SOLUTION: 1. Use table of values and plotting points in polar grid to graph the polar equation r=2/sin(theta). 2. What type of graph did you get? 3. Justify your answer to part 2

Algebra ->  Trigonometry-basics -> SOLUTION: 1. Use table of values and plotting points in polar grid to graph the polar equation r=2/sin(theta). 2. What type of graph did you get? 3. Justify your answer to part 2      Log On


   



Question 1004654: 1. Use table of values and plotting points in polar grid to graph the polar equation r=2/sin(theta).
2. What type of graph did you get?
3. Justify your answer to part 2 by converting polar equation to rectangular equation.
I HAVE FILLED IN THE VALUES BELOW.
θ
0 r=2/sin(theta)
pi/6 undefined
pi/4 4
pi/3 2sqroott2
pi/2 4sqroot3/3
2pi/3 2
3pi/4 4sqroot3/3
5pi/6 4
pi undefined
3pi/2 -2
2pi undefined
When I start looking at it and trying to graph I get so confused. I am seeing everything from a circle on a graphing calculator, to a straight line, and even a parabola. Can anyone help me with the answers and graph? Maybe point out where I went wrong?

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the equation in polar coordinates is r = 2/sin(theta)

in rectangular coordinates, you would get:

x = r * cos(theta)
y = r * sin(theta)

using pythagorus, the hypotenuse of the triangle formed by x and y is equal to:

h^2 = x^2 + y^2 = r^2 * cos^2(theta) + r^2 * sin^2(theta) which becomes:

h^2 = r^2 * (sin^2(theta) + cos^2(theta)) which becomes:

h^2 = r^2 which becomes:

hypotenuse = j = r

you are given that r = 2 / sin(theta)

this means that the hypotenuse is equal to 2 / sin(theta).

the sine of theta is equal to y / r

r is equal to 2 / sin(theta)

this means that r is equal to 2 / (y / r)

since 2 / (y / r) is the same as 2 * (r / y), then:

r is equal to 2 * (r / y)

multiply both sides of this equation by y and divide both sides of this equation by r and you get:

y = 2

what this says is that y is going to be equal to 2 regardless of the value of x.

that means that the graph of the equation of r = 2 / sin(theta) is a straight line that is parallel to the x-axis and has a value of 2.

you can verify this graphically.

just plot the graph of y = 2 / sin(theta) and you will see that it is a straight line at y = 2.

you need graphing software that can graph equations in polar form.

the following calculator can do that:

http://www.desmos.com/calculator

the graph i made is shown below:

$$$

you can see that the graph is a horizontal line at y = 2.

the following picture may help clarify the relationship between the polar coordinate form and the rectangular form of the equation.

$$$

what you see in the pictures is that r = 2 / sin(theta) is the hypotenuse of a right triangle that has theta as the angle and the opposite side as y and the adjacent side as x.

since sin(theta) = y/r, then y = r * sin(theta)

since cos(theta) = x/r, then x = r * cos(theta)

since r is given as being equal to 2/sin(theta), you can replace r in the equations for x and y with 2/sin(theta) and solve for x and y.

you get y = 2

you get x = 2 cot(theta).

by the pythagorean formula, r = sqrt(x^2 + y^2)

replace x^2 with (2cot(theta)^2 and replace y with 2^2 and you wind up with:

r = sqrt(4cot^2(theta) + 4)

this is also equivalent to r = sqrt(4/tan^2(theta) + 4)

that last form is easier to calculate from when you use your calculator.

what i think you were trying to do is manually plot the equation.

that's a chore.

you can do it, but it's much easier to graph it using appropriate graphing software.

also knowing the translations from polar to rectangular helps.

i used graphing software to see that the graph was a line at y = 2.

once i saw that, i was able to figure out how it became that way using algebra.

calculating for y said that y was always going to be equal to 2.

that corresponded with the graph and so i was reasonably confident that the equation of r = 2 / sin(theta) was the same as the equation y = 2.

what you were doing was trying to calculate r which would not be the same each time but would always wind up creating a line where the y value of the coordinate point would be equal to 2.

you would only have been able to see this after connecting all those points.