 1 in the numerator and denominator respectively, for two obvious reasons:

    (i) It gets \sin in the numerator and \cos in the denominator; which is the form for \tan.

    (ii) It gets rid of the pesky 1 \pm so the division becomes straightforward.

    Anywho, that gets you \frac{\sin^2 \frac{\theta}{2}}{\cos^2 \frac{\theta}{2}}\equiv \cdots
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    The obvious way; convert everything to sines and cosines first off: \frac{1 - \cos \theta}{1 + \cos \theta}, then well, everybody should be seeing that we want \frac{\theta}{2} so the only sensible plan of attack is to use \cos \theta = 1 - 2\sin^2 \frac{\theta}{2} and \cos \theta = 2\cos^2 \frac{\theta}{2} - 1 in the numerator and denominator respectively, for two obvious reasons:
    Everybody except me I suppose. I was thinking too much along the lines of the half angle formulae when I saw \tan^2 \frac{\theta}{2}.
    (i) It gets \sin in the numerator and \cos in the denominator; which is the form for \tan.

    (ii) It gets rid of the pesky 1 \pm so the division becomes straightforward.

    Anywho, that gets you \frac{\sin^2 \frac{\theta}{2}}{\cos^2 \frac{\theta}{2}}\equiv \cdots
    Thanks for the help!
    • Thread Starter
    Offline

    2
    ReputationRep:
    M3 question:
    Is there a way that I can visualise \frac{dv}{dx} in my head? I know it means 'the rate of change of velocity with respect to displacement' but it still feels a bit strange when all the questions in mechanics so far are to do with '_____ with respect to time'.
    Online

    3
    ReputationRep:
    (Original post by Palette)
    M3 question:
    Is there a way that I can visualise \frac{dv}{dx} in my head? I know it means 'the rate of change of velocity with respect to displacement' but it still feels a bit strange when all the questions in mechanics so far are to do with '_____ with respect to time'.
    Just write it as \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{\mathrm{d}v}{\mathrm{d}t} \cdot \frac{\mathrm{d}t}{\mathrm{d}x} = \frac{a}{v}.
    • Thread Starter
    Offline

    2
    ReputationRep:
    I am currently studying the mean value theorem, a piece of calculus that is taught in schools in the US but not in the UK.

    I wanted to know some results that can be derived from the mean value theorem and one of them happens to be:

    If f(a,b)\rightarrow\mathbb(R) is differentiable and f'(x)=0 for all x \in (a,b), then f is constant.

    I am confused by what f(a,b) means.

    f: (a,b) \rightarrow \mathbb{R}
    Offline

    3
    ReputationRep:
    (Original post by Palette)
    I am currently studying the mean value theorem, a piece of calculus that is taught in schools in the US but not in the UK.

    I wanted to know some results that can be derived from the mean value theorem and one of them happens to be:

    If f(a,b)\rightarrow \mathbb(R) is differentiable and f'(x)=0 for all x \in (a,b), then f is constant.

    I am confused by f(a,b) .
    It's a function of 2 variables, a and b. (I believe).
    Online

    3
    ReputationRep:
    (Original post by Palette)
    I am confused by what f(a,b) means.
    Wow, where did you get that from? That's a horrible abuse of notation. It should be: the function f: (a, b) \to \mathbb{R} is differentiable, etc... i.e: the domain of f is (a, b) and the co-domain is \mathbb{R}.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Wow, where did you get that from? That's a horrible abuse of notation. It should be: the function f: (a, b) \to \mathbb{R} is differentiable, etc... i.e: the domain of f is (a, b) and the co-domain is \mathbb{R}.
    The TeX messed up when I was copying and pasting it. Was in a rush so I didn't have my TeX sheet with me. I can show you the source- it looks fine there:

    http://math.stackexchange.com/questi...-value-theorem

    It must have been TSR as I copied directly from the TeX source in the link above. Even had I spotted it, I wouldn't have known about the distinction between f: (a,b) and f(a,b) as I'd have thought that they were as interchangeable as f(x) and f: x.

    Please forgive my numerous mathematical heresies over the past year.
    Online

    3
    ReputationRep:
    (Original post by Palette)
    I'd have thought that they were as interchangeable as f(x) and f: x.
    Those aren't interchangeable.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Those aren't interchangeable.
    My C3 textbook states:

    You can write functions in two different ways: f(x)= and f:x \rightarrow.

    Is the textbook misleading or am I making some classic mistake in analysis?

    P.S. I now understand why the TeX went wrong. I now remember that there were initially spacing issues with the TeX when I first copied it so I had to retype parts. Hence \mathbb{R} was accidentally typed as \mathbb(R) and the colon was left out in f: (a, b) as I was rushing it. I think that explains it!
    Offline

    3
    ReputationRep:
    (Original post by Palette)
    My C3 textbook states:

    You can write functions in two different ways: f(x)= and f:x \rightarrow.

    Is the textbook misleading or am I making some classic mistake in analysis?

    P.S. I now understand why the TeX went wrong. I now remember that there were initially spacing issues with the TeX when I first copied it so I had to retype parts. Hence \mathbb{R} was accidentally typed as \mathbb(R) and the colon was left out in f: (a, b) as I was rushing it. I think that explains it!
    You can write a function like this
     \displaystyle f : \mathbb{R} \rightarrow \mathbb{R}
     \displaystyle x\mapsto x+4 .
    As you can see the function is defined completely, you have the domain, codomain and the mapping.
    You won't see it at A-level but it can be quite important to have all these terms defined for a function.
    For example if we have
     \displaystyle f: \mathbb{R} \rightarrow \mathbb{R}^+
     \displaystyle x \mapsto x^2
    This is different to the function
     \displaystyle f: \mathbb{C} \rightarrow \mathbb{R}
     \displaystyle x \mapsto x^2 .
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: November 26, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Today on TSR
Poll
Would you rather have...?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.