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anonymous

  • one year ago

Find the area bounded by the curve x=t-1/t, y=t=1/t, y=65/8

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  1. Astrophysics
    • one year ago
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    Is that \[x= t - \frac{ 1 }{ t }~~~y=t-\frac{ 1 }{ t }~~~y=\frac{ 65 }{ 8 }\]

  2. ali2x2
    • one year ago
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    @Astrophysics Yes it looks like it

  3. ali2x2
    • one year ago
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    y=t=

  4. hartnn
    • one year ago
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    y could be t+1/t since, + and = are on the same key of the keyboard

  5. Astrophysics
    • one year ago
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    I think you're right hartnn! So areas in parametric curves we have \[\int\limits_{a}^{b}g(t)f'(t) dt\]

  6. Astrophysics
    • one year ago
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    |dw:1439044557140:dw| now to see where the curve touches the line we set \[t+\frac{ 1 }{ t } = \frac{ 65 }{ 8 }\] solving for t will tell us at which time our curve will touch the line, then you can make a graph to make your curve. So we get our time to be \[t=\frac{ 1 }{ 8 } ~~~and ~~~ t = 8\]

  7. Astrophysics
    • one year ago
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    To find our values for x, you can make a graph as I suggested (just too keep your values and see what's going on), or you can just do when t = 1/8 \[x= t-\frac{ 1 }{ t } \implies \frac{ 1 }{ 8 } - \frac{ 1 }{ \frac{ 1 }{ 8 } } = ...\] \[x= 8-\frac{ 1 }{ 8 } = ...\] You'll then have your x,y,t values and you should be able to graph it, that's the major part, then setting up the integral shouldn't be too bad.

  8. Astrophysics
    • one year ago
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    I don't plan on doing this whole thing for you, so if you're there, please ask for help.

  9. ganeshie8
    • one year ago
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    Alternatively, if you don't like parametric curves, you may try eliminating the parameter easily by noticing that \[(a+b)^2-(a-b)^2 = 4ab\]

  10. ganeshie8
    • one year ago
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    \[y^2 - x^2 = \left(t+\frac{1}{t}\right)^2 - \left(t-\frac{1}{t}\right)^2=4*t*\frac{1}{t} = 4 \]

  11. ganeshie8
    • one year ago
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    so the problem is equivalent to finding the area between curves : \(y^2-x^2=4\) and \(y=65/8\)

  12. Astrophysics
    • one year ago
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    So much easier xD

  13. ganeshie8
    • one year ago
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    Not really, looks both require almost same effort haha!

  14. Astrophysics
    • one year ago
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    Ah, I see, I just think it's easier to see the curve with your method...but yeah I think it's equal effort hehe.

  15. anonymous
    • one year ago
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    still not sure how to get the right answer.

  16. Astrophysics
    • one year ago
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    Ok well, doing it the method I suggested, what values did you get for x?

  17. Astrophysics
    • one year ago
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    \[x= t-\frac{ 1 }{ t } \implies \frac{ 1 }{ 8 } - \frac{ 1 }{ \frac{ 1 }{ 8 } } = ...\] and \[x= 8-\frac{ 1 }{ 8 } = ...\]

  18. Astrophysics
    • one year ago
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    We'll go over this step by step, so just find what x's are and we can continue :)

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