Notes: `x=6t^2 , y=2t^3 , 1

Tuesday, 13 May 2014

$\displaystyle{x}={6}{{t}}^{{2}},{y}={2}{{t}}^{{3}},{1}$

The formula of arc length of a parametric equation on the interval $\displaystyle{a}\leq{t}\leq{b}$ is:

$\displaystyle{L}={\int_{{a}}^{{b}}}\sqrt{{{{\left(\frac{{\left.{d}{x}}}{{\left.{d}{t}}}\right)}}^{{2}}+{{\left(\frac{{\left.{d}{y}}}{{\left.{d}{t}}}\right)}}^{{2}}}}{\left.{d}{t}\right.}$

The given parametric equation is:

$\displaystyle{x}={6}{{t}}^{{2}}$

$\displaystyle{y}={2}{{t}}^{{3}}$

The derivative of x and y are:

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}={12}{t}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{t}}}={6}{{t}}^{{2}}$

So the integral needed to compute the arc length of the given parametric equation on the interval $\displaystyle{1}\leq{t}\leq{4}$ is:

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{{{{\left({12}{t}\right)}}^{{2}}+{{\left({6}{{t}}^{{2}}\right)}}^{{2}}}}{\left.{d}{t}\right.}$

The simplified form of the integral is:

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{{{144}{{t}}^{{2}}+{36}{{t}}^{{4}}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{\ldots}$

The formula of arc length of a parametric equation on the interval $\displaystyle{a}\leq{t}\leq{b}$ is:

$\displaystyle{L}={\int_{{a}}^{{b}}}\sqrt{{{{\left(\frac{{\left.{d}{x}}}{{\left.{d}{t}}}\right)}}^{{2}}+{{\left(\frac{{\left.{d}{y}}}{{\left.{d}{t}}}\right)}}^{{2}}}}{\left.{d}{t}\right.}$

The given parametric equation is:

$\displaystyle{x}={6}{{t}}^{{2}}$

$\displaystyle{y}={2}{{t}}^{{3}}$

The derivative of x and y are:

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}={12}{t}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{t}}}={6}{{t}}^{{2}}$

So the integral needed to compute the arc length of the given parametric equation on the interval $\displaystyle{1}\leq{t}\leq{4}$ is:

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{{{{\left({12}{t}\right)}}^{{2}}+{{\left({6}{{t}}^{{2}}\right)}}^{{2}}}}{\left.{d}{t}\right.}$

The simplified form of the integral is:

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{{{144}{{t}}^{{2}}+{36}{{t}}^{{4}}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{1}}^{{4}}}\sqrt{{{36}{{t}}^{{2}}{\left({4}+{{t}}^{{2}}\right)}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{1}}^{{4}}}{6}{t}\sqrt{{{4}+{{t}}^{{2}}}}{\left.{d}{t}\right.}$

To take the integral, apply u-substitution method.

$\displaystyle{u}={4}+{{t}}^{{2}}$

$\displaystyle{d}{u}={2}{t}{\left.{d}{t}\right.}$

$\displaystyle\frac{{1}}{{2}}{d}{u}={t}{\left.{d}{t}\right.}$

$\displaystyle{t}={1}$ , $\displaystyle{u}={4}+{{1}}^{{2}}={5}$

$\displaystyle{t}={4}$ , $\displaystyle{u}={4}+{{4}}^{{2}}={20}$

Expressing the integral in terms of u, it becomes:

$\displaystyle{L}={6}{\int_{{1}}^{{4}}}\sqrt{{{4}+{{t}}^{{2}}}}\cdot{t}{\left.{d}{t}\right.}$

$\displaystyle{L}={6}{\int_{{5}}^{{20}}}\sqrt{{u}}\cdot\frac{{1}}{{2}}{d}{u}$

$\displaystyle{L}={3}{\int_{{5}}^{{20}}}\sqrt{{u}}{d}{u}$

$\displaystyle{L}={3}{\int_{{5}}^{{20}}}{{u}}^{{\frac{{1}}{{2}}}}{d}{u}$

$\displaystyle{L}={3}\cdot\frac{{{u}}^{{\frac{{3}}{{2}}}}}{{\frac{{3}}{{2}}}}$ $\displaystyle{{\mid}_{{5}}^{{20}}}$

$\displaystyle{L}={2}{{u}}^{{\frac{{3}}{{2}}}}$ $\displaystyle{{\mid}_{{5}}^{{20}}}$

$\displaystyle{L}={2}{u}\sqrt{{u}}$ $\displaystyle{{\mid}_{{5}}^{{20}}}$

$\displaystyle{L}={2}{\left({20}\right)}\sqrt{{20}}-{2}{\left({5}\right)}\sqrt{{5}}$

$\displaystyle{L}={40}\sqrt{{20}}-{10}\sqrt{{5}}$

$\displaystyle{L}={40}\cdot{2}\sqrt{{5}}-{10}\sqrt{{5}}$

$\displaystyle{L}={80}\sqrt{{5}}-{10}\sqrt{{5}}$

$\displaystyle{L}={70}\sqrt{{5}}$

Therefore, the arc length of the curve is $\displaystyle{70}\sqrt{{5}}$ units.

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