## 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.

1. frank0520

Here is a picture of the question.

2. Loser66

@oldrin.bataku

3. anonymous

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

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

\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

is the $r^{2}$ supposed to be with the 625 or the 9cos(2θ)?

7. anonymous

@frank0520 i multiplied the $$r$$ into the square root

8. IrishBoy123

$$r^2 = 0.36 \ cos 2 \theta$$ $$2 \ r \ r' = (-2) \ 0.36 \ sin \ 2 \theta$$ $$r' = \frac{- 0.36 \ sin \ 2 \theta}{r}$$ $$ds = \sqrt{0.36 \ cos 2 \theta + \frac{(0.36)^2 sin^2 2 \theta}{0.36 \ cos 2 \theta }}$$ $$ds = \frac{0.6}{\sqrt{cos 2 \theta }}$$

9. IrishBoy123

|dw:1436954333151:dw|

10. IrishBoy123

$$dm = \rho ds$$ $$m = \int \rho ds = \int \sqrt{0.36 \ cos 2 \theta} \frac{0.6}{\sqrt cos 2 \theta} d \theta = 0.36 \int d \theta$$ $$dM_o = r cos \theta \rho \ ds = r \ cos \theta \ r \ \frac{0.6}{\sqrt cos 2 \theta} d \theta$$ $$M_o = \int r^2 \ cos \theta \ \frac{0.6}{\sqrt cos 2 \theta} d \theta = 0.6^3 \int \sqrt{cos 2 \theta} \ cos \theta \ d \theta$$ $$\bar x = \frac {M_o}{m}$$

11. frank0520

@IrishBoy123 what limits did you use to get that answer?

12. IrishBoy123

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

@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.

14. IrishBoy123