convert the polar equation to rectangular form.
1.) r sec(theta) = 3
2.) r = 4 cos(theta) - 4 sin(theta)
convert from rectangular equation to polar form.
1.) x^2 + (y-1)^2 = 1
2.) (x-1)^2 + (y+4)^2 = 17
1)r=3cos(theta)
r^2=3rcos(theta)
x^2+y^2=3x
2)r=4cos(theta)-4sin(theta)
r^2=4rcos(theta)-4rsin(theta)
x^2+y^2=4x-4y
x^2-4x +y^2+4y =0
x^2-4x+4+y^2+4y+4=4+4
(x-2)^2 +(y+2)^2=8
1)x^2+(y-1)^2=1
x^2+y^2-2y+1=1
x^2+y^2=2y
r^2=2rsin(theta)
r=2sin(theta)
2)(x-1)^2+(y+4)^2=17
x^2-2x+1+y^2+8y+16=17
x^2+y^2=2x-8y
r^2=2rcos(theta)-8rsin(theta)
r=2cos(theta)-8sin(theta)
thank you soooooooooo much
To convert the given polar equations to rectangular form, and given rectangular equations to polar form, let's take one equation at a time.
1.) r sec(theta) = 3
To convert this equation to rectangular form, we'll use the following relationships:
r = sqrt(x^2 + y^2)
tan(theta) = y/x
Using these relationships, we can rewrite the given equation as:
sqrt(x^2 + y^2) * (1/cos(theta)) = 3
Multiplying both sides by cos(theta), we get:
x^2 + y^2 = 3cos(theta)
This is the rectangular form of the given polar equation.
2.) r = 4 cos(theta) - 4 sin(theta)
Similarly, using the relationships mentioned earlier, we can rewrite this equation as:
sqrt(x^2 + y^2) = 4(cos(theta) - sin(theta))
Squaring both sides, we get:
x^2 + y^2 = 16(cos^2(theta) - 2cos(theta)sin(theta) + sin^2(theta))
Rearranging and using trigonometric identities, we have:
x^2 + y^2 = 16(cos^2(theta) + sin^2(theta) - 2sin(theta)cos(theta))
Since cos^2(theta) + sin^2(theta) = 1, the equation simplifies to:
x^2 + y^2 = 16(1 - 2sin(theta)cos(theta))
This is the rectangular form of the given polar equation.
To convert the given rectangular equations to polar form, let's again take one equation at a time.
1.) x^2 + (y - 1)^2 = 1
To convert this equation to polar form, we use the following relationships:
x = rcos(theta)
y = rsin(theta)
Substituting these values in the given equation, we get:
(rcos(theta))^2 + (rsin(theta) - 1)^2 = 1
Expanding and simplifying, we have:
r^2cos^2(theta) + r^2sin^2(theta) - 2rsin(theta) + 1 = 1
Simplifying further, we get:
r^2(cos^2(theta) + sin^2(theta)) - 2rsin(theta) = 0
Using the identity cos^2(theta) + sin^2(theta) = 1, the equation becomes:
r^2 - 2rsin(theta) = 0
This is the polar form of the given rectangular equation.
2.) (x - 1)^2 + (y + 4)^2 = 17
Similarly, substituting the values of x and y, we get:
(rcos(theta) - 1)^2 + (rsin(theta) + 4)^2 = 17
Expanding and simplifying, we have:
r^2cos^2(theta) - 2rcos(theta) + 1 + r^2sin^2(theta) + 8rsin(theta) + 16 = 17
Simplifying further, we get:
r^2(cos^2(theta) + sin^2(theta)) - 2rcos(theta) + 8rsin(theta) = 0
Using the identity cos^2(theta) + sin^2(theta) = 1, the equation becomes:
r^2 - 2rcos(theta) + 8rsin(theta) = 0
This is the polar form of the given rectangular equation.
To convert a polar equation to rectangular form, you can use the following formulas:
1.) r sec(theta) = 3
First, we need to eliminate the secant term. Since sec(theta) = 1/cos(theta), we can multiply both sides of the equation by cos(theta):
r sec(theta) * cos(theta) = 3 * cos(theta)
Using the trigonometric identity sec(theta) * cos(theta) = 1, the equation simplifies to:
r = 3 * cos(theta)
To convert a rectangular equation to polar form, you can use the following formulas:
1.) x^2 + (y-1)^2 = 1
We can express x and y in terms of polar coordinates r and theta. By substituting x = r cos(theta) and y = r sin(theta), the equation becomes:
(r cos(theta))^2 + (r sin(theta) - 1)^2 = 1
Expanding and simplifying the equation:
r^2 cos^2(theta) + r^2 sin^2(theta) - 2r sin(theta) + 1 - 2r sin(theta) + r^2 cos^2(theta) = 1
Combining like terms, we get:
r^2 (cos^2(theta) + sin^2(theta)) - 2r sin(theta) + 1 - 2r sin(theta) = 1
Using the trigonometric identity cos^2(theta) + sin^2(theta) = 1, the equation simplifies to:
r^2 - 2r sin(theta) = 0
Factoring out r, we get:
r (r - 2sin(theta)) = 0
From this equation, we have two possibilities:
r = 0 and r - 2sin(theta) = 0
For r = 0, we have a single point at the origin.
For r - 2sin(theta) = 0, we divide both sides of the equation by r:
1 - 2sin(theta)/r = 0
Adding 2sin(theta)/r to both sides, we get:
1 = 2sin(theta)/r
Rearranging the equation, we have:
r = 2sin(theta)
Therefore, the polar form of the equation is r = 2sin(theta).
2.) (x-1)^2 + (y+4)^2 = 17
By substituting x = r cos(theta) and y = r sin(theta), the equation becomes:
(r cos(theta) - 1)^2 + (r sin(theta) + 4)^2 = 17
Expanding and simplifying the equation:
r^2 cos^2(theta) - 2r cos(theta) + 1 + r^2 sin^2(theta) + 8r sin(theta) + 16 = 17
Combining like terms, we get:
r^2 (cos^2(theta) + sin^2(theta)) - 2r cos(theta) + 8r sin(theta) + 17 - 16 = 17
Using the trigonometric identity cos^2(theta) + sin^2(theta) = 1, the equation simplifies to:
r^2 - 2r cos(theta) + 8r sin(theta) = 0
Factoring out r, we get:
r (r - 2cos(theta) + 8sin(theta)) = 0
From this equation, we have two possibilities:
r = 0 and r - 2cos(theta) + 8sin(theta) = 0
For r = 0, we have a single point at the origin.
For r - 2cos(theta) + 8sin(theta) = 0, we rearrange the terms:
r = 2cos(theta) - 8sin(theta)
Therefore, the polar form of the equation is r = 2cos(theta) - 8sin(theta).