Notes: `r=1-sintheta` Find the points of horizontal and vertical tangency (if any) to the polar curve.

$\displaystyle{r}={1}-{\sin{\theta}}$

To solve, express the polar equation in parametric form. To convert it to parametric equation, apply the formula

$\displaystyle{x}={r}{\cos{\theta}}$

$\displaystyle{y}={r}{\sin{\theta}}$

Plugging in $\displaystyle{r}={1}-{\sin{\theta}}$ , the formula becomes:

$\displaystyle{x}={\left({1}-{\sin{\theta}}\right)}{\cos{\theta}}={\cos{\theta}}-{\sin{\theta}}{\cos{\theta}}$

$\displaystyle{y}={\left({1}-{\sin{\theta}}\right)}{\sin{\theta}}={\sin{\theta}}-{{\sin}}^{{2}}\theta$

So the equivalent parametric equation of $\displaystyle{r}={1}-{\sin{\theta}}$ is:

$\displaystyle{x}={\cos{\theta}}-{\sin{\theta}}{\cos{\theta}}$

$\displaystyle{y}={\sin{\theta}}-{{\sin}}^{{2}}\theta$

Then, take the derivative of x and y with respect to theta.

$\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}=-{\sin{\theta}}-{\left({\sin{\theta}}\cdot{\left(-{\sin{\theta}}\right)}+{\cos{\theta}}\cdot{\cos{\theta}}\right)}$

$\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}=-{\sin{\theta}}+{{\sin}}^{{2}}\theta-{{\cos}}^{{2}}\theta$

$\displaystyle\frac{{\left.{d}{y}}}{{{d}\theta}}={\cos{\theta}}-{2}{\sin{\theta}}{\cos{\theta}}$

Take note that the slope of the tangent is equal to dy/dx.

$\displaystyle{m}=\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$

To get the dy/dx of a parametric equation, apply the formula:

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{\frac{{\left.{d}{y}}}{{{d}\theta}}}}{{\frac{{\left.{d}{x}}}{{{d}\theta}}}}$

When the tangent line is horizontal, the slope of the tangent is zero.

$\displaystyle{0}=\frac{{\frac{{\left.{d}{y}}}{{{d}\theta}}}}{{\frac{{\left.{d}{x}}}{{{d}\theta}}}}$

This implies that the polar curve will have a horizontal tangent when $\displaystyle\frac{{\left.{d}{y}}}{{{d}\theta}}={0}$ and $\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}\ne{0}$ .

Setting the derivative of y yields:

$\displaystyle\frac{{\left.{d}{y}}}{{{d}\theta}}={0}$

$\displaystyle{\cos{\theta}}-{2}{\sin{\theta}}{\cos{\theta}}={0}$

$\displaystyle{\cos{\theta}}{\left({1}-{2}{\sin{\theta}}\right)}={0}$

$\displaystyle{\cos{\theta}}={0}$

$\displaystyle\theta=\frac{\pi}{{2}},\frac{{{3}\pi}}{{2}}$

$\displaystyle{1}-{2}{\sin{\theta}}={0}$

$\displaystyle-{2}{\sin{\theta}}=-{1}$

$\displaystyle{\sin{\theta}}=\frac{{1}}{{2}}$

$\displaystyle\theta=\frac{\pi}{{6}},\frac{{{5}\pi}}{{6}}$

Take note that at $\displaystyle\theta=\frac{\pi}{{2}}$ , the value of $\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}$ is zero. Since both $\displaystyle\frac{{\left.{d}{y}}}{{{d}\theta}}$ and $\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}$ are zero, the slope at this value of theta is indeterminate.

$\displaystyle{m}=\frac{{0}}{{0}}$ (indeterminate)

So the polar curve has horizontal tangents at:

$\displaystyle\theta_{{1}}=\frac{\pi}{{6}}+{2}\pi{n}$

$\displaystyle\theta_{{2}}=\frac{{{5}\pi}}{{6}}+{2}\pi{n}$

$\displaystyle\theta_{{3}}=\frac{{{3}\pi}}{{2}}+{2}\pi{n}$

where n is any integer.

To determine the points $\displaystyle{\left({r},\theta\right)}$ , plug-in the values of theta to the polar equation.

$\displaystyle{r}={1}-{\sin{\theta}}$

$\displaystyle\theta_{{1}}=\frac{\pi}{{6}}+{2}\pi{n}$

$\displaystyle{r}_{{1}}={1}-{\sin{{\left(\frac{\pi}{{6}}+{2}\pi{n}\right)}}}={1}-{\sin{{\left(\frac{\pi}{{6}}\right)}}}={1}-\frac{{1}}{{2}}=\frac{{1}}{{2}}$

$\displaystyle\theta_{{2}}=\frac{{{5}\pi}}{{6}}+{2}\pi{n}$

$\displaystyle{r}_{{2}}={1}-{\sin{{\left(\frac{{{5}\pi}}{{6}}+{2}\pi{n}\right)}}}={1}-{\sin{{\left(\frac{{{5}\pi}}{{6}}\right)}}}={1}-\frac{{1}}{{2}}=\frac{{1}}{{2}}$

$\displaystyle\theta_{{3}}=\frac{{{3}\pi}}{{2}}+{2}\pi{n}$

$\displaystyle{r}_{{3}}={1}-{\sin{{\left(\frac{{{3}\pi}}{{2}}+{2}\pi{n}\right)}}}={1}-{\sin{{\left(\frac{{{3}\pi}}{{2}}\right)}}}={1}-{\left(-{1}\right)}={2}$

Therefore, the polar curve has horizontal tangent at points

$\displaystyle{\left(\frac{{1}}{{2}},\frac{\pi}{{6}}+{2}\pi{n}\right)}$ , $\displaystyle{\left(\frac{{1}}{{2}},\frac{{{5}\pi}}{{6}}+{2}\pi{n}\right)}$ , and $\displaystyle{\left({2},\frac{{{3}\pi}}{{2}}+{2}\pi{n}\right)}$ .

Moreover, when the tangent line is vertical, the slope is undefined.

$\displaystyle{u}{n}{d}{e}{f{{i}}}{n}{e}{d}=\frac{{\frac{{\left.{d}{y}}}{{{d}\theta}}}}{{\frac{{\left.{d}{x}}}{{{d}\theta}}}}$

This implies that the polar curve will have vertical tangent when $\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}={0}$ and $\displaystyle\frac{{\left.{d}{y}}}{{{d}\theta}}\ne{0}$ .

Setting the derivative of x equal to zero yields:

$\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}={0}$

$\displaystyle-{\sin{\theta}}+{{\sin}}^{{2}}\theta-{{\cos}}^{{2}}\theta={0}$

$\displaystyle-{\sin{\theta}}+{{\sin}}^{{2}}\theta-{\left({1}-{{\sin}}^{{2}}\theta\right)}={0}$

$\displaystyle{2}{{\sin}}^{{2}}\theta-{\sin{\theta}}-{1}={0}$

$\displaystyle{\left({2}{\sin{\theta}}+{1}\right)}{\left({\sin{\theta}}-{1}\right)}={0}$

$\displaystyle{2}{\sin{\theta}}+{1}={0}$

$\displaystyle{\sin{\theta}}=-\frac{{1}}{{2}}$

$\displaystyle\theta=\frac{{{7}\pi}}{{6}},\frac{{{11}\pi}}{{6}}$

$\displaystyle{\sin{\theta}}-{1}={0}$

$\displaystyle{\sin{\theta}}={1}$

$\displaystyle\theta=\frac{\pi}{{2}}$

Take note that at $\displaystyle\theta=\frac{\pi}{{2}}$ , both $\displaystyle\frac{{\left.{d}{y}}}{{{d}\theta}}$ and $\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}$ are zero. So the slope is indeterminate at this value of theta.

$\displaystyle{m}=\frac{{0}}{{0}}$ (indeterminate)

So the polar curve has vertical tangents at:

$\displaystyle\theta_{{1}}=\frac{{{7}\pi}}{{6}}+{2}\pi{n}$

$\displaystyle\theta_{{2}}=\frac{{{11}\pi}}{{6}}+{2}\pi{n}$

where n is any integer.

To determine the points $\displaystyle{\left({r},\theta\right)}$ , plug-in the values of theta to the polar equation.

$\displaystyle{r}={1}-{\sin{\theta}}$

$\displaystyle\theta_{{1}}=\frac{{{7}\pi}}{{6}}+{2}\pi{n}$

$\displaystyle{r}_{{1}}={1}-{\sin{{\left(\frac{{{7}\pi}}{{6}}+{2}\pi{n}\right)}}}={1}-{\sin{{\left(\frac{{{7}\pi}}{{6}}\right)}}}={1}-{\left(-\frac{{1}}{{2}}\right)}=\frac{{3}}{{2}}$

$\displaystyle\theta_{{2}}=\frac{{{11}\pi}}{{6}}+{2}\pi{n}$

$\displaystyle{r}_{{2}}={1}-{\sin{{\left(\frac{{{11}\pi}}{{6}}+{2}\pi{n}\right)}}}={1}-{\sin{{\left(\frac{{{11}\pi}}{{6}}\right)}}}={1}-{\left(-\frac{{1}}{{2}}\right)}=\frac{{3}}{{2}}$

Therefore, the polar curve has vertical tangent at points $\displaystyle{\left(\frac{{3}}{{2}},\frac{{{7}\pi}}{{6}}+{2}\pi{n}\right)}$ and $\displaystyle{\left(\frac{{3}}{{2}},\frac{{{11}\pi}}{{6}}+{2}\pi{n}\right)}$ .

Notes

Monday, 14 August 2017

$\displaystyle{r}={1}-{\sin{\theta}}$ Find the points of horizontal and vertical tangency (if any) to the polar curve.

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

Monday, 14 August 2017

Find the points of horizontal and vertical tangency (if any) to the polar curve.

No comments:

Post a Comment

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

$\displaystyle{r}={1}-{\sin{\theta}}$ Find the points of horizontal and vertical tangency (if any) to the polar curve.

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