Exam Updated Informationap Calculus

AP Calculus: Students may only take one (1) AP Calculus Exam. The AP Program has allowed students to switch from Calculus BC to Calculus AB and vice versa at no charge. There will be no AB subscore this year. The deadline to make this change is extended to APRIL 20th. AP coordinators must call AP Services for Educators to unlock their order to. Exam Format: (2) FRQs 45 minutes total (20 minutes per FRQ) AP Calculus AB Exam Information: AP Calculus AB Exam is May 12 th at 2:00 PM. AP is encouraging students to login at 1:30 PM. Late/Make-up date is June 1 at 4 pm; Exam Format: Two Free Response Questions (each question will have several parts).

Speed
  • Apr 03, 2020 Here is the latest information on the 2020 AP Calculus Exams as of April 3, 2020. Updated 4/29/2020 Update: A message from the AP Program 4/28/2020 Subject: How to Prepare Your Students for the 2020 AP Exams Dear Colleagues, Additional information is now available to help guide you and your students through the exam day experience.
  • Exam questions assess the course concepts and skills outlined in the course framework. For more information on exam weighting, download the AP Calculus AB Course and Exam Description (CED). Encourage your students to visit the AP Calculus AB student page for exam information and exam practice. TUE, MAY 4, 2021, 8 AM LOCAL.
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  • Multiple Choice Questions to Prepare for the AP Calculus AB Exam is your essential tool to scoring well on AP Calculus AB Exam. This book fits the College Board requirements for the 2018 AP Exam, and reflects all the changes in the AP Calculus AB curriculum and the AP Exam format which took place in the 2016-2017 school year.
  • Multiple Choice Questions to Prepare for the AP Calculus AB Exam is your essential tool to scoring well on AP Calculus AB Exam. This book fits the College Board requirements for the 2019 AP Exam, and reflects all the recent changes in the AP Calculus AB curriculum and the AP Exam format.
  • The exam is 3 hours and 15 minutes long. The AP Calculus exam has 2 sections, multiple choice and free response sections. Note: You do now allow taking both AP Calculus AB and Calculus BC exam within the same year.You need a Graphical calculator for this exam.
  • AP Calculus AB Updates and New Resources for 2019-20. To bring you and your students new classroom resources and supports, we’re making updates to AP Calculus AB for the 2019-20 school year.
  • AP Calculus AB Practice Exam Multiple Choice Problem 80 The acceleration of a particle moving along the x-axis is given by a(t) = ln(3^t + 2). At time t=2, the velocity of the particle is 2.
  • The AP Calculus AB exam requires extensive use of a graphing calculator, and students are required to provide their own (see the collegeboard calculator policy). The AP Calculus AB test is comprised of 2 sections each with 2 parts. Each of the 2 sections make up 50% of the overall score. The Multiple Choice section:
  • Multiple Choice Questions to Prepare for the AP Calculus BC Exam is your essential tool to scoring well on AP Calculus BC Exam. This book fits the College Board requirements for the 2019 AP Exam, and reflects all the recent changes in the AP Calculus BC curriculum and the AP Exam format.
  • The AP Calculus BC exam requires extensive use of a graphing calculator, and students are required to provide their own (see the collegeboard calculator policy). The AP Calculus BC test is comprised of 2 sections each with 2 parts. Each of the 2 sections make up 50% of the overall score. The Multiple Choice section:
  • You can use the resources below as you prepare for the AP Exam. Click here for details about the exam format.. Sample Questions. You’ll find sample multiple choice and free-response questions in the AP Calculus AB and BC Course and Exam Description (.pdf/6.4MB).
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  • AP Calculus AB Updates and New Resources for 2019-20. To bring you and your students new classroom resources and supports, we’re making updates to AP Calculus AB for the 2019 …
  • Apr 02, 2019 · AP Calculus AB Practice Exam Multiple Choice Problem 80 The acceleration of a particle moving along the x-axis is given by a(t) = ln(3^t + 2). At time t=2, the velocity of the particle is 2.
  • Multiple Choice Questions to Prepare for the AP Calculus AB Exam is your essential tool to scoring well on AP Calculus AB Exam. This book fits the College Board requirements for the 2019 AP Exam, and reflects all the recent changes in the AP Calculus AB curriculum and the AP Exam format.
  • You can use the resources below as you prepare for the AP Exam. Click here for details about the exam format.. Sample Questions. You’ll find sample multiple choice and free-response questions in the AP Calculus AB and BC Course and Exam Description (.pdf/6.4MB).
  • Multiple Choice Questions to Prepare for the AP Calculus AB Exam is your essential tool to scoring well on AP Calculus AB Exam. This book fits the College Board requirements for the 2018 AP Exam, and reflects all the changes in the AP Calculus AB curriculum and the AP Exam format which took place in the 2016-2017 school year.
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  • The exam is 3 hours and 15 minutes long. The AP Calculus exam has 2 sections, multiple choice and free response sections. Note: You do now allow taking both AP Calculus AB and Calculus BC exam within the same year.You need a Graphical calculator for this exam.
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Calculus I Examples For Exam II

If it is not already on your hard drive, you will need to download the free DPGraph Viewer to view some of the pictures linked to on this page.QuickTime free download.

Here is a Geometer's Sketchpad video with audio (Larger Version) of me describing approximating the slope of a tangent line with a secant line.

Here is a Quicktime Animation showing a secant line approximating a tangent line.

Here is a Winplot demonstration of a secant line that can be used to approach a tangent line. You may need to download the file to your desktop and then use the freeware Winplot to open the file (by opening Winplot, clicking on Window, clicking on 2-dim, clicking on File, clicking on Open, and then opening SecantSlope from your desktop. You can use the sliders to vary the values of A and H. The A-value gives you the x-value at a point on the graph of the function you are looking at. The H-value gives you the x-value of another point on the graph of the function you are looking at in terms of A + H (so H is your delta x). The secant line through the points on the graph of the given function with x-values of A and A + H is graphed and its slope displayed at the top of the graph. The default function is f(x) = 3sin(x). This function is named FN and you can edit the definition of FN by clicking on Equa, choosing User functions ... , highlighting the function and editing its definition. Changing the definition to x^2 for example will produce a different graph.

Computing the derivative using the limit definition of derivative

Derivative Approximation

The following points are the points pictured on the graph of the function at the right: (-0.5,10.875), (-0.4,10.816), (-0.3,10.693), (-0.2,10.512), (-0.1,10.279), (0,10), (0.1,9.681),(0.2,9.328), (0.3,8.947), (0.4,8.544), (0.5,8.125). Approximate the derivative of the function at the points (-0.3,10.693), (0,10), and (0.4,8.544).

Finding the Equation of a Tangent Line

Find an equation of the line tangent to the graph of the function

f(x) = x3 - 2x2 - 3x + 10 at the point (-1,10).

Click on the picture at the right to see a Quicktime animation of tangent lines as x

goes from -2 to 2.4.

Differentiation Example--Product Rule (and Chain Rule)

What would the graph look like for x greater than 4?

Differentiation Example--Quotient Rule

What do you think the graph looks like outside the viewing window shown above?

Differentiation Example--Chain Rule (Twice)

Do you think a graphing calculator could draw an accurate graph of this function over an x-interval of [0,10]?

Use implicit differentiation to find the equations of the tangent and normal lines to the graph of the relation whose equation is given below at the point (4,2).

In the graph below the tangent line is drawn in blue and the normal line is drawn in red.

Use implicit differentiation to find an equation of the line tangent to the graph of the given relation at the indicated point.

Click on the picture to enlarge.

Three more implicit differentiation examples

DPGraph picture of z = xcos(y) + 2y - y2 and z = c.

When c = 0 the intersection of the two graphs will be the graph of this relation.

DPGraph picture of z = x2cos(y) + 2xy - xy2 and z = c.

When c = 0 the intersection of the two graphs will be the graph of this relation.

Compare the graph on the right to the graph in example 1 and note the difficulty atx = 0. This graph was constructed using Maple. Can you tell what the problem is around the y-axis? Click on the graph for a larger Maple image.

DPGraph picture of z = 2xy - y2 + 3 + 3xy3 - cos(xy2) and z = c.

When c = 0 the intersection of the two graphs will be the graph of this relation.

Quicktime AnimationAverage and Instantaneous SpeedAnimation--No Scales

Click on the picture to see an animation.

Just as she is entering a 20 mph speed zone, Emily notices a police car up ahead parked a little way off the road. Emily applies the brakes gently to begin slowing down (she does not want to be too obvious). She first hits the brakes when the front of her car is at position zero in the picture above and she slows down for three seconds with a constant negative acceleration. She travels a total of 111 feet during these three seconds and at the end of the three seconds the front of her car is level with the radar gun being used to clock her speed. It turns out that the radar began clocking her motion at the instant she first started to slow down. The tables below give distance traveled in feet and time in seconds from the time when she first applied the brakes.

Time

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.0

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

2.9

3.0

Distance

0.00

3.99

7.96

11.91

15.84

19.75

23.64

27.51

31.36

35.19

39.00

42.79

46.56

50.31

54.04

57.75

61.44

65.11

68.76

72.39

76.00

79.59

83.16

86.71

90.24

93.75

97.24

100.71

104.16

107.59

111.00

Time

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.10

. . . .

2.90

2.91

2.92

2.93

2.94

2.95

2.96

2.97

2.98

2.99

3.00

Distance

0.0000

0.3999

0.7996

1.1991

1.5984

1.9975

2.3964

2.7951

3.1936

3.5919

3.9900

. . . . . .

107.5900

107.9319

108.2736

108.6151

108.9564

109.2975

109.6384

109.9791

110.3196

110.6599

111.0000

What was Emily's average speed in feet per second and in miles per hour from time t = 0 seconds to time t = 3 seconds? What would you estimate to have been Emily's speed in feet per second and in miles per hour at time t = 0 seconds? What would you estimate to have been Emily's speed in feet per second and in miles per hour at time t = 3 seconds? Estimate Emily's speed in feet per second and in miles per hour at time t = 2 seconds. I will call the position function s(t).

Answers

Average speed from t = 0 to t = 3 seconds:

Speed at t = 0 seconds:

That guess turns out to be pretty good since the radar, measuring to an accuracy of 7 digits to the right of the decimal, indicated that

s(0.0000001) = 0.000004.

Speed at t = 3 seconds:

This guess also seems to be pretty good since the radar, measuring to an accuracy of 7 digits to the right of the decimal, indicated that

s(2.9999999) = 110.9999966. Can you see why this makes the 34 ft/sec estimate of the speed at t = 3 seconds seem pretty accurate?

Speed at t = 2 seconds:

Since we have less precise data around t = 2 seconds it would probably be more accurate to look at s(2.1) and s(1.9).

Do you think Emily got pulled over and if so do you think she got a ticket or a warning? Do we have enough information to make an educated guess at the answer to that question? The idea for this example was inspired by a terrific presentation on speed from Calculus Quest.

Notice that in this example the average speed over the three second time interval is equal to the average of the estimated instantaneous speeds at the beginning and end of the time interval. Would this always be the case? Would it be possible for someone to be traveling at 40 ft/sec at time t = 0 seconds, 34 ft/sec at time t = 3 seconds, and yet the average speed over this 3 second interval be 38 ft/sec or even be 42 ft/sec?

EC Find an equation representation of the position function, s(t), that would nicely fit the data given above.

Average and Instantaneous Speed

The position function shown below gives the position of an object in feet as a function of time in seconds. Determine when the object is traveling to the right (positive direction) and when the object is traveling to the left (negative direction). Determine the average speed from t = 1 to t = 5. Determine the instantaneous speed at t = 1 and at t = 5.

The graph below shows position (not distance traveled) as a function of time. Click here or on the picture to see a linear motion animation. In the linear motion animation the animated point on the left vertical axis represents the elapsed time. Quicktime animation

A Vertically Launched Projectile

Exam Updated Information Ap Calculus Frq

Quicktime Animation (In the animation the point traveling down the vertical axis indicates the speed of the projectile.)

The function above (called a position function) gives the height above the ground (in feet) of a projectile launched vertically (straight up) with an initial velocity of 192 ft/sec and an initial height of 100 feet above the ground (air resistance is being ignored here). The variable 't' represents time in seconds with t = 0 corresponding to the moment the projectile was launched. Here are the types of questions you need to be able to answer.

(A) What is the average speed of the projectile during the first 3 seconds after launch?

(B) What is the average speed of the projectile from t = 1 second to t = 4 seconds?

(C) What is the instantaneous speed of the projectile at t = 3 seconds?

(D) What will be the maximum height attained by the projectile?

(E) How long until the projectile comes back down and hits the ground?

Solutions

(A) We need to find the total distance traveled during the first three seconds and divide this total distance traveled by the time interval (3 seconds). This means we need to compute

Updated

(B)

(C) The derivative of the position function will give us the instantaneous speed. Thus we need to evaluate the derivative at t = 3 seconds (i.e., substitute 3 for t).

(D) At some point in time the projectile is going to stop going up and start coming back down. At the instant it stops going up its speed will be zero ft/sec. When it starts coming back down its speed (velocity in the sense that direction is now being considered) will be negative. At the instant when the speed is zero the projectile will be at its highest point (maximum height). We need to find the value of time 't' when the derivative (which gives velocity) is zero. We can then evaluate the position function (which gives height above the ground) at that value of t to find the maximum height.

Exam Updated Information Ap Calculus Notes

Updated

(E) The projectile will hit the ground when the value of the position function is zero.

Exam Updated Informationap Calculus 2

The t-value of -0.5 would correspond to before the projectile was launched and is not in the domain of our position function describing this motion. The answer must come from the positive solution to the equation above. Thus the projectile comes back down and hits the ground in 12.5 seconds.

Section 2.6 Number 2a
In the picture above the blue point is the point where x = 3. Click on the picture (Quicktime version) to see an animation of a point moving along the graph of the parabola as a function of time (t) where