Notes: `r=2csctheta+3` Find the points of horizontal tangency (if any) to the polar curve.

Saturday, 11 July 2015

$\displaystyle{r}={2}{\csc{\theta}}+{3}$ Find the points of horizontal tangency (if any) to the polar curve.

$\displaystyle{r}={2}{\csc{\theta}}+{3}$

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 $\displaystyle{r}={2}{\csc{\theta}}+{3}$ , the formula becomes:

$\displaystyle{x}={\left({2}{\csc{\theta}}+{3}\right)}{\cos{\theta}}$

$\displaystyle{x}={2}{\cot{\theta}}+{3}{\cos{\theta}}$

$\displaystyle{y}={\left({2}{\csc{\theta}}+{3}\right)}{\sin{\theta}}$

$\displaystyle{y}={2}+{3}{\sin{\theta}}$

So the equivalent parametric equation of $\displaystyle{r}={2}{\csc{\theta}}+{3}$ is:

$\displaystyle{x}={2}{\cot{\theta}}+{3}{\cos{\theta}}$

$\displaystyle{y}={2}+{3}{\sin{\theta}}$

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

$\displaystyle{r}={2}{\csc{\theta}}+{3}$

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 $\displaystyle{r}={2}{\csc{\theta}}+{3}$ , the formula becomes:

$\displaystyle{x}={\left({2}{\csc{\theta}}+{3}\right)}{\cos{\theta}}$

$\displaystyle{x}={2}{\cot{\theta}}+{3}{\cos{\theta}}$

$\displaystyle{y}={\left({2}{\csc{\theta}}+{3}\right)}{\sin{\theta}}$

$\displaystyle{y}={2}+{3}{\sin{\theta}}$

So the equivalent parametric equation of $\displaystyle{r}={2}{\csc{\theta}}+{3}$ is:

$\displaystyle{x}={2}{\cot{\theta}}+{3}{\cos{\theta}}$

$\displaystyle{y}={2}+{3}{\sin{\theta}}$

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

$\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}={2}\cdot{\left(-{{\csc}}^{{2}}\theta\right)}+{3}\cdot{\left(-{\sin{\theta}}\right)}$

$\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}=-{2}{{\csc}}^{{2}}\theta-{3}{\sin{\theta}}$

$\displaystyle\frac{{\left.{d}{y}}}{{{d}\theta}}={3}{\cos{\theta}}$

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 numerator is zero. So set the derivative of y equal to zero.

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

$\displaystyle{3}{\cos{\theta}}={0}$

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

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

So the polar curve have a horizontal tangents at:

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

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

where n is any integer.

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

$\displaystyle{r}={2}{\csc{\theta}}+{3}$

$\displaystyle\theta=\frac{\pi}{{2}}+{2}\pi{n}$ ,

$\displaystyle{r}={2}{\csc{{\left(\frac{\pi}{{2}}+{2}\pi{n}\right)}}}+{3}={2}{\csc{{\left(\frac{\pi}{{2}}\right)}}}+{3}={2}\cdot{1}+{3}={5}$

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

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

Therefore, the polar curve has horizontal tangents at points $\displaystyle{\left({5},\frac{\pi}{{2}}+{2}\pi{n}\right)}$ and $\displaystyle{\left({1},\frac{{{3}\pi}}{{2}}+{2}\pi{n}\right)}$ .