Stitz-Zeager_College_Algebra_e-book

Weve seen this matrix sin cos most recently in the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5π 3 θ= x 5π 4 x Q P √ P has rectangular coordinates (2, −2 3) π P has polar coordinates 4, 53 Q has rectangular coordinates (−3, −3) √π Q has polar coordinates 3 2, 54 3. The point R(0, −3) lies along the negative y -axis. While we could go through the usual computations4 to find the polar form of R, in this case we can find the polar coordinates of R using the definition. Since the pole is identified with the origin, we can easily tell the point R is 3 units from the pole, which means in the polar representation (r, θ) of R we know r = ±3. π Since we require r ≥ 0, we choose r = 3. Concerning θ, the angle θ = 32 satisfies 0 ≤ θ < 2π 3π with its terminal side along the negative y -axis, so our answer is 3, 2 . To check, we note π π x = r cos(θ) = 3 cos 32 = (3)(0) = 0 and y = r sin(θ) = 3 sin 32 = 3(−1) = −3. 4. The point S (−3, 4) lies in Quadrant II. With x = −3 and y = 4, we get r2 = (−3)2 +(4)2 = 25 so r = ±5. As usual, we choose r = 5 ≥ 0 and proceed to determine θ. We have tan(θ) = 4 Since x = 0, we would have to determine θ ge...
View Full Document

This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

Ask a homework question - tutors are online