Question #335070

Find the length of cardioid r = 1+sinθ\theta


1
Expert's answer
2022-05-05T14:34:44-0400

This is the equation of acardioid. The polar equation is expressed in terms of a periodic function, so we have to know its period to get the exact length (and not just a portion or some excess of it).

r=1+sinθr=1+\sin\theta has period T=2π (the function can be obtained as a vertical translation of the sine function in the plane of coordinates (θ,r)).The length of a periodic polar curve can be computed by integrating the arc length on a complete period of the function, i.e. on an interval I of length T=2π:

L=IdsL=\int_Ids

where ds=r2+(drdθ)2dθds=\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta

So we have to compute the derivative:

drdθ=ddθ(1+sinθ)=cosθ\frac{dr}{d\theta}=\frac{d}{d\theta}(1+\sin{\theta})=cos\theta

and this implies

ds=(1+sinθ)2+cosθ2dθ=1+2sinθ+sin2θ+cos2θdθds=\sqrt{(1+sin\theta)^2+cos\theta^2}d\theta=\sqrt{1+2sin\theta+sin^2\theta+cos^2\theta}d\theta


By the Pythagorean trigonometric identity cos

cos2θ+sin2θ=1cos^2\theta+sin^2\theta=1

ds=2+2sinθdθ=2(1+sinθ)dθds=\sqrt{2+2sin\theta}d\theta=\sqrt{2(1+sin\theta)}d\theta

Integrating that square root is not easy... we'd better take advantage of trigonometric identities. We have 1+sinθ\sqrt{1+sin\theta}

 (multiplied by a numerical coefficient) and we would like to get rid of that square root. The half angle formula for cosine states that

 cos2(α2)=1+cosα2cos^2(\frac{\alpha}{2})=\frac{1+cos\alpha}{2}.

We can use this with sine too, because if

α=θπ2\alpha=\theta-\frac{\pi}{2}

 then

 cosα=cos(θπ2)=sinθcos\alpha=cos(\theta-\frac{\pi}{2})=sin\theta.

So we substitute in the formula and get cos2θπ22=1+cos(θπ2)2cos^2\frac{\theta-\frac{\pi}{2}}{2}=\frac{1+cos (\theta-\frac{ \pi}{2})}{2}

and then

2cos2(θ4π4)=1+sinθ2cos^2 {(\frac{\theta}{4}-\frac{\pi}{4})}=1+sin\theta

ds=4cos2(θ4π4)dθ=2cos(θ4π4)dθds=\sqrt{4 cos^2 {(\frac{\theta}{4}-\frac{\pi}{4})}}d\theta= 2| cos {(\frac{\theta}{4}-\frac{\pi}{4})} |d\theta

Now we can integrate this on an interval of length 2π. To do that, we would like to get rid of the absolute value.

We know that

cos(x2π4)0    π2+2πkx2π4π2+2πkcos {(\frac{x}{2}-\frac{\pi}{4})} \eqslantgtr0 \iff -\frac{\pi}{2}+2\pi k \eqslantless \frac{x}{2}-\frac{\pi}{4}\eqslantless \frac{\pi}{2}+2\pi k

for k any integer value, so we get that −

π2π+4πkx32π+4πk-\frac{\pi}{2}\pi+4\pi k \eqslantless x\eqslantless \frac{3}{2}\pi+4\pi k

In the end, we can integrate the arc length on 

I=[π2;32π]I=[-\frac{\pi}{2}; \frac{3}{2}\pi], since this is an interval of length 2π. We can remove the absolute value, since cos(x2π2)0cos (\frac{x}{2}-\frac{\pi}{2}) \eqslantgtr 0 on I. π23π22cos(θ2π4)dθ=4sinπ24sinπ2=4+4=8\int_{-\frac{\pi}{2}}^\frac{3\pi}{2} 2cos (\frac{\theta}{2}-\frac{\pi}{4})d\theta=4sin\frac{\pi}{2}-4sin\frac{-\pi}{2}=4+4=8



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