How do you convert #(-2, 5)# to polar form?

Question

Samantha Abate · Accepted Answer

#(sqrt(29), pi-tan^(-1)(5/2))# or #(-sqrt(29), tan^(-1)(-5/2))# To convert rectangular #(a,b)# to polar we use the two formulas: #r=sqrt(a^2+b^2)# and #hat theta=tan^(1)(|b/a|)# then we have to "fix" the quadrant for #theta#. #r=sqrt((-2)^2+5^2)=sqrt(29)# #hat theta = tan^(-1)(5/2)# the point #(-2,5)# is in QII, so #theta = pi-hat theta# therefore #theta = pi-tan^(-1)(5/2)#. An alternate, but equivalent representation is #(-sqrt(29), tan^(-1)(-5/2))#, which uses a QIV angle but a negative #r# to end up in QII.

Abigail Armstrong · Answer

undefined To convert the point (-2, 5) to polar form, you can use the following steps: 1. Calculate the distance from the origin (r) using the formula: $$ r = \sqrt{x^2 + y^2} $$ 2. Determine the angle (θ) using the formula: $$ \theta = \arctan\left(\frac{y}{x}\right) $$ 3. Ensure the angle is within the appropriate range: - If $ x > 0 $, $ \theta = \arctan\left(\frac{y}{x}\right) $ - If $ x < 0 $, $ \theta = \arctan\left(\frac{y}{x}\right) + \pi $ - If $ x = 0 $ and $ y > 0 $, $ \theta = \frac{\pi}{2} $ - If $ x = 0 $ and $ y < 0 $, $ \theta = -\frac{\pi}{2} $ 4. Express the point in polar form as: $$ (-2, 5) = (r, \theta) $$