anonymous
  • anonymous
Challenge: Compute the are of the loop of curve \[y^2=x^2\frac{1+x}{1-x}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Anyone
anonymous
  • anonymous
Lets do brainstorming everyone give all ideas coming to mind
anonymous
  • anonymous
integrate?

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anonymous
  • anonymous
I got it ans is \[2\int\limits_{-1}^{0}x^2\sqrt{\frac{1+x}{1-x}}\]
anonymous
  • anonymous
\[2-\frac{\pi}{2}\]
TuringTest
  • TuringTest
\[y=2\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}dx\]isn't it? why is yours still x^2 squared?
TuringTest
  • TuringTest
A=... sorry
anonymous
  • anonymous
Sorry
TuringTest
  • TuringTest
is that a no? is should be x^2
TuringTest
  • TuringTest
or should not?
anonymous
  • anonymous
not x^2
anonymous
  • anonymous
turing yours was correct
anonymous
  • anonymous
Should be
TuringTest
  • TuringTest
\[A=\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}dx\]\[u=1-x\]groovy :)
TuringTest
  • TuringTest
but my sub wont' work
anonymous
  • anonymous
that won't work.. i don't think
anonymous
  • anonymous
Easier if you rationalise miltiply sqrt(1-x) in numerator and denominator
TuringTest
  • TuringTest
oh yeah, ok
anonymous
  • anonymous
I thought it was arc....I assume everyone else is looking at area...lol
anonymous
  • anonymous
Sorry 1+x
anonymous
  • anonymous
compute the area of the loop of the curve
anonymous
  • anonymous
=area under curve
anonymous
  • anonymous
Simplifies to \[\int\limits\limits_{-1}^{2}(\frac{x}{\sqrt{1-x^2}}-\sqrt{1-x^2}+\frac{1}{{\sqrt1-x^2}})dx\]
TuringTest
  • TuringTest
where did the 2 come from? I have something else
anonymous
  • anonymous
Typing error i menat it to be outside
anonymous
  • anonymous
which 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
  • TuringTest
ok, I still have a different expression though I'm trying to figure out if they're the same...
TuringTest
  • TuringTest
am I doing something wrong?\[A=\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}\cdot\sqrt{\frac{1+x}{1+x}}dx=\int_{-1}^{0}x\frac{1+x}{\sqrt{1-x}}dx\]\[=\int_{-1}^{0}\frac{x}{\sqrt{1-x}}+\frac{x^2}{\sqrt{1-x}}dx\]because my expression looks different ...or am I just tired?
TuringTest
  • TuringTest
2 times all that*
anonymous
  • anonymous
yeah (1-x)(1+x)=1-x^2 you look tired
TuringTest
  • TuringTest
ok that's just a typo though
TuringTest
  • TuringTest
I meant that the expression is different as far as I seeam I doing something wrong?\[A=2\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}\cdot\sqrt{\frac{1+x}{1+x}}dx=2\int_{-1}^{0}x\frac{1+x}{\sqrt{1-x^2}}dx\]\[=2\int_{-1}^{0}\frac{x}{\sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}}dx\]is what I meant
anonymous
  • anonymous
If that is the case add 1and subtract 1 from x^2 to arrive at mine
TuringTest
  • TuringTest
oh gotcha :P
nikvist
  • nikvist
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anonymous
  • anonymous
Nice!
anonymous
  • anonymous
I feel double integrals no worth this problem it is not that difficult

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