How do you find the length of the curve $x = 3 t - t^{3}$ $x=3t-t^3$ , $y = 3 t^{2}$ $y=3t^2$ , where $0 \leq t \leq \sqrt{3}$ $0<=t<=sqrt(3)$ ?

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How do you find the length of the curve $x = 3 t - t^{3}$ $x=3t-t^3$ , $y = 3 t^{2}$ $y=3t^2$ , where $0 \leq t \leq \sqrt{3}$ $0<=t<=sqrt(3)$ ?

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Answer 1 · 2014-08-15T05:09:05

$6 \sqrt{3}$ $6sqrt3$ .

Explanation:

The answer is $6 \sqrt{3}$ $6sqrt3$ .

The arclength of a parametric curve can be found using the formula: $L = \int_{t_{i}}^{t_{f}} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t$ $L=int_(t_i)^(t_f)sqrt(((dx)/(dt))^2+((dy)/(dt))^2)dt$ . Since $x$ $x$ and $y$ $y$ 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: $L = \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$ $L=int_a^b sqrt(1+((dy)/(dx))^2)dx$ . If you need the derivation of the parametric formula, please ask it as a separate question.

We find the 2 derivatives:
$\frac{d x}{d t} = 3 - 3 t^{2}$ $(dx)/(dt)=3-3t^2$
$\frac{d y}{d t} = 6 t$ $(dy)/(dt)=6t$

And we substitute these into the integral:
$L = \int_{0}^{\sqrt{3}} \sqrt{{(3 - 3 t^{2})}^{2} + {(6 t)}^{2}} d t$ $L=int_0^(sqrt3)sqrt((3-3t^2)^2+(6t)^2)dt$

And solve:
$= \int_{0}^{\sqrt{3}} \sqrt{9 - 18 t^{2} + 9 t^{4} + 36 t^{2}} d t$ $=int_0^(sqrt3)sqrt(9-18t^2+9t^4+36t^2)dt$
$= \int_{0}^{\sqrt{3}} \sqrt{9 + 18 t^{2} + 9 t^{4}} d t$ $=int_0^(sqrt3)sqrt(9+18t^2+9t^4)dt$
$= \int_{0}^{\sqrt{3}} \sqrt{{(3 + 3 t^{2})}^{2}} d t$ $=int_0^(sqrt3)sqrt((3+3t^2)^2)dt$
$= \int_{0}^{\sqrt{3}} (3 + 3 t^{2}) d t$ $=int_0^(sqrt3)(3+3t^2)dt$
$= 3 t + t^{3} {∣ ∣}_{0}^{\sqrt{3}}$ $=3t+t^3|_0^(sqrt3)$
$= 3 \sqrt{3} + 3 \sqrt{3}$ $=3sqrt3+3sqrt3$
$= 6 \sqrt{3}$ $=6sqrt3$

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.

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How do you find the length of the curve $x = 3 t - t^{3}$ $x=3t-t^3$ , $y = 3 t^{2}$ $y=3t^2$ , where $0 \leq t \leq \sqrt{3}$ $0<=t<=sqrt(3)$ ?

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How do you find the length of the curve $x = 3 t - t^{3}$ $x=3t-t^3$ , $y = 3 t^{2}$ $y=3t^2$ , where $0 \leq t \leq \sqrt{3}$ $0<=t<=sqrt(3)$ ?