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This blog is dedicated to helping students have a good understanding of what to expect on the AP Calculus Free Response Questions. Use the Topics menu on the right to find the solution you're looking for.
Topics
Area and Volume:
2002 1 BC,2002 3b BC,2003 1 BC,2003 1b BC,2004 2 BC ,2004 5b BC,2005 1 BC,2005 5b BC,2006 1 BC,2006 1b BC
Rates:
2002 2 BC,2002 2b BC,2003 5 BC,2004 1 BC,2005 2b BC,2005 5 BC,2006 2 BC,2006 4b BC
Differential Equations:
Series:
Parametric:
FTC:
Polar:
2003 3 BC, 2003 2b BC, 2005 2 BC
Other: Email me at Kaleb_Allinson@lkstevens.wednet.edu Please comment if you have a question about a solution.
2002 1 BC,2002 3b BC,2003 1 BC,2003 1b BC,2004 2 BC ,2004 5b BC,2005 1 BC,2005 5b BC,2006 1 BC,2006 1b BC
Rates:
2002 2 BC,2002 2b BC,2003 5 BC,2004 1 BC,2005 2b BC,2005 5 BC,2006 2 BC,2006 4b BC
Differential Equations:
Series:
Parametric:
FTC:
Polar:
2003 3 BC, 2003 2b BC, 2005 2 BC
Other: Email me at Kaleb_Allinson@lkstevens.wednet.edu Please comment if you have a question about a solution.
2 comments:
I got all of this down to part c, and then when you are finding the length of the line .75x, where does the 1.455 come from? Could you write out the formula you used really quick?
The x value of the end of the section you need to find the length of is 1.9 something or other. I called it A in the problem. The y coordinate is found by plugging that into .75x. That's how I got 1.455. These are then essentially the a and b values of a right triangle and you do the Pythagorean theorem to find the hypotenuse.
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