Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The curve is $r = e^{-b\theta}$ where $b > 0$ and $θ \in [0, \infty)$.

I got that the arc length is $\frac{\sqrt{b^2 + 1}}{b}$ (is this correct?), but computing the centroid $(x, y)$ looks awful. I'm not sure where to start.

share|improve this question
    
Now, wouldn't x = (1/L) lim_T-->∞ ∫[0 to T] e^(-bθ) ds? But I don't know how to compute that integral!! –  Mark T Apr 12 '13 at 16:52
    
What is the centroid of a curve? A region in the plane has a centroid. –  Stefan Smith Apr 13 '13 at 1:16
add comment

2 Answers 2

up vote 1 down vote accepted

Your curve has the parametric representation $$\gamma:\quad \theta\mapsto(e^{-b\theta}\cos\theta, e^{-b\theta}\sin\theta)\qquad(0\leq\theta<\infty)\ .$$ It follows that $$ds=\sqrt{(x'(\theta))^2+(y'(\theta))^2}\ d\theta=\sqrt{1+b^2}\ e^{-b\theta}\ d\theta\ .$$ The centroid $(\xi,\eta)$ of $\gamma$ is characterized by the so-called moment equations $$\xi \ L(\gamma)=\int_\gamma x\ ds,\qquad \eta\ L(\gamma)=\int_\gamma y\ ds\ .$$ You already have computed $L(\gamma)={\sqrt{1+b^2}\over b}$. In addition we need $$\int_\gamma x\ ds=\sqrt{1+b^2}\int_0^\infty e^{-2b\theta}\ \cos\theta\ d\theta=\ldots={2b\sqrt{1+b^2}\over 1+4b^2}$$ and $$\int_\gamma y\ ds=\sqrt{1+b^2}\int_0^\infty e^{-2b\theta}\ \sin\theta\ d\theta=\ldots={\sqrt{1+b^2}\over 1+4b^2}\ .$$ It follows that $$\xi={2b^2\over 1+4b^2},\qquad \eta={b\over1+4b^2}\ .$$

share|improve this answer
add comment

I must admit I've never seen it asked that a student find the centroid of a polar curve, so I'm not sure of the context for your problem. I would guess, since the curve is a one-dimensional object in a two-dimensional space (the plane), that you are being asked to find the point at which one-half of the total arclength (which is finite) is reached, passing inward from $\theta = 0$. (And I'm leaving this as an answer, rather than a comment, since I don't have enough rep yet...)

EDIT: I concur with your arclength result, so you want to find the value of $\theta$ at which you reach an arclength of $\frac{\sqrt{1+b^2}}{2b}$, then evaluate the radius of the curve there.* If you're asked to give the Cartesian coordinates, you can then also carry out the requisite transformation.

*Interestingly, this result is independent of $b$.

share|improve this answer
    
I don't know how to do the centroid though! help! –  Mark T Apr 12 '13 at 17:29
    
You presumably calculated the arclength using –  RecklessReckoner Apr 12 '13 at 17:38
    
You presumably calculated the arclength using $\sqrt{1+b^2}\int^{\infty}_{0} e^{-b\theta} d\theta $ to get the answer you quoted. Now set the upper limit of the arclength integral to some unknown angle $\Theta$ ; you will get an arclength function which you set equal to $\frac{\sqrt{1+b^2}}{2b}$. You would then solve this for the angle $\Theta$; this is the angle at which the midpoint of the arclength occurs. You can put this into the polar curve equation to get $r$. With the polar coordinates in hand, you can convert these to rectangular coordinates. –  RecklessReckoner Apr 12 '13 at 17:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.