# Homework Help: Arc lengths of polar curves

1. Apr 5, 2010

### JOhnJDC

1. The problem statement, all variables and given/known data
Find the total length of the cardioid r=a(1-cos theta)

2. Relevant equations

ds2=r2dtheta2+dr2

ds= integral from beta to alpha sqrt[r2 + (dr/d theta)2]dtheta

3. The attempt at a solution

dr=a(sin theta)d theta

ds2=a2(1-cos theta)2d theta2 + a2sin2theta (d theta2), which reduces to:

ds=21/2a(1-cos theta)1/2(d theta)

I'm good up to this point. My book says that the above equation for ds simplifies to:

2a|sin1/2theta|d theta

I don't see how to arrive at this simplification. Can someone please explain? Many thanks.

2. Apr 5, 2010

### JOhnJDC

I have a separate, basic question related to arc lengths of polar curves. I'm trying to derive the formula ds2=r2dtheta2+dr2, which enables me to compute arc lengths of polar curves by integration, from the rectangular equation for the differential element of arc length, namely ds2=dx2 + dy2.

I know that I need to use the transformation equations:

x=r cos theta and y= r sin theta

My book says that I need to differentiate with respect to theta using the product rule to obtain:

dx/dtheta = -r sin theta + cos theta dr/dtheta

When I differentiate, I get dx/dtheta = -r sin theta. I'm not sure how the product rule fits in, which I presume is where does the second half come from (+ cos theta dr/dtheta). I think I'm missing something basic here.

3. Apr 5, 2010

### Dick

It's a trig trick. 1-cos(theta)=2*cos^2(theta/2). Half angle formula.

4. Apr 5, 2010

### JOhnJDC

I'm not so strong in trig, but did you mean to say 1-cos(theta)=2*sin2theta/2?

If 1-cos2theta = 2sin2theta, then
1-cos theta = 2sin2theta/2 because we halved the angle cos2theta on the left side?

5. Apr 5, 2010

### Dick

Sure. Sorry. That's what I meant to say. Must be getting late.

6. Apr 5, 2010

### JOhnJDC

Thanks, Dick.

Any ideas on my proof question from above (my 2nd post)?

7. Apr 6, 2010

### Dick

I would use the product rule on the differentials. dx=d(r*cos(t))=dr*cos(t)+r*d(cos(t))=dr*cos(t)-r*sin(t)dt. Do the same with dy and evaluate dx^2+dy^2.

8. Apr 6, 2010

### Dick

I would use the product rule on the differentials. dx=d(r*cos(t))=dr*cos(t)+r*d(cos(t))=dr*cos(t)-r*sin(t)dt. Do the same with dy and evaluate dx^2+dy^2.