Prove the ellipses:
where k is an arbitrary real number
have the same area as the ellipse
Last edited by Paradoxica; 20 Nov 2017 at 12:53 AM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Letting the value of the solution to x = cos(x) be r, the following area expressions are obtained through integration:
A₁ = sin(r) - r²/2
A₂ = 1 - sin(r) + r²/2
A₂ - A₁ = 1 - 2sin(r) + r²
r > sin(r) (proof is trivial and left as an exercise)
- 2sin(r) > - 2r
1 - 2sin(r) + r² > 1 - 2r + r² = (1-r)² > 0
A₂ - A₁ > 0
A₂ > A₁
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
These pls
Capture.JPG
Capture 2.JPG
Could show me some working pls. Sorry for the trouble
Never mind, I got it. Cheers, the advice made things easier.
This one pls
Capture 3.JPG
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Appreciate some help with this: Find the range of values of |z| and arg(z) for |z-4-4i|=2sqrt(2)
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