## anonymous one year ago Find the area bounded by the curve x=t-1/t, y=t=1/t, y=65/8

1. Astrophysics

Is that $x= t - \frac{ 1 }{ t }~~~y=t-\frac{ 1 }{ t }~~~y=\frac{ 65 }{ 8 }$

2. ali2x2

@Astrophysics Yes it looks like it

3. ali2x2

y=t=

4. hartnn

y could be t+1/t since, + and = are on the same key of the keyboard

5. Astrophysics

I think you're right hartnn! So areas in parametric curves we have $\int\limits_{a}^{b}g(t)f'(t) dt$

6. Astrophysics

|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

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

I don't plan on doing this whole thing for you, so if you're there, please ask for help.

9. ganeshie8

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

$y^2 - x^2 = \left(t+\frac{1}{t}\right)^2 - \left(t-\frac{1}{t}\right)^2=4*t*\frac{1}{t} = 4$

11. ganeshie8

so the problem is equivalent to finding the area between curves : $$y^2-x^2=4$$ and $$y=65/8$$

12. Astrophysics

So much easier xD

13. ganeshie8

Not really, looks both require almost same effort haha!

14. Astrophysics

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

still not sure how to get the right answer.

16. Astrophysics

Ok well, doing it the method I suggested, what values did you get for x?

17. Astrophysics

$x= t-\frac{ 1 }{ t } \implies \frac{ 1 }{ 8 } - \frac{ 1 }{ \frac{ 1 }{ 8 } } = ...$ and $x= 8-\frac{ 1 }{ 8 } = ...$

18. Astrophysics

We'll go over this step by step, so just find what x's are and we can continue :)