Question

Find the length of the curve, l. y^2 = 4(x+5)^3 text(, ) 0<=x<=1 text(, ) y>0 g

Answer 1

Parameterize the curve by


`\mathbf r(t)=\begin{cases}x(t)=t-5\\y(t)=2t^(3/2)\end{cases}`

Then


`\mathbf r'(t)=\langle1,3t^(1/2)\implies\|\mathbf r'(t)\|=√(1+9t)`

The length of the curve
`\mathcal C` is then


`\displaystyle\int_(\mathcal C)\mathrm dS=\int_(t=5)^(t=6)\left\|\mathbf r'(t)\right\|\,\mathrm dt=\int_5^6√(1+9t)\,\mathrm dt`

`=\displaystyle\frac19\int_5^6√(1+9t)\,\mathrm d(1+9t)=\frac2{27}(55^(3/2)-46^(3/2))\approx7.10398`