Algebra & Calculus questions

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
TSR launches Learn Together! - Our new subscription to help improve your learning 16-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
Search Thread
Advanced Search
Sign in to Reply
  1. raheem94's Avatar
    21 26-04-2012  20:37
    • TSR Demigod
    • Posts: 5,512
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by SubAtomic)


    4. \dfrac{dy}{dx} = \dfrac{(1-y^2)x}{\sqrt{1-x^2}}

    Hmmm had a go but think I may be going wrong somewhere.

    So I multiplied by the denominator, did reciprocal, multiplied by x and that is where stopped as am not sure if am going wrong

    \displaystyle (1 – x^2)^{1/2} \dfrac{dy}{dx} = (1 – y^2)x

    \displaystyle \dfrac{x}{\sqrt{1–x^2}} dx = \dfrac{1}{\sqrt{1–y^2}} dy
    \displaystyle \dfrac{dy}{dx} = \dfrac{(1-y^2)x}{\sqrt{1-x^2}} \implies \int \frac1{1-y^2} \ dy = \int \frac{x}{\sqrt{1-x^2}} \ dx \\ \implies \int \frac1{(1-y)(1+y)} \ dy = \int \frac{x}{\sqrt{1-x^2}} \ dx

    Use partial fractions for LHS.

    For RHS, \displaystyle \int \frac{x}{\sqrt{1-x^2}} \ dx \implies \int x(1-x^2)^{-\frac12} \ dx

    To integrate this, try differentiating, \displaystyle z=(1-x^2)^{\frac12}
  2. f1mad's Avatar
    22 26-04-2012  20:49
    • TSR Demigod
    • Posts: 5,423
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by raheem94)
    =

    To integrate this, try differentiating, \displaystyle z=(1-x^2)^{\frac12}
    I would let z= 1-x^2.

    It's even easier to manipulate it so that it's of the form: f'(x)/sqrtf(x)=

    2sqrtf(x)+c.
  3. f1mad's Avatar
    23 26-04-2012  20:54
    • TSR Demigod
    • Posts: 5,423
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by SubAtomic)

    Edit: f1mad -2x
    That's right, and you already have an x present in the integrand.

    So?
  4. f1mad's Avatar
    24 26-04-2012  20:56
    • TSR Demigod
    • Posts: 5,423
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by SubAtomic)
    Lol no idea why I put the root sign in the (1-y) expression, had a mini latex meltdown

    \displaystyle\dfrac {d}{dx} (1-x^2)^{1/2} = - \dfrac{x}{\sqrt{1-x^2}}
    Divide both sides by -1, and integrate both sides.
  5. raheem94's Avatar
    25 26-04-2012  20:56
    • TSR Demigod
    • Posts: 5,512
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by SubAtomic)
    Lol no idea why I put the root sign in the (1-y) expression, had a mini latex meltdown

    \displaystyle\dfrac {d}{dx} (1-x^2)^{1/2} = - \dfrac{x}{\sqrt{1-x^2}}
    This shows us, \displaystyle \int \frac{x}{\sqrt{1-x^2}} \ dx = -(1-x^2)^{\frac12} + C
    Last edited by raheem94; 26-04-2012 at 21:01.
  6. f1mad's Avatar
    26 26-04-2012  21:00
    • TSR Demigod
    • Posts: 5,423
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by raheem94)
    This shows us, \displaystyle \int \frac{x}{\sqrt{1-x^2}} \ dx = -(1-x^2) + C
    It's not.
  7. raheem94's Avatar
    27 26-04-2012  21:02
    • TSR Demigod
    • Posts: 5,512
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by SubAtomic)
    Put -1/2 outside integrand and -2x as numerator?

    Lost
    Sorry, i made a mistake, forgot to type the power, i have corrected it now.
  8. raheem94's Avatar
    28 26-04-2012  21:04
    • TSR Demigod
    • Posts: 5,512
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by f1mad)
    It's not.
    Thanks for correcting me, i was replying to multiple threads, along with making some grade boundary calculations, hence i was doing everything very quickly. I did it correctly myself but while typing i forgot to write the power.
  9. f1mad's Avatar
    29 26-04-2012  21:09
    • TSR Demigod
    • Posts: 5,423
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by SubAtomic)
    Will go work on what have got and come back later as feel a bit lost:rolleyes:
    I x/(1-x^2)^1/2 dx

    Perhaps following on with the substitution would better sense.

    Let u= 1-x^2

    du/dx= -2x

    du= -2x dx

    We already have x in the integrand, so divide by -2

    du/-2= xdx

    Then it becomes: I -1/2 * 1/(u^1/2) du

    You will see that this is a standard integral result:

    I f'(x)/sqrtf(x) dx=

    2sqrtf(x)+c.
    Last edited by f1mad; 26-04-2012 at 21:11.
    Post rating:
    1
  10. raheem94's Avatar
    30 26-04-2012  21:13
    • TSR Demigod
    • Posts: 5,512
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by f1mad)
    I would let z= 1-x^2.

    It's even easier to manipulate it so that it's of the form: f'(x)/sqrtf(x)=

    2sqrtf(x)+c.
    At first i solved this by using a trigonometric substitution, x=sin\theta .

    Though it will be difficult for the OP to understand this.
  11. SubAtomic's Avatar
    31 26-04-2012  22:27
    • Exalted and Worshipped Member
    • Posts: 1,316
    Re: Probability, Algebra, Calculus etc etc, questions
    So am back and as I suspected I have not done anything partial fraction. I have been using the Integral formulas.

    So

    \displaystyle\int x(1-x^2)^{-1/2} \ dx = - \dfrac{1}{2} \int (1-x^2)^{-1/2}(-2x) \ dx

    So this becomes

    \displaystyle - \dfrac{1}{2} \times \dfrac{2}{1} (1-x^2)^{1/2} + C

    So will start again

    \displaystyle\int (1-y^2)^{-1} \ dy =

    Have never done partial fractions is this the only way? If so I need to go learn the method.
    Last edited by SubAtomic; 26-04-2012 at 23:01.
  12. raheem94's Avatar
    32 26-04-2012  22:46
    • TSR Demigod
    • Posts: 5,512
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by SubAtomic)
    So am back and as I suspected I have not done anything partial fraction. I have been using the Integral formulas.

    So

    \displaystyle\int x(1-x^2)^{-1/2} \ dx = - \dfrac{1}{2} \int (1-x^2)^{-1/2}(-2x) \ dx

    So this becomes

    \displaystyle - \dfrac{1}{2} \times \dfrac{2}{1} (1-x^2)^{1/2} + C

    And for the y term the same scenario

    \displaystyle - \dfrac{1}{2} \int (1-y^2)^{-2}(-2y) \ dy

    And then

    \displaystyle - \dfrac{1}{2} \times -1 (1-y^2)^{-1} \ = \ \dfrac{1}{2} \cdot \dfrac{1}{(1-y^2)}

    So

    \displaystyle\dfrac{1}{2(1-y^2)} \ = \ -(1-x^2)^{1/2} + C

    Will edit this in a sec back to the paper for a min
    Your 'x' terms are correct, but 'y' is wrong.

    Here is how to deal with the 'y' part.

    \displaystyle \int \frac1{1-y^2} \ dy = \int \frac1{(1-y)(1+y)} \ dy

    We can't integrate this unless we use partial fractions.

    \displaystyle \frac1{(1-y)(1+y)} \equiv \frac{A}{1-y} + \frac{B}{1+y} \\ \displaystyle \frac1{(1-y)(1+y)} \equiv \frac{A(1+y)+B(1-y)}{(1-y)(1+y)}

    So solving the above expression gives, A=\frac12 \text{ and } B=\frac12

    Hence, \displaystyle \frac1{(1-y)(1+y)} \equiv \frac{\frac12}{1-y} + \frac{\frac12}{1+y}

    So the integral becomes,
    \displaystyle \int \frac1{1-y^2} \ dy = \int \frac1{(1-y)(1+y)} \ dy = \int \left( \frac{\frac12}{1-y} + \frac{\frac12}{1+y} \right) \ dy =\frac12 \int \left( \frac{1}{1-y} + \frac{1}{1+y} \right) \ dy = \frac12(-ln(1-y)+ln(1+y))

    Do you get it now?
    Post rating:
    1
  13. raheem94's Avatar
    33 26-04-2012  23:09
    • TSR Demigod
    • Posts: 5,512
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by SubAtomic)
    Kinda. Are partial fractions the only way because if this is the case then I need to go learn the method.
    Yes, you need to know partial fractions for it. There might be some other way, but i can't think of any other way.

    Do you really don't know partial fractions?

    You should learn it before doing C4 integration.
  14. raheem94's Avatar
    34 26-04-2012  23:21
    • TSR Demigod
    • Posts: 5,512
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by SubAtomic)
    Am not doing a standard A-level it is a supposed equivalent (although it seems to be leaving me ****ed on questions like this). Need to go learn it then.

    Examsolutions?
    Which qualification are you doing?

    Examsolutions is a good website to learn A-Level maths, here is a link to partial fractions, http://examsolutions.co.uk/maths-rev...ntoduction.php
  15. raheem94's Avatar
    35 27-04-2012  00:00
    • TSR Demigod
    • Posts: 5,512
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by SubAtomic)
    Well I'm supposedly done lol :rolleyes: I have near enough completed the course, just on with the hypothesis tests and that's it.

    Here have a look at a taster for yourself here and then tell me what you think. I don't think it quite cuts it as an A-level equivalent.

    Cheers for the link
    I may have a look at the link later, i am currently busy studying Hypothesis testing, it is a bit difficult to understand at first.
  16. SubAtomic's Avatar
    36 30-04-2012  15:50
    • Exalted and Worshipped Member
    • Posts: 1,316
    Re: Probability, Algebra, Calculus etc etc, questions
    Quick question.

    So a differential equation I have to find

    \displaystyle \frac {dy}{dx} = (y^{something} \cdot something^{x})^\frac{something}{ something}

    And am given y = 50 \ when \ x = 0 \ so \ y(0) = 50

    Do I integrate and simplify as much as possible before subbing x = 0 or not? Bit confused with an assignment question still because I get the same answer over and over again and cannot seem to manipulate my answer to look like the choices I have.

    lol sorry for all the somethings:rolleyes:
    Last edited by SubAtomic; 30-04-2012 at 16:16.
  17. raheem94's Avatar
    37 30-04-2012  16:04
    • TSR Demigod
    • Posts: 5,512
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by SubAtomic)
    Quick question.

    So a differential equation I have to find

    \displaystyle \frac {dy}{dx} = (y^{something} \cdot something^{x})^\frac{something}{ something}

    And am given y = (something) \ when \ x = 0 \ so \ y(0) = something

    Do I integrate and simplify as much as possible before using x = 0 or not? Bit confused with an assignment question still because I get the same answer over and over again and cannot seem to manipulate my answer to look like the choices I have.

    lol sorry for all the somethings:rolleyes:
    Can you post the exact question.

    I can't assume what 'something' is.
  18. f1mad's Avatar
    38 30-04-2012  16:14
    • TSR Demigod
    • Posts: 5,423
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by SubAtomic)
    ...
    You can only plug in what x and y are, when you have got the GS of the DE.

    You don't really need to simplify it once that's done, unless they ask for it. Although usually you would get it into the form y(x)= something.

    Once you have got the GS, plug in x=0, y=50 and solve for the arbitrary constant to get a particular solution.
    Last edited by f1mad; 30-04-2012 at 16:16.
  19. SubAtomic's Avatar
    39 30-04-2012  16:17
    • Exalted and Worshipped Member
    • Posts: 1,316
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by f1mad)
    You can only plug in what x and y are, when you have got the GS of the DE.

    You don't really need to simplify it once that's done, unless they ask for it. Although usually you would get it into the form y(x)= something.

    Once you have got the GS, plug in x=0, y=50 and solve for the arbitrary constant to get a particular solution.
    Cheers get ya:cool:

    (Original post by f1mad)
    ...

    So I would use x = 0 in the general solution then simplify to y = something after? Because I definitely have to get it as y = something in the end, although this is what have been doing so I need to manipulate and manipulate till I get the answer then
    Last edited by SubAtomic; 30-04-2012 at 16:35.
  20. f1mad's Avatar
    40 30-04-2012  16:49
    • TSR Demigod
    • Posts: 5,423
    Re: Probability, Algebra, Calculus etc etc, questions
    (Original post by SubAtomic)
    Cheers get ya:cool:




    So I would use x = 0 in the general solution then simplify to y = something after? Because I definitely have to get it as y = something in the end, although this is what have been doing so I need to manipulate and manipulate till I get the answer then
    You could do either.
Sign in to Reply
Search Thread
Advanced Search
Share this discussion:  
Useful resources

Articles:

Study Help rules and posting guidelinesStudy Help Member of the Month nominationsGuide to MathematicsMathematics resourcesMathematics websitesMathematics revision notes in the TSR wikiCambridge Assessment admissions tests (including STEP)How to use LaTeXCareers in Mathematics

Useful Dates:

Quick Link:

Unanswered Maths Threads

Groups associated with this forum:

View associated groups
Article updates
Moderators

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

This forum is moderated by:

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

Shortcuts

Get Started

TSR Group

Using TSR

Info

Connect with TSR

Reputation gems:
The Reputation gems seen here indicate how well reputed the user is, red gem indicate negative reputation and green indicates a good rep.
Post rating score:
These scores show if a post has been positively or negatively rated by our members.