videoĢ014 Multiple Choice AB Exam MC Parts 1&2 (pg 18-49 MC key pg. MC #1-5.video #Ģ012 Multiple Choice AB Exam MC Parts 1&2 (pg. MC #1-9.video MC # MC #16.22.video MC # MC # MC # MC #ġ998 Multiple Choice AB Exam MC Part 1 Solutions MC Part 2 Solutions MC #1-4.video MC #5-9, MC # MC # MC # MC # MC # MC # 2008 Multiple Choice AB Exam MC Part 1 Solutions MC Part 2 Solutions
#Ap calculus ab response 2008 free#
AP Calc Formula Practice WS Derivative/Integral RulesĬollege Board Released FRQs from 2000-2018 (Solution Keys under "Scoring Guidelines")ĪB Calculus Free Response Notebook (Collection of 1990-2010 AB FRQs & Solutions Organized by Topics/Category)Ģ019 AP FRQ Questions 2019 FRQs Solution Key 2021 AB FRQ Questions 2021 FRQ Solution Key 2003 Multiple Choice AB Exam MC Part 1 Solutions MC Part 2 Solutions We will be reviewing applications of derivatives (related rates and optimization), applications of integrals (density and volume), and techniques of antidifferentiating (integration by parts).Fall Semester Summary Sheet Spring Semester Summary Sheet.Three AP calculus exam free-response per night.Developing Expert Voices projects due soon.Wiki constructive modification due Sunday midnight.Related Rates: Distance and Velocity and Spongebob! To obtain dP/dt, we differentiate P = b + s. The rate of the tip of the man's shadow is dP/dt. The exam would never ask you to do part B because you need to do part B anyways to answer part A. We plug in the numbers to obtain ds/dt.Ī) Note this question is underlined in blue. To differentiate, we use The Product Rule and The Chain Rule-Refer to Related Rates: Area, Perimeter, and Diagonal of a Rectangle for reference. We simplify our proportions and differentiate to determine the rates of change (That's the definition of a derivative!). Since the man is walking toward the lamppost, by convention, the rate is negative.ī) Using similar triangles, we can see that the ratio of L to M equals the ratio of s to b+s. We know the height of the lamppost (L = 16), the height of the man (M = 6), the rate at which the man walks toward the streetlight (db/dt = -5), and the length of the man's shadow from the base of the lamppost (b = 10).
Related Rates: Shadows Including Similar Triangles Differentiate the formula using The Power Rule and The Chain Rule. Differentiate the formula using The Chain Rule, since P, L, and W are with respect to time-a function within a function! Then plug and chug.Ĭ) We know that the diagonal (D), the length (L), and the width (W) are related in The Pythagorean Theorem: the square of two sides (in this case, L and W) equals the square of the hypotenuse (D), so D^2 = L^2 + W^2. Since area, length, and width are all with respect to time-meaning that area, length, and width are functions of time-we must use The Chain Rule to differentiate.ī) We know the formula for the perimeter of a rectangle as P = 2L + 2W. Since L * W is a product, we use The Product Rule to differentiate. Since we're looking for the rates of change (That's the definition of a derivative!) we differentiate the formula, with respect to time. The length (L), the width (W), and the rates of which the lengths (dL/dt) and widths (dW/dt) are changing.īy convention, we're going to designate an increasing rate as a positive rate and a decreasing rate as a negative rate.Ī) We know the formula for the area of a rectangle as A = L * W, where A is area, L is length, and W is width.
Related Rates: Area, Perimeter, and Diagonal of a Rectangle Today's class we decided to do #2, reviewing related rates. Doing old exam free-response questions and study how each question evolved throughout the years and do 3 questions a night.Doing questions in class related to rusty topics.Related Rates: Distance and Velocity and Spongebob!.Related Rates: Shadows Including Similar Triangles.Related Rates: Area, Perimeter, and Diagonal of a Rectangle.Scribe again!!!!!OK, we start with finishing the h.How are you keeping first things first?.
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