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Differential Equation question - C4 watch

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    Just working through my revision guide and I can't do this question.

    ~ At time t minutes an ink stain has area Acm^2. When t=1, A=1, and the rate of increase of A is given by t^2\frac{dA}{dt}= A\sqrt {A}

    a) Find an expression for A in terms of t.

    I got \int \frac{1}{A\sqrt {A}}  dA = \int \frac{1}{t^2}  dt

    but I'm not sure if this is right, and I'm a bit unsure as to how you integrate \frac{1}{A\sqrt {A}}


    The answer is A = \frac{4t^2}{(1+t)^2}

    Thanks
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    (Original post by allyourbase)
    Just working through my revision guide and I can't do this question.

    ~ At time t minutes an ink stain has area Acm^2. When t=1, A=1, and the rate of increase of A is given by t^2\frac{dA}{dt}= A\sqrt {A}

    a) Find an expression for A in terms of t.

    I got \int \frac{1}{A\sqrt {A}}  dA = \int \frac{1}{t^2}  dt

    but I'm not sure if this is right, and I'm a bit unsure as to how you integrate \frac{1}{A\sqrt {A}}


    The answer is A = \frac{4t^2}{(1+t)^2}

    Thanks
    You need to write it as a power.

    A^{-1}\times A^{-\frac{1}{2}}=?

    Simplify, and integrate from there.

    Quote me if you'd like any more help
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    well

    1/(A square root A) = A ^ (-3/2) becausre square root of A is A^1/2,
    does that answer your question?

    beaten to it again lol
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    (Original post by lordcrusade9)
    well

    1/(A square root A) = A ^ (-3/2) becausre square root of A is A^1/2,
    does that answer your question?

    beaten to it again lol
    Just glad mine's right - look at my thread history!
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    (Original post by Muppety_Kid)
    You need to write it as a power.

    A^{-1}\times A^{-\frac{1}{2}}=?

    Simplify, and integrate from there.

    Quote me if you'd like any more help
    Ok, I get

     \int A^\frac {-3}{2} dA = \int t^{-2} dt

     -2A^\frac{-1}{2} = -t^{-1} + c

    -2 = -1 + c \to c = -1


     -2A^\frac{-1}{2}= -t^{-1} - 1

    2A^\frac{-1}{2}= t^{-1} + 1

    A^\frac{-1}{2} = \frac {1}{2} (t^{-1} - 1)

    \sqrt {A} = \frac {2}{(t^{-1} - 1)}

     A = \frac {4}{(t^{-1}-1)^2}


    Anyone see where I'm going wrong? Sorry for being crap at maths
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    (Original post by allyourbase)
    Ok, I get

     \int A^\frac {-3}{2} dA = \int t^{-2} dt

     -2A^\frac{-1}{2} = -t^{-1} + c

    -2 = -1 + c \to c = -1


     -2A^\frac{-1}{2}= -t^{-1} - 1

    2A^\frac{-1}{2}= t^{-1} + 1

    A^\frac{-1}{2} = \frac {1}{2} (t^{-1} - 1)

    \sqrt {A} = \frac {2}{(t^{-1} - 1)}

     A = \frac {4}{(t^{-1}-1)^2}


    Anyone see where I'm going wrong? Sorry for being crap at maths
    Ugh, you're doing Edexcel aren't you? Me too

    Give me 5 minutes to work it through myself...
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    (Original post by allyourbase)
    Ok, I get

     \int A^\frac {-3}{2} dA = \int t^{-2} dt

     -2A^\frac{-1}{2} = -t^{-1} + c

    -2 = -1 + c \to c = -1


     -2A^\frac{-1}{2}= -t^{-1} - 1

    2A^\frac{-1}{2}= t^{-1} + 1

    A^\frac{-1}{2} = \frac {1}{2} (t^{-1} - 1)

    \sqrt {A} = \frac {2}{(t^{-1} - 1)}

     A = \frac {4}{(t^{-1}-1)^2}


    Anyone see where I'm going wrong? Sorry for being crap at maths
    That's right (except the minus sign you seem to have picked up). Multiply top and bottom by t^2.
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    (Original post by generalebriety)
    That's right (except the minus sign you seem to have picked up). Multiply top and bottom by t^2.
    Just doing it now. I see where the minus sign has come in - look at my 5th and 6th lines lol.
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    (Original post by allyourbase)

    2A^\frac{-1}{2}= t^{-1} + 1

    A^\frac{-1}{2} = \frac {1}{2} (t^{-1} - 1)

    You see, where you halved both sides, you've taken t^{-1} - 1 instead of t^{-1} + 1
    Is that right?
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    (Original post by generalebriety)
    That's right (except the minus sign you seem to have picked up). Multiply top and bottom by t^2.
    Would you mind showing me how that works?

    When I times top + bottom by t^2 I get

    \sqrt {A} = \frac {2t^2}{t+t^2}
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    (Original post by allyourbase)
    Would you mind showing me how that works?

    When I times top + bottom by t^2 I get

    \sqrt {A} = \frac {2t^2}{t+t^2}
    :confused:

    That wasn't the line I wanted you to work with. Try multiplying the line below that by t^2/t^2. :p:
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    (Original post by DevonHowell)
    You see, where you halved both sides, you've taken t^{-1} - 1 instead of t^{-1} + 1
    Is that right?
    Is that rhetoric? I'm genuinely confused :confused:

    If you're asking because you're unsure yourself, it's in the spoiler below.

    Spoiler:
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    No, it's not right. It should be t^{-1}+1 - he's forgotten to take out the - all the way through on the right hand side.
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    (Original post by Muppety_Kid)
    Is that rhetoric? I'm genuinely confused :confused:

    If you're asking because you're unsure yourself, it's in the spoiler below.

    Spoiler:
    Show
    No, it's not right. It should be t^{-1}+1 - he's forgotten to take out the - all the way through on the right hand side.
    For some reason I'd changed a + to a - . Dunno why.. Just a mistake.
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    (Original post by generalebriety)
    :confused:

    That wasn't the line I wanted you to work with. Try multiplying the line below that by t^2/t^2. :p:
    Yeh, it's a crafty trick, but sometimes they multiply the top and bottom by the same (so the fraction's in the same "ratio"). It's not strictly in its simplest form then, but it's a lot easier to work with than what you've got currently, especially if it asked you to find the area when t is a certain value.
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    (Original post by allyourbase)
    For some reason I'd changed a + to a - . Dunno why.. Just a mistake.
    Sorry, I wasn't having a go at you. The other guy asked if you were right when you did that, and I didn't know if he was trying to give you a hint, or he didn't understand.
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    I have tried to help by highlighting a possible mistake. I was not 100% and I'd written it before I saw someone had already mentioned the mistake.
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    (Original post by generalebriety)
    :confused:

    That wasn't the line I wanted you to work with. Try multiplying the line below that by t^2/t^2. :p:
    ****.


    \frac {4t^2}{t^2(t^{-2}+2t^{-1}+1)}


    \frac {4t^2}{1+2t+t^2)}


    \frac {4t^2}{(t+1)^2}

    thanks <3 x23532
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    (Original post by DevonHowell)
    I have tried to help by highlighting a possible mistake. I was not 100% and I'd written it before I saw someone had already mentioned the mistake.
    I wasn't having a go at you either.

    Both of you, please, DON'T HURT ME! :bawling:
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    (Original post by allyourbase)
    ****.


    \frac {4t^2}{t^2(t^{-2}+2t^{-1}+1)}


    \frac {4t^2}{1+2t+t^2)}


    \frac {4t^2}{(t+1)^2}

    thanks <3 x23532
    For what it's worth:

    a^2b^2 = (ab)^2, and so

    t^2(t^{-1} + 1)^2 = [t(t^{-1} + 1)]^2 = (1 + t)^2.
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    (Original post by generalebriety)
    For what it's worth:

    a^2b^2 = (ab)^2, and so

    t^2(t^{-1} + 1)^2 = [t(t^{-1} + 1)]^2 = (1 + t)^2.
    I blame my maths teachers of course

    Thanks
 
 
 
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