Question

Find the arclength of the curve \mathbf r(t = \langle 6 t^2, 2\sqrt{6} t, \ln t\rangle, 1 \le t \le 4

Answer 1

`\mathbf r(t)=\langle6t^2,2\sqrt6 t,\ln t\rangle`

`\implies(\mathrm d\mathbf r)/(\mathrm dt)=\left\langle12t,2\sqrt6,\frac1t\right\rangle`

The arc length is then given by


`\displaystyle\int_1^4\sqrt{(\mathrm d\mathbf r)/(\mathrm dt)\cdot(\mathrm d\mathbf r)/(\mathrm dt)}\,\mathrm dt`

`\displaystyle\int_1^4\sqrt{144t^2+24+\frac1{t^2}}\,\mathrm dt`

`\displaystyle\int_1^4√(\left(12t+\frac1t\right)^2)\,\mathrm dt`

`\displaystyle\int_1^4\left(12t+\frac1t\right)\,\mathrm dt`

`=90+\ln4`