Notes: Find the arc length from (0,3) clockwise to `(2,sqrt(5))` along the circle `x^2+y^2 = 9`

Sunday, 21 September 2014

Find the arc length from (0,3) clockwise to $\displaystyle{\left({2},\sqrt{{{5}}}\right)}$ along the circle $\displaystyle{{x}}^{{2}}+{{y}}^{{2}}={9}$

Use the arc length formula,

$\displaystyle{L}=\int{d}{s}$

$\displaystyle{d}{s}=\sqrt{{{1}+{{\left(\frac{{\left.{d}{y}}}{{\left.{d}{x}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$ , if y=f(x), a $\displaystyle\le$ x $\displaystyle\le$ b

Given $\displaystyle{{x}}^{{2}}+{{y}}^{{2}}={9}$

$\displaystyle\Rightarrow{{y}}^{{2}}={9}-{{x}}^{{2}}$

$\displaystyle{y}={{\left({9}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{1}}{{2}}{{\left({9}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}-{1}}}\cdot{\left(-{2}{x}\right)}$

$\displaystyle=-\frac{{x}}{\sqrt{{{9}-{{x}}^{{2}}}}}$

Plug in the above in ds,

$\displaystyle{d}{s}={\sqrt{{{1}+{\left(-\frac{{x}}{\sqrt{{{9}-{{x}}^{{2}}}}}\right)}}}}^{{2}}{\left.{d}{x}\right.}$

$\displaystyle{d}{s}=\sqrt{{{1}+\frac{{{x}}^{{2}}}{{{9}-{{x}}^{{2}}}}}}{\left.{d}{x}\right.}$

$\displaystyle{d}{s}=\sqrt{{\frac{{{9}-{{x}}^{{2}}+{{x}}^{{2}}}}{{{9}-{{x}}^{{2}}}}}}{\left.{d}{x}\right.}$

$\displaystyle{d}{s}=\frac{{3}}{\sqrt{{{9}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

The limits are x=0 and x=2,

$\displaystyle{L}={\int_{{0}}^{{2}}}\frac{{3}}{\sqrt{{{9}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

$\displaystyle={3}{\int_{{0}}^{{2}}}\frac{{1}}{\sqrt{{{9}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

Now let's first evaluate the indefinite integral by using integral substitution,

Let $\displaystyle{x}={3}{\sin{{\left({u}\right)}}}$

$\displaystyle{\left.{d}{x}\right.}={3}{\cos{{\left({u}\right)}}}{d}{u}$

$\displaystyle\int\frac{{1}}{\sqrt{{{9}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}=\int\frac{{1}}{\sqrt{{{9}-{{\left({3}{\sin{{\left({u}\right)}}}\right)}}^{{2}}}}}{3}{\cos{{\left({u}\right)}}}{d}{u}$

$\displaystyle=\int\frac{{{3}{\cos{{\left({u}\right)}}}}}{\sqrt{{{9}-{9}{{\sin}}^{{2}}{\left({u}\right)}}}}{d}{u}$

$\displaystyle=\int\frac{{{3}{\cos{{\left({u}\right)}}}}}{{\sqrt{{{9}}}\sqrt{{{1}-{{\sin}}^{{2}}{\left({u}\right)}}}}}{d}{u}$

$\displaystyle=\int\frac{{{3}{\cos{{\left({u}\right)}}}}}{{{3}{\cos{{\left({u}\right)}}}}}{d}{u}$

$\displaystyle=\int{1}{d}{u}$

$\displaystyle={u}$

substitute back u and add a constant C to the solution,

$\displaystyle={\arcsin{{\left(\frac{{x}}{{3}}\right)}}}+{C}$

$\displaystyle{L}={3}{{\left[{\arcsin{{\left(\frac{{x}}{{3}}\right)}}}\right]}_{{0}}^{{2}}}$

$\displaystyle{L}={3}{\left[{\arcsin{{\left(\frac{{2}}{{3}}\right)}}}-{\arcsin{{\left({0}\right)}}}\right]}$

$\displaystyle{L}={3}{\arcsin{{\left(\frac{{2}}{{3}}\right)}}}$

$\displaystyle\approx{2.19}$

Use the arc length formula,

$\displaystyle{L}=\int{d}{s}$

$\displaystyle{d}{s}=\sqrt{{{1}+{{\left(\frac{{\left.{d}{y}}}{{\left.{d}{x}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$ , if y=f(x), a $\displaystyle\le$ x $\displaystyle\le$ b

Given $\displaystyle{{x}}^{{2}}+{{y}}^{{2}}={9}$

$\displaystyle\Rightarrow{{y}}^{{2}}={9}-{{x}}^{{2}}$

$\displaystyle{y}={{\left({9}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{1}}{{2}}{{\left({9}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}-{1}}}\cdot{\left(-{2}{x}\right)}$

$\displaystyle=-\frac{{x}}{\sqrt{{{9}-{{x}}^{{2}}}}}$

Plug in the above in ds,

$\displaystyle{d}{s}={\sqrt{{{1}+{\left(-\frac{{x}}{\sqrt{{{9}-{{x}}^{{2}}}}}\right)}}}}^{{2}}{\left.{d}{x}\right.}$

$\displaystyle{d}{s}=\sqrt{{{1}+\frac{{{x}}^{{2}}}{{{9}-{{x}}^{{2}}}}}}{\left.{d}{x}\right.}$

$\displaystyle{d}{s}=\sqrt{{\frac{{{9}-{{x}}^{{2}}+{{x}}^{{2}}}}{{{9}-{{x}}^{{2}}}}}}{\left.{d}{x}\right.}$

$\displaystyle{d}{s}=\frac{{3}}{\sqrt{{{9}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

The limits are x=0 and x=2,

$\displaystyle{L}={\int_{{0}}^{{2}}}\frac{{3}}{\sqrt{{{9}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

$\displaystyle={3}{\int_{{0}}^{{2}}}\frac{{1}}{\sqrt{{{9}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

Now let's first evaluate the indefinite integral by using integral substitution,

Let $\displaystyle{x}={3}{\sin{{\left({u}\right)}}}$

$\displaystyle{\left.{d}{x}\right.}={3}{\cos{{\left({u}\right)}}}{d}{u}$

$\displaystyle\int\frac{{1}}{\sqrt{{{9}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}=\int\frac{{1}}{\sqrt{{{9}-{{\left({3}{\sin{{\left({u}\right)}}}\right)}}^{{2}}}}}{3}{\cos{{\left({u}\right)}}}{d}{u}$

$\displaystyle=\int\frac{{{3}{\cos{{\left({u}\right)}}}}}{\sqrt{{{9}-{9}{{\sin}}^{{2}}{\left({u}\right)}}}}{d}{u}$

$\displaystyle=\int\frac{{{3}{\cos{{\left({u}\right)}}}}}{{\sqrt{{{9}}}\sqrt{{{1}-{{\sin}}^{{2}}{\left({u}\right)}}}}}{d}{u}$

$\displaystyle=\int\frac{{{3}{\cos{{\left({u}\right)}}}}}{{{3}{\cos{{\left({u}\right)}}}}}{d}{u}$

$\displaystyle=\int{1}{d}{u}$

$\displaystyle={u}$

substitute back u and add a constant C to the solution,

$\displaystyle={\arcsin{{\left(\frac{{x}}{{3}}\right)}}}+{C}$

$\displaystyle{L}={3}{{\left[{\arcsin{{\left(\frac{{x}}{{3}}\right)}}}\right]}_{{0}}^{{2}}}$

$\displaystyle{L}={3}{\left[{\arcsin{{\left(\frac{{2}}{{3}}\right)}}}-{\arcsin{{\left({0}\right)}}}\right]}$

$\displaystyle{L}={3}{\arcsin{{\left(\frac{{2}}{{3}}\right)}}}$

$\displaystyle\approx{2.19}$

Notes

Sunday, 21 September 2014

Find the arc length from (0,3) clockwise to $\displaystyle{\left({2},\sqrt{{{5}}}\right)}$ along the circle $\displaystyle{{x}}^{{2}}+{{y}}^{{2}}={9}$

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

Sunday, 21 September 2014

Find the arc length from (0,3) clockwise to along the circle

No comments:

Post a Comment

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

Find the arc length from (0,3) clockwise to $\displaystyle{\left({2},\sqrt{{{5}}}\right)}$ along the circle $\displaystyle{{x}}^{{2}}+{{y}}^{{2}}={9}$

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