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anonymous
 4 years ago
Challenge:
Compute the are of the loop of curve
\[y^2=x^2\frac{1+x}{1x}\]
anonymous
 4 years ago
Challenge: Compute the are of the loop of curve \[y^2=x^2\frac{1+x}{1x}\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lets do brainstorming everyone give all ideas coming to mind

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got it ans is \[2\int\limits_{1}^{0}x^2\sqrt{\frac{1+x}{1x}}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0\[y=2\int_{1}^{0}x\sqrt{\frac{1+x}{1x}}dx\]isn't it? why is yours still x^2 squared?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0is that a no? is should be x^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0turing yours was correct

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0\[A=\int_{1}^{0}x\sqrt{\frac{1+x}{1x}}dx\]\[u=1x\]groovy :)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0but my sub wont' work

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that won't work.. i don't think

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Easier if you rationalise miltiply sqrt(1x) in numerator and denominator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I thought it was arc....I assume everyone else is looking at area...lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0compute the area of the loop of the curve

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Simplifies to \[\int\limits\limits_{1}^{2}(\frac{x}{\sqrt{1x^2}}\sqrt{1x^2}+\frac{1}{{\sqrt1x^2}})dx\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0where did the 2 come from? I have something else

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Typing error i menat it to be outside

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0which should give the anwer approx.....using winplot .4292 But of course the mathematics is more fun than letting the computer do all of the work.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0ok, I still have a different expression though I'm trying to figure out if they're the same...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0am I doing something wrong?\[A=\int_{1}^{0}x\sqrt{\frac{1+x}{1x}}\cdot\sqrt{\frac{1+x}{1+x}}dx=\int_{1}^{0}x\frac{1+x}{\sqrt{1x}}dx\]\[=\int_{1}^{0}\frac{x}{\sqrt{1x}}+\frac{x^2}{\sqrt{1x}}dx\]because my expression looks different ...or am I just tired?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah (1x)(1+x)=1x^2 you look tired

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0ok that's just a typo though

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0I meant that the expression is different as far as I seeam I doing something wrong?\[A=2\int_{1}^{0}x\sqrt{\frac{1+x}{1x}}\cdot\sqrt{\frac{1+x}{1+x}}dx=2\int_{1}^{0}x\frac{1+x}{\sqrt{1x^2}}dx\]\[=2\int_{1}^{0}\frac{x}{\sqrt{1x^2}}+\frac{x^2}{\sqrt{1x^2}}dx\]is what I meant

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If that is the case add 1and subtract 1 from x^2 to arrive at mine

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I feel double integrals no worth this problem it is not that difficult
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