arc length

[logistic_guy]

Find the arc-length of the curve

\(\displaystyle \boldsymbol{\gamma}(t) = (3t^2, t - 3t^3)\)

starting at \(\displaystyle t = 0\).

 

[khansaheb]

Find the arc-length of the curve

\(\displaystyle \boldsymbol{\gamma}(t) = (3t^2, t - 3t^3)\)

starting at \(\displaystyle t = 0\).

Please show us what you have tried and exactly where you are stuck.

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[logistic_guy]

Find the arc-length of the curve

\(\displaystyle \boldsymbol{\gamma}(t) = (3t^2, t - 3t^3)\)

starting at \(\displaystyle t = 0\).

The arc-length is:

\(\displaystyle s = \int_0^t \left\| \dot{\boldsymbol{\gamma}}(u) \right\| \ du = \int_0^t \sqrt{\dot{x}^2 + \dot{y}^2} \ du\)


\(\displaystyle \dot{x} = 6u\)


\(\displaystyle \dot{y} = 1 - 9u^2\)

Then,

\(\displaystyle s = \int_0^t \sqrt{(6u)^2 + (1 - 9u^2)^2} \ du\)


\(\displaystyle = \int_0^t \sqrt{36u^2 + 1 - 18u^2 + 81u^4} \ du\)


\(\displaystyle = \int_0^t \sqrt{18u^2 + 1 + 81u^4} \ du\)


\(\displaystyle = \int_0^t \sqrt{(9u^2 + 1)(9u^2 + 1)} \ du\)


\(\displaystyle = \int_0^t \sqrt{(9u^2 + 1)^2} \ du\)


\(\displaystyle = \int_0^t (9u^2 + 1) \ du\)

Then, the arc length is:

\(\displaystyle s = \textcolor{blue}{3t^3 + t}\)