{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

test1_2009

# test1_2009 - 1(15 marks Determine the area interior to both...

This preview shows pages 1–3. Sign up to view the full content.

1. (15 marks) Determine the area interior to both curves whose equations in polar coordinates are given by C 1 : r = sin ( θ 2 ) θ [0 , 2 π ] C 2 : r = sin ( θ 2 ) θ [2 π, 4 π ] (A sketch is enough to argue about the intersection of the two curves.) Solution. A sketch of the two curves is as follows (the first curve is plotted in blue): -1 1 -1 1 x y They are both symmetric with respect to the x -axis (because r ( θ ) = r (2 π - θ ) for the first one, r (2 π + θ ) = r (4 π - θ ) for the second). Moreover, the second curve is obtained from the first by reflecting along the origin (because r 2 (2 π + θ ) = - r 1 ( θ )). 1 From this we deduce that to compute the area of the intersection we can just take the area inside the sector delimited by the first curve in the interval [0 , π/ 2] and multiply by 4. Thus A = 4 · 1 / 2 Z π 2 0 sin 2 ( θ/ 2) Recall that cos ( θ ) = cos 2 θ/ 2 - sin 2 ( θ/ 2) = 1 - 2 sin 2 ( θ/ 2). Then A = Z π 2 0 (1 - cos ( θ )) = π 2 - 1 1 If you used the correct symmetries without proving them rigorously, you will not lose any points. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2. (15 marks) Show that the lines L 1 and L 2 given by L 1 : x = y - 1 2 = z - 2 3 L 2 : x - 3 - 4 = y - 2 - 3 = z - 1 - 2 intersect, find the point of intersection and the angle at which they intersect. Solution. A point of intersection is a point which satisfies the equa- tions of both lines, that is: x = y - 1 2 = z - 2 3 x - 3 - 4 = y - 2 - 3 = z - 1 - 2 ⇐⇒ x = y - 1 2 y - 1 2 - 3 - 4 = y - 2 - 3 y - 1 2 = z - 2 3 y - 2 - 3 = z - 1 - 2 ⇐⇒ x = - 1 , y = - 1 , z = - 1 .
This is the end of the preview. Sign up to access the rest of the 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