Tag: Polar Coordinates

Sketch a circle tangent to the y-axis and compute its area from -π/2 to π/2

by RoRi

August 7, 2015

Define a circle tangent to the $y$ -axis by:

$f(\theta) = 2 \cos \theta, \qquad - \pi/2 \leq \theta \leq \pi/2.$

Sketch this graph in polar coordinates and compute the area of the radial set.

The sketch is as follows:

Rendered by QuickLaTeX.com

Then, we compute the area,

$\begin{align*} a(R) = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 4 \cos^2 \theta \, d \theta &= 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 - \sin^2 \theta) \, d \theta \\ &= 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d \, \theta + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 - 2 \sin^2 \theta - 1) \, d \theta \\ &= 2 \pi + \left( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos (2 \theta) \, d\theta \right) - \pi \\ &= \pi + \frac{1}{2} \int_{-\pi}^{\pi} \cos \theta \, d \theta \\ &= \frac{1}{2} \left( \sin \theta \biggr \rvert_{-\pi}^{\pi} \right) + \pi \\ &= \pi. \end{align*}$

Sketch the Spiral of Archimedes and compute its area from 0 to 2 π

by RoRi

August 7, 2015

Define
Spiral of Archimedes: $f(\theta) = \theta$ for $0 \leq \theta \leq 2 \pi$ .

Sketch this graph in polar coordinates and compute the area of the radial set.

The sketch is as follows:

Rendered by QuickLaTeX.com

Then, we compute the area,

$a(R) = \frac{1}{2} \int_0^{2 \pi} \theta^2 \, d \theta = \frac{1}{2} \frac{(2 \pi)^3}{3} = \frac{4 \pi^3}{3}.$

Show an equivalence between given sets in Cartesian and polar coordinates

by RoRi

August 7, 2015

Show that the sets of points satisfying

$(x^2 + y^2)^2 = |x^2-y^2|; \qquad \qquad r = \sqrt{|\cos (2\theta)|}.$

are equal.

Letting $x = r \cos \theta$ and $y = r \sin \theta$ we plug in to the given Cartesian equation,

$\begin{align*} (x^2 + y^2)^2 = | x^2 - y^2 | && \implies && (r^2 \cos^2 \theta + r^2 \sin^2 \theta)^2 &= | r^2 \cos^2 \theta - r^2 \sin^2 \theta| \\ && \implies && r^4 &= r^2 | \cos (2 \theta) | & (r^2 > 0, \text{ so } |r^2| = r^2)\\ && \implies && r^2 &= | \cos (2 \theta) | \\ && \implies && r &= \sqrt{ | \cos (2 \theta)|}. \end{align*}$

Show an equivalence between given sets in Cartesian and polar coordinates

by RoRi

August 7, 2015

Show that the sets of points satisfying

$(x^2 + y^2)^2 = x^2-y^2, \quad y^2 \leq x^2; \qquad \qquad r = \sqrt{\cos (2\theta)}, \quad \cos (2 \theta) \geq 0.$

are equal.

Letting $x = r \cos \theta$ and $y = r \sin \theta$ we plug in to the given Cartesian equation,

$\begin{align*} (x^2+y^2)^2 = x^2 - y^2 && \implies && (r^2 \cos^2 \theta + r^2 \sin^2 \theta)^2 &= r^2 \cos^2 \theta - r^2 \sin^2 \theta \\ && \implies && r^4 &= r^2 \cos (2 \theta) \\ && \implies && r^2 &= \cos (2 \theta) \\ && \implies && r &= \sqrt{\cos (2 \theta)}. \end{align*}$

Show an equivalence between given sets in Cartesian and polar coordinates

by RoRi

August 6, 2015

Show that the sets of points satisfying

$x^2 + y^2 - x = \sqrt{x^2 + y^2}; \qquad r = 1 + \cos \theta.$

are equal.

Letting $x = r \cos \theta$ and $y = r \sin \theta$ we plug in to the given Cartesian equation,

$\begin{align*} x^2 + y^2 - x = \sqrt{x^2 + y^2} && \implies && r^2 \cos^2 \theta + r^2 \sin^2 \theta - r \cos \theta &= \sqrt{r^2 \cos^2 \theta + r^2 \sin^2 \theta} \\ && \implies && r^2 - r \cos \theta &= r \\ && \implies && 1 + \cos \theta &= r. \qquad \blacksquare \end{align*}$

Show an equivalence between given sets in Cartesian and polar coordinates

by RoRi

August 6, 2015

Show that the sets of points satisfying

$(x-1)^2 + y^2 = 1 \qquad \text{and} \qquad r = 2 \cos \theta, \quad \cos \theta > 0$

are equal.

Letting $x = r \cos \theta$ and $y = r \sin \theta$ we plug in to the given Cartesian equation,

$\begin{align*} (x-1)^2 + y^2 = 1 && \implies && (r \cos \theta - 1)^2 + r^2 \sin^2 \theta &= 1 \\ && \implies && r^2 \cos^2 \theta - 2 r \cos \theta + 1 + r^2 \sin^2 \theta &=1 \\ && \implies && r^2 (\cos^2 \theta + \sin^2 \theta) - 2 r \cos \theta &= 0 \\ && \implies && r &= 2 \cos \theta, \end{align*}$

where $\cos \theta > 0$ so $r > 0$ (hence, we can divide by $r$ ) in the second to last line.