AP Calculus ABFRQ#1 2011 By: Kamil Davis Period: 6 and 7
By looking at the question number we can tell that this is a calculator question. We are given both an equation for the velocity and acceleration, and told that the position of the particle when t=0 is 2. We are also given a domain from t=0 to t=6, so that interval is our main conern. Our first manner of business is to graph the functions. Please remember which is which.
A Quick Review of the Concepts • The relation between position, velocity, and acceleration. • The knowledge of understanding how to analyze the graph of velocity. • The use of derivative and integral as inverse operations. • Learning to visualize and deduce based on information in the given.
Getting to know your faithful partner. When you see and equation that it's exactly pretty to look at, don't panic. Remain calm and see if you can simplify or manipulate the equation with any identities, properties, or rules. And If it's a calculator question, use your calculator. I've personally battled many equations with Lucifer, my faithful TI-83.
Analysis of the Graphs 2sin(e^(x/4))+1 and 0.5(e^(x/4))cos((e^(x/4)) are shown above and the graphs show a familiar form that I'll go into on the next slide.
See the cat below? It's flight path is an oscillation. But that refers to the shape of its flight path. Its velocity graph would actually be a straight line, because it's moving at a constant speed. Oscillating Behavior Oscillation refers to a graph that has a wave pattern, generally seen among sine and cosine graphs and their variants.
Don't be intimidated! • AP test writers love to try to scare you away with weird particle graphs, and laugh at you as you try to understand the motion path. Oscillation on a velocity graph means that a particle is moving back and forth. It can go right, then left, then right again. If the graph is below the x-axis, it's moving left, if it's above the x-axis, it's moving right. So the particle is moving like the AP Test writer in the example.
Doing Part A To find whether or not speed is increasing or decreasing at t=5.5, we need to look at both graphs of 2sin(e^(x/4))+1 and 0.5(e^(x/4))cos((e^(x/4))once more. At t=5.5, both the velocity and acceleration are below the x-axis and negative. Since the sign of both the velocity and acceleration are same, which is negative, the speed is increasing. Think of this as the particle traveling left, and getting faster as it travels left.
Further Evaluating Part A Graphs aren't always easy to read like this one. So in such cases, go to your faithful TI-83 or TI-84 or TI- OVER 9000! graphing calculator and use the Calc button to find the value of each function when t=5.5. You should find that v(5.5)= -0.45337 and a(5.5)= -1.35851. One can also use the table function of the graphing calculator to look at the values. It's important to learn each method because this question's concepts could show up on a non-calculator question.
Part B The average velocity of the function is what we need to find. The formula below is average velocity, and the interval from t=0 to t=6 will represent our lower and upper limit. v(t) will be the velocity function. Substitute into the integral and solve. Which should be about 1.949. Round to at least 3 places.
Net Distance is the same as displacement. It only factors the distance from the starting position to the finish. So if I take 10 paces left, then 10 paces right, my net distance is 0 paces. Total Distance is a measurement of total distance traveled. So if I take 10 paces left, then 10 paces right, my total distance is 20 paces. Understanding Part CNet Distance vs Total Distance
Answering Part C • Total distance traveled from t=0 to t=6 is simply the integral of the absolute value of velocity from 0 to 6. The absolute value forces the calculator to factor the negative area of the curve, and include that distance traveled. Input the formula below into your calculator to find the answer, which is 12.573. Once again emphasizing that we round to at least 3 significant figures.
Cross-Examining Part D To find a change of direction, one must look at the graph of velocity function and find the point in which the function crosses the x-axis. A change in sign, be it a positive velocity to a negative velocity or vice versa, reflects a change in direction. We are also finding the position of the partcile at the point that is changes direction. Which is the integral if velocity from t=0 to the point it changes direction, plus the initial postion.
Putting Part D To Rest Once again, the graphing calculator can be used to make the job quicker. Use the trace function to find a point in the interval (0,6) that crosses the x-axis. Then use the calc button and find the zero of the function to find the point that v(t) changes sign. You will see that the point is at t = 5.19552. So we know that our bounds are lower limit 0 and upper limit 5.19552. We aren't rounding anything until the final step to mitigate the possiblity of errors. Since velocity is the derivative of position, we can integrate to solve in reverse. The postion in time at t=0 is 2, so that will be added to the integral of v(t) dt from 0 to 5.19552.
The Fun in Learning Concepts • Calculus is about understanding concepts. The concepts in relation to position, velocity, and acceleration on a graph are the same as the relation between a function and its higher order derivatives. We also see the topic in curve sketches. Once you can apply these concepts, Calculus is much less intimidating.
Citations • http://www.mszhao.com/calculus.cfm?subpage=1306470 • my.hrw.com/math06_07/nsmedia/tools • www.imommies.com • www.tumblr.com • www.bulbapedia.net • forgot-to-remember.deviantart.com • memebase.cheezburger.com