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Calculus I Examples For Exam II
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). 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 Exam Dates and Information – AP Students – The.
How students will be successful in AP Calculus How caregivers can help students be successful Students will be successful if:. Students devote 30+ min/day to reviewing for the AP exam. Students complete the problems under named time/calculator conditions. Students review correct answers when available. 2020 AP Exam Schedule and Information – AP Coronavirus Updates College Board 2020 AP Exam Schedule and Information The 2020 AP Exam administration is complete. A small number of students whose exam submissions were incomplete will be registered to take exams in the August 24–31 window.
| 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. |
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| 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 |
Exam Updated Information Ap Calculus Frq
| 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
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?
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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
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(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.
(E) The projectile will hit the ground when the value of the position function is zero.
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 |