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Polar Coordinates and Equations in Polar Form
Polar Coordinates and Equations in Polar Form: Problems with Solutions
Problem 1
Convert $(0,\frac{\pi}{2})$ from polar to Cartesian coordinates.
$(0,\frac{\pi}{2})\equiv (0,1)$
$(0,\frac{\pi}{2})\equiv (1,0)$
$(0,\frac{\pi}{2})\equiv (1,\sqrt{2})$
$(0,\frac{\pi}{2})\equiv (0,0)$
Solution:
We use the transformation equations
$x=r\cos\theta$
$y=r\sin \theta $
where $r=0$ $\theta =\frac{\pi}{2}$
$x=0\cdot \cos(\frac{\pi}{2})=0$
$y=0\cdot \sin (\frac{\pi}{2})=0$
So $(0,\frac{\pi}{2})$ is equivalent to $(0,0)$ in Cartesian Coordinates.
Problem 2
Convert $(-\sqrt{2},\frac{\pi}{4})$ from polar to Cartesian coordinates.
$(-\sqrt{2},\frac{\pi}{4})\equiv (0,-\frac{1}{4})$
$(-\sqrt{2},\frac{\pi}{4})\equiv (-1,-1)$
$(-\sqrt{2},\frac{\pi}{4})\equiv (1,-1)$
$(-\sqrt{2},\frac{\pi}{4})\equiv (1,1)$
Solution:
Answer: $(-\sqrt{2},\frac{\pi}{4})\equiv (-1,-1)$
We use the transformation equations
$x=r\cos\theta$
$y=r\sin\theta $ where $r=-\sqrt{2}\qquad \theta =\pi /4$
$x=-\sqrt{2}\cdot \cos(\pi /4)=-\sqrt{2}.\frac{\sqrt{2}}{2}=-1$
$y=-\sqrt{2}\cdot \sin (\pi /4)=-\sqrt{2}.\frac{\sqrt{2}}{2}=-1$
The point in Cartesian coordinates is $(-1,-1)$.
Problem 3
Convert the equation $y=10$ to polar form.
$r=10\sin\theta$
$r=\tan \frac{10}{\pi }$
$r=\frac{10}{\sin\theta}$
$r=10\tan \frac{10}{\pi }$
Solution:
$y=10\Longrightarrow y=r\sin\theta =10\Longrightarrow r=\frac{10}{\sin\theta }=10\csc \theta $
Problem 4
Convert the equation $x^{2}-y^{2}=4$ to polar form.
$r^{2}\left( \cos \theta \right) =4$
$r\left( \cos 2\theta \right) =4$
$r\left( \cos 2\theta \right) =-4$
$r^{2}\left( \cos 2\theta \right) =4$
Solution:
Answer: $r^{2}\left( \cos 2\theta \right) =4$
We use the transformation equations
$x=rcos\theta \qquad y=rsin\theta$
$x^{2}-y^{2}=4\Longrightarrow \left( rcos\theta \right)^{2}-\left(rsin\theta \right) ^{2}=4\Longrightarrow r^{2}\left( \cos ^{2}\theta -\sin^{2}\theta \right) =4$
$r^{2}\left( \cos 2\theta \right) =4$
This is a hyperbola equation, and here is its graph.
Problem 5
Convert the equation $y^{2}=4x$ to polar form.
$r=4\sin \theta \cos \theta $
$r=4\cot \theta $
$r=4\frac{\cos \theta }{sin^{2}\theta }$
$r=\cot 4\theta \cos 4\theta $
Solution:
We use the transformation equations
$x=r\cos\theta \qquad y=r\sin\theta $
$y^{2}=4x$ $\Longrightarrow \left( r\sin\theta \right) ^{2}=4r\cos\theta \Longrightarrow r^{2}\sin^{2}\theta =4r\cos\theta \Longrightarrow r=4\frac{\cos \theta }{sin^{2}\theta }$
This is the equation of a parabola.
Problem 6
How do we represent the orange ray in polar coordinates?
$\theta =2\frac{\pi }{3};r\leq -2$
$\theta =2\frac{\pi }{3};2\leq r<\infty $
$\theta =2\frac{\pi }{3};0\leq r\leq -2$
$\theta =2\frac{\pi }{3};r\geq -2$
Solution:
$\theta =2\frac{\pi }{3};r\leq -2$
This is the graph of a ray that makes an angle of $2\frac{\pi}{3}$ with the positive $x-axis$, but goes in the opposite direction starting at $-2.$
Problem 7
Calculate the equation in polar coordinates of this semicircle.
$-\pi \leq \theta \leq 2\pi \qquad r=-1$
$0\leq \theta \leq \pi \qquad r=1$
$-\pi \leq \theta \leq 2\pi \qquad r=1$
$0\leq \theta \leq \pi \qquad r=-1$
Solution:
Answer: $0\leq \theta \leq \pi \qquad r=-1$
This is half a circle that starts at $\pi$ and goes to $2\pi $. I know that the interval starts at $0$, but $r$ is negative.
Problem 8
What is the equation in polar coordinates of the blue region?
$\frac{3\pi }{4}\leq \theta \leq \frac{5\pi }{4}\qquad 0\leq r\leq 1$
$-\frac{\pi }{4}\leq \theta \leq \frac{5\pi }{4}\qquad 0\leq r\leq 1$
$-\frac{\pi }{4}\leq \theta \leq \frac{3\pi }{4}\qquad -1\leq r\leq 1$
$-\frac{\pi }{4}\leq \theta \leq \frac{\pi }{4}\qquad -1\leq r\leq 1$
Solution:
Answer: $-\frac{\pi }{4}\leq \theta \leq \frac{\pi }{4}\qquad -1\leq r\leq 1$
The graph of this set of inequalities is two wedges cut out of the circle with radius of $-1$ and $1$, and all points in blue that are between those two values by the lines $\pm \frac{\pi }{4}$.
Problem 9
Convert the polar equation $r\sin\theta =4$ to rectangular equation.
$y=4x$
$y=4$
$y=4x+1$
$y=4x-1$
Solution:
Answer: $y=4$
$r\sin\theta =4\Longrightarrow y=4$
Since $y=r\sin \theta $ is the transformation equation.
This is an equation of a horizontal line through point $(0,4)$.
Problem 10
Convert the polar equation $r\sin\theta =r\cos\theta +4$ to rectangular equation.
$y=4x$
$y=x+4$
$y+x=4$
$y=4x+r$
Solution:
Answer: $y=x+4$
We use the transformation equations
$x=r\cos\theta \qquad y=r\sin\theta$
So $r\sin\theta =r\cos\theta +4\Longrightarrow y=x+4$
This is a straight line with slope $1$ and $y$-intercept at $(0,4)$
Problem 11
The points of intersection of the graphs of the functions $r=\sin \theta $ and $r=\sin 2\theta $ are:
$\left( \sqrt{3},3\pi \right) ,\left( -\sqrt{3},5\pi \right) $
$\left(\frac{\sqrt{3}}{2},\frac{1}{3}\pi \right) ,\left( -\frac{\sqrt{3}}{2},\frac{5}{3}\pi \right)$
$\left(-\frac{\sqrt{3}}{2},\frac{1}{3}\pi \right),\left(\frac{\sqrt{3}}{2},\frac{5}{3}\pi \right)$
$\left(\frac{\sqrt{3}}{2},-\frac{1}{3}\pi \right),\left(-\frac{\sqrt{3}}{2},-\frac{5}{3}\pi \right)$
Solution:
We match both equations
$r=\sin \theta $ and $r=\sin 2\theta $
then $\sin \theta =\sin 2\theta =2\sin \theta \cos \theta \Longrightarrow 2\cos \theta =1\Longrightarrow \cos \theta =\frac{1}{2}$
So, $\theta =\arccos \frac{1}{2}= \frac{1}{3}\pi ,\frac{5}{3}\pi $
when $\theta =\frac{1}{3}\pi \Longrightarrow $ $r=\sin \left( \frac{1}{3}\pi \right) =\frac{\sqrt{3}}{2}$ and $r=\sin \left( \frac{2}{3}\pi \right) =\frac{\sqrt{3}}{2}$
and when $\theta =\frac{5}{3}\pi \Longrightarrow r=\sin \left( \frac{5}{3}\pi \right) =-\frac{\sqrt{3}}{2}$ and $r=\sin \left( \frac{10}{3}\pi \right) = -\frac{\sqrt{3}}{2}$
The points of intersection are $\left( \frac{\sqrt{3}}{2},\frac{1}{3}\pi \right) ,\left( -\frac{\sqrt{3}}{2},\frac{5}{3}\pi \right) $
Problem 12
The following equations represent $r=\frac{2}{1-2\sin \theta },r=\frac{3}{4+\cos \theta }$
parabola, ellipse
hyperbola, parabola
hyperbola, ellipse
parabola, circle
Solution:
Answer: Hyperbola, ellipse.
$r=\frac{2}{1-2\sin \theta }$
If we compare each of the terms in the given equation in polar form $r=\frac{ep}{1-e\sin\theta }$
we can see that $e=2$.
Therefore, the conic section is a hyperbola.
$r=\frac{3}{4+\cos \theta }=\frac{3/4}{1+1/4\cos \theta }$
and we compare with $r=\frac{ep}{1+e\sin\theta }$ then $e=\frac{1}{4}$
So, the conic section is an ellipse.
Problem 13
Identify the conic section represented by the equation
$r=\frac{4}{3-2\sin\theta}$
Parabola
Hyperbola
Circle
Ellipse
Solution:
$r=\frac{4}{3-2\sin \theta }=\frac{4/3}{1-2/3\sin \theta }$
We see that $e=\frac{2}{3}$ then
The equation $r=\frac{4}{3-2\sin \theta}$ represents an ellipse.
Problem 14
Identify the conic section represented by the equation
$r=\frac{1}{1-\cos\theta }$
Parabola
Hyperbola
Circle
Ellipse
Solution:
The conic section represents a parabola which axis is horizontal, since $r$ is not defined when $\theta=0$
The vertex of the parabola is $\theta =\pi $
Vertex: $(\frac{1}{2},\pi )$
$y-axis$ intersection: $(1,\frac{\pi }{2});$ $(1,\frac{3\pi }{2})$
Problem 15
Identify the conic section represented by the equation
$r=\frac{2}{1+2\cos\theta }$
Parabola
Hyperbola
Circle
Ellipse
Solution:
We see that $e=2$.
So the equation $r=\frac{2}{1+2\cos \theta }$ represents a hyperbola,
which axis is horizontal, along the x-axis. The vertices, which are the edges of the transverse axis of the hyperbola, are $\theta =0$ and $\theta =\pi $
Vertices: $(\frac{2}{3},0);\ \ (-2,\pi )$
$y-axis$ intersection at: $(2,\frac{\pi }{2});\ \ (2,\frac{3\pi }{2})$
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