Convert the equation y =#-(sqrt(3)/(3))x# to polar form?

Question

Anna Andrews · Accepted Answer

#theta=(5pi)/6+pin# where #n# is an element of all integers

This is my first time attempting converting from rectangular to polar, so please feel free to correct If I am wrong The conversion from Rectangular to Polar: #x=rcostheta# #y=rsintheta# And the point is solve for #theta# in this case as the #r# cancels: 1. Substitute into the Rectangular form: #rsintheta=-sqrt3/3*rcostheta# 2. Subtract #rcostheta# from both sides: #(rsintheta)/(rcostheta)= -sqrt3/3# 3. Recall #sintheta/costheta= tantheta#: #tantheta= -sqrt3/3# 4. General solution #theta=(5pi)/6+pin# where #n# is an element of all integers

Zoe Austin · Answer

undefined To convert the equation $ y = -\frac{\sqrt{3}}{3}x $ to polar form, we use the relationships between Cartesian (x, y) and polar (r, θ) coordinates:

$$ x = r \cos(\theta) $$
$$ y = r \sin(\theta) $$

First, substitute $ x $ and $ y $ in the equation $ y = -\frac{\sqrt{3}}{3}x $ with their polar forms:

$$ r \sin(\theta) = -\frac{\sqrt{3}}{3} r \cos(\theta) $$

Next, divide both sides by $ r $ to isolate $ \theta $:

$$ \frac{\sin(\theta)}{\cos(\theta)} = -\frac{\sqrt{3}}{3} $$

Now, simplify the left side using the tangent function:

$$ \tan(\theta) = -\frac{\sqrt{3}}{3} $$

Finally, find the polar form by taking the arctangent (inverse tangent) of both sides:

$$ \theta = \arctan\left(-\frac{\sqrt{3}}{3}\right) $$

This gives the polar form of the equation.