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frank0520

  • one year ago

Let C be one lobe of the lemniscate 100r^2 =36cos(2θ), -π/4≤θ≤π/4. Suppose a wire in the shape of C has a linear density function ρ(x,y)=sqrt(x^2 +y^2). Find the center of mass of the wire. Show all your computations but round your final answers to two decimal places. Hint: Use implicit differentiation on the equation above; in polar coordinates, ds=sqrt(r^2 + (r')^2) dθ is particularly simple. In fact, ρ ds=c dθ for some constant c.

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  1. frank0520
    • one year ago
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    Here is a picture of the question.

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  2. Loser66
    • one year ago
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    @oldrin.bataku

  3. anonymous
    • one year ago
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    so \((x,y)=(r\cos\theta,r\sin\theta)\implies (dx,dy)=(\cos\theta\,dr -r\sin\theta\,d\theta,\sin\theta\,dr +r\cos\theta\,d\theta)\) so it follows: $$dx^2=\cos^2\theta\, dr^2-2r\cos\theta\sin\theta\,dr\,d\theta+r^2\sin^2\theta\,d\theta^2\\dy^2=\sin^2\theta\, dr^2+2r\cos\theta\sin\theta\,dr\,d\theta+r^2\cos^2\theta\,d\theta^2$$ so we get \(ds=\sqrt{dx^2+dy^2}=\sqrt{(\cos^2\theta+\sin^2\theta)dr^2+r^2(\cos^2\theta+\sin^2\theta)d\theta^2}\) so $$ds=d\theta\sqrt{\left(\frac{dr}{d\theta}\right)^2+r^2}$$

  4. anonymous
    • one year ago
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    now $$100r^2=36\cos(2\theta)\\200r\frac{dr}{d\theta}=-72\sin(2\theta)\\\frac{dr}{d\theta}=-\frac{72\sin(2\theta)}{200r}=-\frac{9\sin(2\theta)}{25r}$$ so $$ds=d\theta\sqrt{\frac{81\sin^2(2\theta)}{625r^2}+\frac{9\cos(2\theta)}{25}}$$ so $$m_s=M\int_C\rho\,ds=M\int_{-\pi/4}^{\pi/4}r \sqrt{\frac{81\sin^2(2\theta)}{625r^2}+\frac{9\cos(2\theta)}{25}}\,d\theta$$

  5. anonymous
    • one year ago
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    $$\begin{align*}\sqrt{\frac{81\sin^2(2\theta)}{625}+r^2\frac{9\cos(2\theta)}{25}}&=\frac35\sqrt{\frac{9}{25}\sin^2(2\theta)+r^2 \cos^2(2\theta)}\\&=\frac3{25}\cos(2\theta)\sqrt{9\tan^2(2\theta)+25r^2}\\&=\frac3{25}\cos(2\theta)\sqrt{9\tan^2(2\theta)+9\cos(2\theta)}\end{align*}$$

  6. frank0520
    • one year ago
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    is the \[r^{2}\] supposed to be with the 625 or the 9cos(2θ)?

  7. anonymous
    • one year ago
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    @frank0520 i multiplied the \(r\) into the square root

  8. IrishBoy123
    • one year ago
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    \(r^2 = 0.36 \ cos 2 \theta \) \(2 \ r \ r' = (-2) \ 0.36 \ sin \ 2 \theta \) \(r' = \frac{- 0.36 \ sin \ 2 \theta}{r}\) \(ds = \sqrt{0.36 \ cos 2 \theta + \frac{(0.36)^2 sin^2 2 \theta}{0.36 \ cos 2 \theta }}\) \(ds = \frac{0.6}{\sqrt{cos 2 \theta }}\)

  9. IrishBoy123
    • one year ago
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    |dw:1436954333151:dw|

  10. IrishBoy123
    • one year ago
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    \(dm = \rho ds\) \(m = \int \rho ds = \int \sqrt{0.36 \ cos 2 \theta} \frac{0.6}{\sqrt cos 2 \theta} d \theta = 0.36 \int d \theta \) \(dM_o = r cos \theta \rho \ ds = r \ cos \theta \ r \ \frac{0.6}{\sqrt cos 2 \theta} d \theta\) \(M_o = \int r^2 \ cos \theta \ \frac{0.6}{\sqrt cos 2 \theta} d \theta = 0.6^3 \int \sqrt{cos 2 \theta} \ cos \theta \ d \theta \) \(\bar x = \frac {M_o}{m}\)

  11. frank0520
    • one year ago
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    @IrishBoy123 what limits did you use to get that answer?

  12. IrishBoy123
    • one year ago
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    they give the limits in question as -π/4≤θ≤π/4 so those, though you could go 0≤θ≤π/4 as its symmetrical about x axis there could easily be some silly algebraic glitch in there so interested in seeing if you agree

  13. frank0520
    • one year ago
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    @oldrin.bataku @IrishBoy123 can verily if my work is correct? mass=\[.36\int\limits_{-\pi/4}^{\pi/4}1d \theta\]=.565487 My=\[.6^3 \int\limits_{-\pi/4}^{\pi/4}\sqrt{\cos(2\theta)}\cos(\theta)d \theta\] =.239916 Mx=\[.6^3\int\limits_{-\pi/4}^{\pi/4}\sqrt{\cos(2\theta)}\sin(\theta)d\] =0 and I get (.42, 0) for the center of mass.

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