# 10.4 Polar coordinates: graphs  (Page 6/16)

 Page 6 / 16

## Sketching the graph of a rose curve ( n Even)

Sketch the graph of $\text{\hspace{0.17em}}r=2\mathrm{cos}\text{\hspace{0.17em}}4\theta .$

Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the line $\text{\hspace{0.17em}}\theta =\frac{\pi }{2}\text{\hspace{0.17em}}$ and the pole.

Now we will find the zeros. First make the substitution $\text{\hspace{0.17em}}u=4\theta .$

$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0=2\mathrm{cos}\text{\hspace{0.17em}}4\theta \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0=\mathrm{cos}\text{\hspace{0.17em}}4\theta \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0=\mathrm{cos}\text{\hspace{0.17em}}u\\ {\mathrm{cos}}^{-1}0=u\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}u=\frac{\pi }{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\theta =\frac{\pi }{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta =\frac{\pi }{8}\end{array}$

The zero is $\text{\hspace{0.17em}}\theta =\frac{\pi }{8}.\text{\hspace{0.17em}}$ The point $\text{\hspace{0.17em}}\left(0,\frac{\pi }{8}\right)\text{\hspace{0.17em}}$ is on the curve.

Next, we find the maximum $\text{\hspace{0.17em}}|r|.\text{\hspace{0.17em}}$ We know that the maximum value of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}u=1\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}\theta =0.\text{\hspace{0.17em}}$ Thus,

$\begin{array}{l}\\ \begin{array}{l}r=2\mathrm{cos}\left(4\cdot 0\right)\hfill \\ r=2\mathrm{cos}\left(0\right)\hfill \\ r=2\left(1\right)=2\hfill \end{array}\end{array}$

The point $\text{\hspace{0.17em}}\left(2,0\right)\text{\hspace{0.17em}}$ is on the curve.

The graph of the rose curve has unique properties, which are revealed in [link] .

 $\theta$ 0 $\frac{\pi }{8}$ $\frac{\pi }{4}$ $\frac{3\pi }{8}$ $\frac{\pi }{2}$ $\frac{5\pi }{8}$ $\frac{3\pi }{4}$ $r$ 2 0 −2 0 2 0 −2

As $\text{\hspace{0.17em}}r=0\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}\theta =\frac{\pi }{8},\text{\hspace{0.17em}}$ it makes sense to divide values in the table by $\text{\hspace{0.17em}}\frac{\pi }{8}\text{\hspace{0.17em}}$ units. A definite pattern emerges. Look at the range of r -values: 2, 0, −2, 0, 2, 0, −2, and so on. This represents the development of the curve one petal at a time. Starting at $\text{\hspace{0.17em}}r=0,\text{\hspace{0.17em}}$ each petal extends out a distance of $\text{\hspace{0.17em}}r=2,\text{\hspace{0.17em}}$ and then turns back to zero $\text{\hspace{0.17em}}2n\text{\hspace{0.17em}}$ times for a total of eight petals. See the graph in [link] .

Sketch the graph of $\text{\hspace{0.17em}}r=4\mathrm{sin}\left(2\theta \right).$

The graph is a rose curve, $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ even

## Sketching the graph of a rose curve ( n Odd)

Sketch the graph of $\text{\hspace{0.17em}}r=2\mathrm{sin}\left(5\theta \right).$

The graph of the equation shows symmetry with respect to the line $\text{\hspace{0.17em}}\theta =\frac{\pi }{2}.\text{\hspace{0.17em}}$ Next, find the zeros and maximum. We will want to make the substitution $\text{\hspace{0.17em}}u=5\theta .$

$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0=2\mathrm{sin}\left(5\theta \right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0=\mathrm{sin}\text{\hspace{0.17em}}u\\ {\mathrm{sin}}^{-1}0=0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}u=0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5\theta =0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta =0\end{array}$

The maximum value is calculated at the angle where $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is a maximum. Therefore,

$\begin{array}{l}\begin{array}{l}\\ r=2\mathrm{sin}\left(5\cdot \frac{\pi }{2}\right)\end{array}\hfill \\ r=2\left(1\right)=2\hfill \end{array}$

Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ odd yields the same number of petals as $\text{\hspace{0.17em}}n,\text{\hspace{0.17em}}$ there will be five petals on the graph. See [link] .

Create a table of values similar to [link] .

 $\theta$ 0 $\frac{\pi }{6}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\frac{5\pi }{6}$ $\pi$ $r$ 0 1 −1.73 2 −1.73 1 0

Sketch the graph of $r=3\mathrm{cos}\left(3\theta \right).$

Rose curve, $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ odd

## Investigating the archimedes’ spiral

The final polar equation we will discuss is the Archimedes’ spiral, named for its discoverer, the Greek mathematician Archimedes (c. 287 BCE - c. 212 BCE), who is credited with numerous discoveries in the fields of geometry and mechanics.

## Archimedes’ spiral

The formula that generates the graph of the Archimedes’ spiral    is given by $\text{\hspace{0.17em}}r=\theta \text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}\theta \ge 0.\text{\hspace{0.17em}}$ As $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ increases, $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ increases at a constant rate in an ever-widening, never-ending, spiraling path. See [link] .

Given an Archimedes’ spiral over $\text{\hspace{0.17em}}\left[0,2\pi \right],$ sketch the graph.

1. Make a table of values for $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ over the given domain.
2. Plot the points and sketch the graph.

## Sketching the graph of an archimedes’ spiral

Sketch the graph of $\text{\hspace{0.17em}}r=\theta \text{\hspace{0.17em}}$ over $\text{\hspace{0.17em}}\left[0,2\pi \right].$

As $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ is equal to $\text{\hspace{0.17em}}\theta ,\text{\hspace{0.17em}}$ the plot of the Archimedes’ spiral begins at the pole at the point (0, 0). While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Further, there is no maximum value, unless the domain is restricted.

Create a table such as [link] .

 $\theta$ $\frac{\pi }{4}$ $\frac{\pi }{2}$ $\pi$ $\frac{3\pi }{2}$ $\frac{7\pi }{4}$ $2\pi$ $r$ 0.785 1.57 3.14 4.71 5.50 6.28

Notice that the r -values are just the decimal form of the angle measured in radians. We can see them on a graph in [link] .

Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
hii
Amit
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Dorbor
well
Biswajit
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Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1