Question

Help me with these polar coordinate problems please!

I filled some in, could I get those checked? For the rest, please explain how I can solve them. I have formulas to use, but I don't know how to use them. They are:

`x=rcos(\theta)\\y=rsin(\theta)\\x^2+y^2=\theta^2\\tan(\theta)=(y)/(x)`


The sooner the better please, because I have a quiz soon. I'll put this at a good amount of points so if you don't have a real answer...please don't answer. Thank you so much.

Answer 1

Convert the rectangular equation to polar form

3x-y+2=0

To convert the rectangular equation 3x - y + 2 = 0 to polar form, we can substitute x = r cos(theta) and y = r sin(theta), where r is the radius and theta is the angle in polar coordinates. This gives:

3(r cos(theta)) - (r sin(theta)) + 2 = 0

Simplifying the equation, we get:

r(3 cos(theta) - sin(theta)) = -2

Dividing both sides by the expression in the parentheses:

r = -2 / (3 cos(theta) - sin(theta))

This is the polar form of the equation.

Convert the rectangular equation to polar form

12.3x-y+2=0

13. xy=4

14. (x+y)-9(x-2)=0 72-36=0

15. y²-8x-16-0 (4-4)(3-4)=0

For the rectangular equation xy = 4, we can convert it to polar form using the substitution x = r cos(theta) and y = r sin(theta), which gives:

r cos(theta) * r sin(theta) = 4

r^2 sin(theta) cos(theta) = 4

Using the identity 2 sin(theta) cos(theta) = sin(2theta), we can simplify the equation to:

r^2 sin(2theta) = 8

r = sqrt(8 / sin(2theta))

This is the polar form of the equation.

For the rectangular equation (x+y) - 9(x-2) = 0, we can simplify it to:

x - 8y + 18 = 0

Then, we can convert it to polar form using the substitution x = r cos(theta) and y = r sin(theta), which gives:

r cos(theta) - 8r sin(theta) + 18 = 0

Simplifying the equation, we get:

r = 18 / (cos(theta) - 8sin(theta))

This is the polar form of the equation.

For the rectangular equation y^2 - 8x - 16 = 0, we can complete the square to get:

y^2 - 8x - 16 = (y - 0)^2 - 16 - 8x

Simplifying the equation, we get:

(y - 0)^2 = 8x + 16

Using the substitution x = r cos(theta) and y = r sin(theta), we get:

r^2 sin^2(theta) = 8r cos(theta) + 16

r^2 sin^2(theta) - 8r cos(theta) - 16 = 0

This equation does not simplify nicely into a standard form of polar equation, but it is still a valid polar form.

16. r = 4 sin θ

17.θ= (π/6)

18. r² = sin 2θ

19. r = 6/(2-3 sin θ)

To convert the polar equation r = 4 sin(theta) to rectangular form, we can use the following trigonometric identities:

sin(theta) = y / r

cos(theta) = x / r

Substituting these into the polar equation, we get:

r = 4 sin(theta)

r = 4 y / r

r^2 = 4 y

x^2 + y^2 = 4 y

This is the rectangular form of the equation.

The equation theta = pi/6 represents a line at an angle of pi/6 radians (30 degrees) from the positive x-axis in the polar coordinate system. In rectangular coordinates, this line has the equation y = x tan(pi/6) = x/sqrt(3).

To convert the polar equation r^2 = sin(2theta) to rectangular form, we can use the following trigonometric identities:

sin(2theta) = 2 sin(theta) cos(theta)

sin(theta) = y / r

cos(theta) = x / r

Substituting these into the polar equation, we get:

r^2 = sin(2theta)

r^2 = 2 sin(theta) cos(theta)

r^2 = 2 (y / r) (x / r)

x^2 + y^2 = 2xy

This is the rectangular form of the equation.

To convert the polar equation r = 6 / (2 - 3 sin(theta)) to rectangular form, we can first substitute sin(theta) = y / r and cos(theta) = x / r, giving:

r = 6 / (2 - 3y / r)

r(2 - 3y / r) = 6

2r - 3y = 6

This is the rectangular form of the equation.

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