Question

Find length of curve the r = sin^2(θ/2),0<=θ<=μ, a>0.

Answer 1

The length of a curve C is given by the integral,

`\displaystyle\int_C\mathrm ds`

where the line element ds is

`\mathrm ds=\sqrt{\left((\mathrm dx)/(\mathrm dt)\right)^2+\left((\mathrm dy)/(\mathrm dt)\right)^2}\,\mathrm dt`

where
`x=x(t)` and
`y=y(t)` are parameterizations of C.

In this case, we have

`x(\theta)=r(\theta)\cos\theta`

`y(\theta)=r(\theta)\sin\theta`

Differentiate with respect to
`\theta` to get

`(\mathrm dx)/(\mathrm d\theta)=(\mathrm dr)/(\mathrm d\theta)\cos\theta-r(\theta)\sin\theta`

`(\mathrm dy)/(\mathrm d\theta)=(\mathrm dr)/(\mathrm d\theta)\sin\theta+r(\theta)\cos\theta`

`(\mathrm dr)/(\mathrm d\theta)=\sin\left(\frac\theta2\right)\cos\left(\frac\theta2\right)=\frac12\sin\theta`

So the arc length is

`\displaystyle\int_C\mathrm ds=\int_0^a√(\left(\frac12\sin\theta\cos\theta-\sin^2\left(\frac\theta2\right)\sin\theta\right)^2+\left(\frac12\sin\theta\sin\theta+\sin^2\left(\frac\theta2\right)\cos\theta\right)^2)\,\mathrm d\theta`

`=\displaystyle\int_0^a√(\sin^2\left(\frac\theta2\right))\,\mathrm d\theta`

`=\displaystyle\int_0^a\sqrt{\frac{1-\cos\theta}2}\,\mathrm d\theta`