Notes: `x=e^(-t)cost , y=e^(-t)sint , 0

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:

$\displaystyle{x}={{e}}^{{-{t}}}{\cos{{t}}}$

$\displaystyle{y}={{e}}^{{-{t}}}{\sin{{t}}}$

The derivative of x and y are with respect to t are:

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}={{e}}^{{-{t}}}\cdot{\left({\cos{{t}}}\right)}'+{\left({{e}}^{{-{t}}}\right)}'\cdot{\cos{{t}}}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}={{e}}^{{-{t}}}\cdot{\left(-{\sin{{t}}}\right)}+{{e}}^{{-{t}}}\cdot{\left(-{1}\right)}{\cos{{t}}}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}=-{{e}}^{{-{t}}}{\sin{{t}}}-{{e}}^{{-{t}}}{\cos{{t}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{t}}}={{e}}^{{-{t}}}\cdot{\left({\sin{{t}}}\right)}'+{\left({{e}}^{{-{t}}}\right)}'\cdot{\sin{{t}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{t}}}={{e}}^{{-{t}}}{\cos{{t}}}+{{e}}^{{-{t}}}\cdot{\left(-{1}\right)}{\sin{{t}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{t}}}={{e}}^{{-{t}}}{\cos{{t}}}-{{e}}^{{-{t}}}{\sin{{t}}}$

Plugging them to the formula, the integral needed to...

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:

$\displaystyle{x}={{e}}^{{-{t}}}{\cos{{t}}}$

$\displaystyle{y}={{e}}^{{-{t}}}{\sin{{t}}}$

The derivative of x and y are with respect to t are:

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}={{e}}^{{-{t}}}\cdot{\left({\cos{{t}}}\right)}'+{\left({{e}}^{{-{t}}}\right)}'\cdot{\cos{{t}}}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}={{e}}^{{-{t}}}\cdot{\left(-{\sin{{t}}}\right)}+{{e}}^{{-{t}}}\cdot{\left(-{1}\right)}{\cos{{t}}}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}=-{{e}}^{{-{t}}}{\sin{{t}}}-{{e}}^{{-{t}}}{\cos{{t}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{t}}}={{e}}^{{-{t}}}\cdot{\left({\sin{{t}}}\right)}'+{\left({{e}}^{{-{t}}}\right)}'\cdot{\sin{{t}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{t}}}={{e}}^{{-{t}}}{\cos{{t}}}+{{e}}^{{-{t}}}\cdot{\left(-{1}\right)}{\sin{{t}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{t}}}={{e}}^{{-{t}}}{\cos{{t}}}-{{e}}^{{-{t}}}{\sin{{t}}}$

Plugging them to the formula, the integral needed to compute the arc length of the given parametric equation on the interval $\displaystyle{0}\leq{t}\leq\frac{\pi}{{2}}$ is:

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{2}}}}}\sqrt{{{{\left(-{{e}}^{{-{t}}}{\sin{{t}}}-{{e}}^{{-{t}}}{\cos{{t}}}\right)}}^{{2}}+{{\left({{e}}^{{-{t}}}{\cos{{t}}}-{{e}}^{{-{t}}}{\sin{{t}}}\right)}}^{{2}}}}{\left.{d}{t}\right.}$

The simplified form of the integral is:

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{2}}}}}\sqrt{{{{\left(-{{e}}^{{-{t}}}{\left({\sin{{t}}}+{\cos{{t}}}\right)}\right)}}^{{2}}+{{\left({{e}}^{{-{t}}}{\left({\cos{{t}}}-{\sin{{t}}}\right)}\right)}}^{{2}}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{2}}}}}\sqrt{{{{e}}^{{-{2}{t}}}{{\left({\sin{{t}}}+{\cos{{t}}}\right)}}^{{2}}+{{e}}^{{-{2}{t}}}{{\left({\cos{{t}}}-{\sin{{t}}}\right)}}^{{2}}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{2}}}}}\sqrt{{{{e}}^{{-{2}{t}}}{\left({{\left({\sin{{t}}}+{\cos{{t}}}\right)}}^{{2}}+{{\left({\cos{{t}}}-{\sin{{t}}}\right)}}^{{2}}\right)}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{2}}}}}{{e}}^{{-{t}}}\sqrt{{{{\left({\sin{{t}}}+{\cos{{t}}}\right)}}^{{2}}+{{\left({\cos{{t}}}-{\sin{{t}}}\right)}}^{{2}}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{2}}}}}{{e}}^{{-{t}}}\sqrt{{{{\sin}}^{{2}}{t}+{2}{\sin{{t}}}{\cos{{t}}}+{{\cos}}^{{2}}{t}+{{\cos}}^{{2}}{t}-{2}{\sin{{t}}}{\cos{{t}}}+{{\sin}}^{{2}}{t}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{2}}}}}{{e}}^{{-{t}}}\sqrt{{{2}{{\sin}}^{{2}}{t}+{2}{{\cos}}^{{2}}{t}}}$

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{2}}}}}{{e}}^{{-{t}}}\sqrt{{{2}{\left({{\sin}}^{{2}}{t}+{{\cos}}^{{2}}{t}\right)}}}$

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{2}}}}}{{e}}^{{-{t}}}\sqrt{{{2}\cdot{\left({1}\right)}}}{\left.{d}{t}\right.}$

$\displaystyle{L}={\int_{{0}}^{{\frac{\pi}{{2}}}}}{{e}}^{{-{t}}}\sqrt{{{2}}}{\left.{d}{t}\right.}$

$\displaystyle{L}=\sqrt{{2}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{e}}^{{-{t}}}{\left.{d}{t}\right.}$

$\displaystyle{L}=-\sqrt{{2}}{{e}}^{{-{t}}}{{\mid}_{{0}}^{{\frac{\pi}{{2}}}}}$

$\displaystyle{L}=-\sqrt{{2}}{\left({{e}}^{{-\frac{\pi}{{2}}}}-{{e}}^{{0}}\right)}$

$\displaystyle{L}=-\sqrt{{2}}{\left({{e}}^{{-\frac{\pi}{{2}}}}-{1}\right)}$

$\displaystyle{L}=\sqrt{{2}}-\sqrt{{2}}{{e}}^{{-\frac{\pi}{{2}}}}$

Therefore, the arc length of the curve is $\displaystyle\sqrt{{2}}-\sqrt{{2}}{{e}}^{{-\frac{\pi}{{2}}}}$ units.

Notes

Sunday, 30 July 2017

$\displaystyle{x}={{e}}^{{-{t}}}{\cos{{t}}},{y}={{e}}^{{-{t}}}{\sin{{t}}},{0}$

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Sunday, 30 July 2017

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How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?