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starwarsfan
Registered User Posts: **166** Junior Member

In the paper 2007, question 3, part (a), I don't understand the solution at all. I'm confused about the areas of polar curves, and prep books aren't helping. (i'm self study...)

Post edited by starwarsfan on

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## Replies to: Calc BC help!

166Junior Member676MemberAP Calculus BC

Student Grade Distributions

AP Examinations - May 2008

Examination Grade Calculus BC

5 - 30,045 students or 43.5%

4 - 12,008 students or 17.4%

3 - 13,408 students or 19.4%

2 - 4,664 students or 6.7%

1 - 8,978 students or 13.0%

Number of Students: 69,103

You can also find the grade distributions for other subjects by going here: http://www.collegeboard.com/student/testing/ap/subjects.html. Click on the subject and on the left-hand side, click on 'Grade Distribution'.

2,303Senior Memberi think BC curve shouldn't be too diff from AB's curve,

probably 60~65+ pts would be the cutoff for 5?

OT

i'd like to know this as well..i suck at polar curves and series~~

where's themathprof? hehe

810MemberThat being said, here's my understanding of it, taking a quickie glance at what's there:

Let @ = theta. If you wanted to calculate the area of the shaded region, the integral expression represents only the area of the section of the curve from @ = 2pi/3 to @ = 4pi/3. In order to understand what's going on there, imagine a straight line connecting the origin to each of the intersection points. That divides the region into two sections: (1) a Pac-man like section consisting of the circle (r = 2) from @ = 2pi/3 traced around clockwise to the other intersection point at @ = 4pi/3, and (2) the rest of the shaded region that's sitting in "Pac-man's mouth".

The integral calculates the area of what's in Pac-man's mouth. The other piece (the 2/3pi(2)^2) of the area calculation is just finding the fraction of the circle that's there. In this case, there's 2/3rds of the circle there (an entire circle encompasses 2pi, and this circle encompasses 4pi/3: 2pi/3 from @ = 0 to @ = 2pi/3, and 2pi/3 from @ = 4pi/3 back to @ = 2pi).

Hope that helps.

2,303Senior Member2,303Senior MemberPauls Online Notes : Calculus II - Area with Polar Coordinates

^there he explains the polar curves etc..

its pretty useful~

2,303Senior Memberits pretty easy now cuz i know how the limits sweep counterclockwise etc

16,060Senior MemberFWIW, I found PR to be very helpful. You'll get a good feeling for what's on the test and its explanations are quite good too.

2,303Senior Member