Question

Find the length of the curve.

3. \(r(t) = \,i + {t^2}\;j + \;{t^3}\;k,\;0 \le t \le 1\)

Short answer

The length of the given curve is approximately \(\frac{1}{{27}}\left( {{{(13)}^{\frac{3}{2}}} - 8} \right)\)

Step-by-step solution

Step 1: Given data and magnitude of derivative function

We have to find the length of the given curve. To do this, first of all, let us calculate the derivative of each of the components for the given vector function we get

\(r'(t) = 0i + 2tj + 3{t^2}k\)

Now let us find the magnitude of the derivative function that is,\(\begin{aligned}{l}\left| {r'(t)} \right| = \sqrt {{{(0)}^2} + {{(2t)}^2} + {{(3{t^2})}^2}} \\ = \sqrt {4{t^2} + 9{t^4}} \\ = t\sqrt {4 + 9{t^2}} \end{aligned}\)

Step 2: Length of the arc

Plug into the arc length formula

\(\begin{aligned}{l}L = \int\limits_0^1 {\left| {r'(t)} \right|} \;dt\\ = \int\limits_0^1 {t\sqrt {4 + 9{t^2}} dt} \\ = \int\limits_4^{13} {\frac{1}{{18}}\sqrt u \;du} \end{aligned}\)

\(\begin{aligned}{l}let\;u = 4 + 9{t^2},du = 18t\;dt\\ = \frac{1}{{18}}\frac{2}{3}\left( {{u^{\frac{3}{2}}}} \right)_4^{13}\\ = \frac{1}{{18}}\frac{2}{3}\left( {{{(4 + 9{t^2})}^{\frac{3}{2}}}} \right)_4^{13}\end{aligned}\)

\(\begin{aligned}{l} = \frac{1}{{27}}\left( {{{(13)}^{\frac{3}{2}}} - {{(4)}^{\frac{3}{2}}}} \right)\\ = \frac{1}{{27}}\left( {{{(13)}^{\frac{3}{2}}} - 8} \right)\end{aligned}\)

The length of the given curve is approximately \(\frac{1}{{27}}\left( {{{(13)}^{\frac{3}{2}}} - 8} \right)\)