# Determining the Length of a Parametric Curve (Parametric Form)

M8-7: arc length (part I)

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 2 videos by Calculus V.

## Key Questions

• Please see how I solved this problem, it has a full explanation.

$6 \sqrt{3}$.

#### Explanation:

The answer is $6 \sqrt{3}$.

The arclength of a parametric curve can be found using the formula: $L = {\int}_{{t}_{i}}^{{t}_{f}} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$. Since $x$ and $y$ are perpendicular, it's not difficult to see why this computes the arclength.

It isn't very different from the arclength of a regular function: $L = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$. If you need the derivation of the parametric formula, please ask it as a separate question.

We find the 2 derivatives:
$\frac{\mathrm{dx}}{\mathrm{dt}} = 3 - 3 {t}^{2}$
$\frac{\mathrm{dy}}{\mathrm{dt}} = 6 t$

And we substitute these into the integral:
$L = {\int}_{0}^{\sqrt{3}} \sqrt{{\left(3 - 3 {t}^{2}\right)}^{2} + {\left(6 t\right)}^{2}} \mathrm{dt}$

And solve:
$= {\int}_{0}^{\sqrt{3}} \sqrt{9 - 18 {t}^{2} + 9 {t}^{4} + 36 {t}^{2}} \mathrm{dt}$
$= {\int}_{0}^{\sqrt{3}} \sqrt{9 + 18 {t}^{2} + 9 {t}^{4}} \mathrm{dt}$
$= {\int}_{0}^{\sqrt{3}} \sqrt{{\left(3 + 3 {t}^{2}\right)}^{2}} \mathrm{dt}$
$= {\int}_{0}^{\sqrt{3}} \left(3 + 3 {t}^{2}\right) \mathrm{dt}$
$= 3 t + {t}^{3} {|}_{0}^{\sqrt{3}}$
$= 3 \sqrt{3} + 3 \sqrt{3}$
$= 6 \sqrt{3}$

Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.

• By taking the derivative with respect to $t$,

$\left\{\begin{matrix}x ' \left(t\right) = 6 t \\ y ' \left(t\right) = 6 {t}^{2}\end{matrix}\right.$

Let us now find the length $L$ of the curve.

$L = {\int}_{0}^{1} \sqrt{{\left[x ' \left(t\right)\right]}^{2} + {\left[y ' \left(t\right)\right]}^{2}} \mathrm{dt}$

$= {\int}_{0}^{1} \sqrt{{6}^{2} {t}^{2} + {6}^{2} {t}^{4}} \mathrm{dt}$

by pulling $6 t$ out of the square-root,

$= {\int}_{0}^{1} 6 t \sqrt{1 + {t}^{2}} \mathrm{dt}$

by rewriting a bit further,

$= 3 {\int}_{0}^{1} 2 t {\left(1 + {t}^{2}\right)}^{\frac{1}{2}} \mathrm{dt}$

by General Power Rule,

$= 3 {\left[\frac{2}{3} {\left(1 + {t}^{2}\right)}^{\frac{3}{2}}\right]}_{0}^{1} = 2 \left({2}^{\frac{3}{2}} - 1\right)$

I hope that this was helpful.

## Questions

• · Yesterday
• · 2 days ago
• · 6 days ago
• · 1 week ago
• · 1 week ago
• · 1 week ago
• · 2 weeks ago
• · 2 weeks ago
• · 2 weeks ago
• · 2 weeks ago
• · 2 weeks ago
• · 2 weeks ago
• · 2 weeks ago
• · 3 weeks ago
• · 3 weeks ago
• · 3 weeks ago
• · 3 weeks ago
• · 3 weeks ago
• · 3 weeks ago
• · 3 weeks ago