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Math Help - Expressions

  1. #16
    Super Member malaygoel's Avatar
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    Quote Originally Posted by nath_quam
    i spotted an error we don't want to integrate answer should be 2x - 4
    Please clarify:
    Which integration you are talking about
    what should be 2x-4

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    Malay
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  2. #17
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    For part i) of the question c original post, captain black has integrated to give a in terms of x we actually want to go the other way therefore answer should be 2x - 4
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  3. #18
    Super Member malaygoel's Avatar
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    Quote Originally Posted by nath_quam
    For part i) of the question c original post, captain black has integrated to give a in terms of x we actually want to go the other way therefore answer should be 2x - 4
    He used the expression a=dv/dt =dv/dx * dx/dt to calculate acceleration. He never used integration(he used integration only for calculating x in terms of t).

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    Malay
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  4. #19
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    Well his answer is wrong do the question
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  5. #20
    Super Member malaygoel's Avatar
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    I don't know whether the question is wrong or not but I found that acc. is 2x -4 when v=1m/s

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    Malay
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  6. #21
    Forum Admin topsquark's Avatar
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    I have just reworked the whole question. The answers given for a) and b) in the original post are incorrect. As I mentioned before, the answers CaptainBlack came up with originally are correct.

    -Dan
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  7. #22
    Grand Panjandrum
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    Quote Originally Posted by nath_quam
    i spotted an error we don't want to integrate answer should be 2x - 4
    We obtain:

    <br />
a=\frac{dv}{dt}=\frac{d}{dt}(2-x)^2=2(2-x)(-1)\frac{dx}{dt}\ \ \ ..(1)<br />

    By differentiation. Now we know that:

    <br />
v=(2-x)^2<br />

    but v=\frac{dx}{dt} which we may substitute back into (1) to get:

    <br />
a=\frac{dv}{dt}=2(2-x)(-1)\frac{dx}{dt}=-2(2-x)^3<br />
.

    RonL
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  8. #23
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    Quote Originally Posted by CaptainBlack
    We obtain:

    <br />
a=\frac{dv}{dt}=2(2-x)(-1)\frac{dx}{dt}=-2(2-x)^3<br />
.

    RonL
    i believe it should be to the power of 1 not 3
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  9. #24
    Grand Panjandrum
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    Quote Originally Posted by nath_quam
    i believe it should be to the power of 1 not 3
    Then please show us your working, so we can see where I have gone wrong.

    RonL
    Last edited by CaptainBlack; July 17th 2006 at 12:21 AM.
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  10. #25
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    Working

    v = (2-x)^2
    expanding gives
    v = 4 - 4x + x^2
    a= dv/dx = 2x -4
    or
    v = (2-x)^2
    a = dv/dx = using d/dx[f(x)^n] = f'(x)n[f(x)]^(n-1)
    a = ((-1)(2))(2-x)^(2-1)
    a = -2(2-x)
    a = 2x - 4
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  11. #26
    Super Member malaygoel's Avatar
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    Quote Originally Posted by nath_quam
    v = (2-x)^2
    expanding gives
    v = 4 - 4x + x^2
    a= dv/dx = 2x -4
    or
    v = (2-x)^2
    a = dv/dx = using d/dx[f(x)^n] = f'(x)n[f(x)]^(n-1)
    a = ((-1)(2))(2-x)^(2-1)
    a = -2(2-x)
    a = 2x - 4
    a is not dv/dx.
    a is dv/dt

    Malay
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  12. #27
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    ah true i'll get back to you i have to think

    ummm..... isn't a = the derivative of v
    Last edited by nath_quam; July 17th 2006 at 12:54 AM.
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  13. #28
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    Just Quickly

    Quote Originally Posted by CaptainBlack
    Like this:

    <br />
v=\frac{dx}{dt}=(2-x)^2<br />

    which is a differential equation of variables seperable type, and so:

    <br />
\int\frac{dx}{(2-x)^2}=\int dt<br />
,

    which may be integrated to give:

    <br />
\frac{1}{2-x}=t+const<br />

    and as x=0 when t=0,\ const=1/2, then rearranging:

    x=\frac{4t}{2t+1}

    RonL
    I believe the constant is -1/2 making the answer the same

    why can't this part

    <br />
\frac{1}{2-x}=t+const<br />

    be

    <br />
\frac{1}{2-x}+const=t<br />
    Last edited by nath_quam; July 17th 2006 at 05:05 AM.
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  14. #29
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by nath_quam
    ah true i'll get back to you i have to think

    ummm..... isn't a = the derivative of v
    As malaygoel stated, a is the time derivative of v.

    As to the previous post, yes the constant can be on either side of the equation, but they would be different constants...

    -Dan
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