3r=-6cosx
I need help converting to rectangular form
also:change to polar co-ordinate form
3xy=5
Unusual to have an x in a polar form, was expecting
3r = -6cosØ , will assume that.
recall that cosØ = x/r
so
3r = -6x/r
3r^2 = -6x
x^2 + y^2 = -6x
for 3xy = 5
cosØ = x/r ---> x = rcosØ
and sinØ = y/r --> y = sinØ
3(rcosØ)(rsinØ) = 5
3r^2 sinØcosØ = 5
To convert the equation 3r = -6cos(x) to rectangular form, we need to use the following conversions:
1. r = sqrt(x^2 + y^2)
2. x = rcos(theta)
3. y = rsin(theta)
Let's begin by converting 3r = -6cos(x) into rectangular form:
1. Divide both sides of the equation by 3 to isolate r:
r = -2cos(x)
2. Replace r in terms of x and y:
sqrt(x^2 + y^2) = -2cos(x)
3. Square both sides of the equation to eliminate the square root:
x^2 + y^2 = 4cos^2(x)
By applying the conversion formulas, we have successfully converted the equation to rectangular form.
Now, let's move on to converting the equation 3xy = 5 to polar coordinate form:
1. Divide both sides of the equation by 3:
xy = 5/3
2. Convert x and y to polar coordinates:
x = rcos(theta)
y = rsin(theta)
3. Substitute these values back into the equation:
rcos(theta) * rsin(theta) = 5/3
4. Simplify the equation:
r^2 * cos(theta) * sin(theta) = 5/3
This is the equation in polar coordinate form.