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

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

Mathematics
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Is that \[x= t - \frac{ 1 }{ t }~~~y=t-\frac{ 1 }{ t }~~~y=\frac{ 65 }{ 8 }\]
@Astrophysics Yes it looks like it
y=t=

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y could be t+1/t since, + and = are on the same key of the keyboard
I think you're right hartnn! So areas in parametric curves we have \[\int\limits_{a}^{b}g(t)f'(t) dt\]
|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 \[t=\frac{ 1 }{ 8 } ~~~and ~~~ t = 8\]
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.
I don't plan on doing this whole thing for you, so if you're there, please ask for help.
Alternatively, if you don't like parametric curves, you may try eliminating the parameter easily by noticing that \[(a+b)^2-(a-b)^2 = 4ab\]
\[y^2 - x^2 = \left(t+\frac{1}{t}\right)^2 - \left(t-\frac{1}{t}\right)^2=4*t*\frac{1}{t} = 4 \]
so the problem is equivalent to finding the area between curves : \(y^2-x^2=4\) and \(y=65/8\)
So much easier xD
Not really, looks both require almost same effort haha!
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.
still not sure how to get the right answer.
Ok well, doing it the method I suggested, what values did you get for x?
\[x= t-\frac{ 1 }{ t } \implies \frac{ 1 }{ 8 } - \frac{ 1 }{ \frac{ 1 }{ 8 } } = ...\] and \[x= 8-\frac{ 1 }{ 8 } = ...\]
We'll go over this step by step, so just find what x's are and we can continue :)

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