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Algebra & Calculus questions

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Applying to Uni? Let Universities come to you. Click here to get your perfect place 20-10-2014
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    (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}
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    (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.
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    (Original post by SubAtomic)

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

    So?
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    (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.
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    (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
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    (Original post by raheem94)
    This shows us,  \displaystyle \int \frac{x}{\sqrt{1-x^2}} \ dx = -(1-x^2) + C
    It's not.
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    (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.
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    (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.
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    (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.
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    (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.
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    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.
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    (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?
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    (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.
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    (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
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    (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.
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    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:
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    (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.
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    (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.
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    (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
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    (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.
Updated: May 7, 2012
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