Stitz-Zeager_College_Algebra_e-book

# to answer questions like these which involve both a

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: r = 3 cos 2 . Using the techniques presented in Example 11.5.2, we ﬁnd that we need to plot both functions as θ ranges from 0 to 4π to obtain the complete graph. To our surprise and/or delight, it appears as if these two equations describe the same curve! y 3 −3 3 x −3 θ r = 3 sin 2 and r = 3 cos θ 2 appear to determine the same curve in the xy -plane To verify this incredible claim,16 we need to show that, in fact, the graphs of these two equations intersect at all points on the plane. Suppose P has a representation (r, θ) which θ θ satisﬁes both r = 3 sin 2 and r = 3 cos 2 . Equating these two expressions for r gives θ θ the equation 3 sin 2 = 3 cos 2 . While normally we discourage dividing by a variable θ expression (in case it could be 0), we note here that if 3 cos 2 = 0, then for our equation θ to hold, 3 sin 2 = 0 as well. Since no angles have both cosine and sine equal to zero, θ θ θ we are safe to divide both sides of the equation 3 sin 2 = 3 cos 2 by 3 cos 2 to get θ tan 2 = 1 which gives θ = π + 2πk for integers k . From these solutions, however, we 2 14 Again, we could have easily chosen to substitute these into r = 3 which would give −r = 3, or r = −3. We obtain t...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern