Please note that knowledge of calculus is not required for AP B or AICE students. It is highly recommended that students taking AP C also take a calculus course, or have previously taken a calculus course. This page is intended for students who are currently in calculus but have not yet reached the relevant concepts, or who need refreshment on these concepts.
In algebra, you were taught how to find the slope of a straight line, both by interpretation of a linear function (for example, we know that y = 2x + 3 has a slope of 2 because the slope is the coefficient of the x term) and with the slope formula, (y2 - y1)/(x2 - x1). However, for non-linear functions, the first method cannot be used, and the second can only provide and average slope over an interval.
Calculus provides us with a way to find the "slope" at any point along any function, whether it is linear or not. This is done by measuring the slope of a line tangent to the curve at that point. An example of a few tangent lines is shown below.
Each of the red lines is tangent to the graph of f(x) at the corresponding blue point. Notice that the lines seem to follow where the function would be if it continued in either direction with a constant slope. In calculus, we use a method known as differentiation to allow us to find the slopes of these tangent lines.
Before we begin learning how to differentiate, it is important to understand a few things. Looking at a linear function, slope is a rate of change. More specifically, it is the rate of change of y with respect to x. In simpler terms, the slope of a line tells us how much the y value changes for each increment that x changes. For linear functions, this value remains constant. However, the rate of change of a non-linear function can't be a constant, leaving only one possibility: it is another function.
This brings us to differentiation. Differentiation is the process of using a given function to find the function representing its rate of change, called its derivative. For a function f(x), the derivative of f(x) is denoted f `(x). This may sound complicated, but it's actually a simple process carried out by using a set of simple rules based on the structure of the function. Here, we will only cover simple polynomials, as they are the only types of functions on the AP Physics C exam.Rule #1: The Constant Rule
If f(x) = c, where c is any constant, f `(x) = 0
Example: f(x) = 1, so f `(x) = 0
Example: f(x) = -10, so f `(x) = 0
Example: f(x) = 3p, so f `(x) = 0
Remember that p is a constant, and so any constant multiplied by p is also a constant.
Example: f(x) = 4(27 + 32), so f `(x) = 0
Note that what 4(27 + 32) evaluates to is irrelevant in finding the derivative because it is still a constant.
Rule #2: The Power Rule
If f(x) = xn, then f `(x) = nxn-1
Example: f(x) = x3, so f `(x) = 3x2
Example: f(x) = x2, so f `(x) = 2x
Note that when there is no exponent written on x, it is to the 1st power.
Example: f(x) = x, so f `(x) = x0 = 1
Note that any number to the power of 0 is 1, so x0 = 1
Rule #3: The Constant Multiple Rule
If f(x) = cg(x), where c is some constant, then f `(x) = cg`(x)
Example: f(x) = 2x2, so f `(x) = 2*2x = 4x
Example: f(x) = 5x, so f `(x) = 5
Remember that the derivative of x by the power rule is 1, and 1*5 = 5
Rule #4: The Sum and Difference Rules
If f(x) = g(x) + h(x), then f `(x) = g`(x) + h`(x)
If f(x) = g(x) - h(x), then f `(x) = g`(x) - h`(x)
Example: f(x) = x3 + x2, so f `(x) = 3x2 + 2x
Example: f(x) = 2x2 - 3x, so f `(x) = 4x - 3
Example: f(x) = 5x4 + 2x3 - x2 - 3x + 5, so f `(x) = 20x3 + 6x2 - 2x - 3
Remember that the derivative of a constant is 0, so any constant terms in f(x) become 0 in f `(x)
Using just these four simple rules, it is possible to take the derivative of any simple polynomial containing only one variable, like those in the examples for rule #4. Also note that if the function has parenthesis, it can be turned into a simple polynomial by distributing.
The most basic way to apply this to physics is through kinematics. In kinematics, we cover position and displacement, velocity, and acceleration. Upon analyzing these three properties of motion, you may notice that velocity is the rate of change of position over time, and acceleration is the rate of change of velocity over time. This means that velocity is the derivative of position with respect to time, and acceleration is the derivative of velocity with respect to time.
Knowing this information, we see that for any position function x(t), x`(t) = v(t), the velocity function, and v`(t) = a(t), the acceleration function. Below, we will look at the fourth Big 5 equation, which actually takes the form of a position function. Note that for any problem using this equation, a, vi, and xi are all constants, only t is variable.x = x(t) = 1/2at2 + vit + xi
x`(t) = v(t) = 2 * 1/2at + vi = at + vi
v'(t) = a(t) = a
Notice that v(t) is the same as the third Big 5 equation, and that a(t) is nothing but a constant. Numbers can be plugged into either of the first two equations to find a missing value, depending which equation is appropriate to the situation.
In an actual calculus class, you will learn many more rules for differentiating more complex functions, but these are all the rules you should need to know for the AP Physics C exam. But you're not done yet, derivatives aren't the only thing you need to know about calculus.Return to Top
Basic Indefinite Integration
Differentiation is a useful tool for physics, however it cannot solve all of our problems. For example, what if you want to go backwards, such as from a velocity function to its position function? Differentiation cannot go backwards like this, however, calculus has another tool for this task: integration. There are two broad types of integration: indefinite and definite. Here, we will start with indefinite integration and its applications, then continue to definite integration and its applications.
Like differentiation, indefinite integration is a process that follows certain rules. Integration also has it's own special notation, which looks something like the following: ∫ f(x)dx. The symbol int; tells you that you need to integrate, f(x) is the function to be integrated, and dx tells you that you are integrating with respect to x. As long as the variable after d matches the variable in the function and there are no other variables in the function, the following rules apply exactly.Rule #1: Constant Rule of Integration
∫ kdx = kx + c, where k and c are constants.
Example: ∫ 5dx = 5x + c
Note the + c that appears after integration. This is called the constant of integration, and must be added at the end of the solution to any indefinite integration problem. The reason for this is because the derivative of any constant is 0, and since you are essentially going from the derivative back to the original function, it is possible for any constant to appear at the end of the function. So for the above integral, the actual solution could be 5x, 5x + 3, 5x - 10, etc., as all these functions have a derivative of 5.
Example: ∫ -7dx = -7x + c
Example: ∫ 4pdx = 4px + c
Remember, p is a constant, so this rule applies to it.
Example: ∫ dx = ∫ 1dx = x + c
Rule #2: Power Rule of Integration
∫ xndx = (xn+1)/(n+1) + c
Example: ∫ x2dx = x3/3 + c
Example: ∫ x3dx = x4/4 + c
Example: ∫ xdx = ∫ x1dx = x2/2 + c
Note that this rule does not work for integrating x-1 = 1/x, as it would result in x0/0, which is undefined. There is a separate rule for this, but it should not be necessary for AP Physics.
Rule #3: Zero Rule of Integration
∫ 0dx = c
Rule #4: Constant Multiple Rule of Integration
∫ kf(x)dx = k∫ f(x)dx
Example: ∫ 2xdx = 2∫ xdx = 2x2/2 + c = x2 + c
Example: ∫ 5x2dx = 5∫ x2dx = 5x3/3 + c
Example: ∫ 2/3x4dx = 2/3∫ x4dx = 2/3(x5/5) + c = 2x5/15 + c
Rule #5: Sum and Difference Rules of Integration
∫ f(x) + g(x)dx = ∫ f(x)dx + ∫ g(x)dx
∫ f(x) - g(x)dx = ∫ f(x)dx - ∫ g(x)dx
Example: ∫ 3x2 + 5dx = 3x3/3 + 5x + c = x3 + 5x + c
Example: ∫ 2x3 + 4x2 + 3x + 1dx = 2x4/4 + 4x3/3 + 3x2/2 + x + c = x4/2 + 4x3/3 + 3x2/2 + x + c
Using these rules, it should be possible to integrate any function used in the AP C curriculum. However, there are two questions that remain: "what do we do with the + c?" and "How do we apply this to physics?"
The first question is answered using an initial condition. For example, the integration problem might be to integrate f `(x) = 3x + 2, and the initial condition might be f(1) = 3. So in order to solve this, we first integrate f `(x) indefinitely, as follows:f `(x) = 3x + 2
f(x) = ∫ 3x + 2dx
f(x) = 3x2/2 + 2x + c
Now we plug the initial condition, f(1) = 3, into the function, replacing x with 1 and f(x) with 3, and solve for c:3 = (3*12)/2 + 2*1 + c
3 = 3/2 + 2 + c
3 = 7/2 + c
c = 3 - 7/2 = -1/2
Now that we know what c equals, we simply plug it back into f(x):f(x) = 3x2/2 + 2x - 1/2
Now we know exactly what f(x) is, with only one variable. Now, we must apply this to physics. The most basic application is again in kinematics. Remember that the derivative of position is velocity, and the derivative of velocity is acceleration. Now, notice that integration is, essentially, the reverse of differentiation, so that means that the integral of acceleration is velocity, and the integral of velocity is position. See the appropriate section's practice problems for more details.
Now that we have covered basic differentiation and indefinite integration, we have only one concept remaining to explain: definite integration and its physical applications.Return to Top
Basic Definite Integration
Now that we have learned indefinite integration, we can begin learning definite integration. Definite integration uses the same set of rules as indefinite integration, but the notation and method have some additions. Also, the answer to a definite integral is not a function, but a number. We will talk more about the meaning of this number later. First, take a look at the notation for a definite integral:f(x)dx
Notice that the only change in notation is the addition of x1 and x2 at the top and bottom of the integral symbol. These are called the limits of integration, and define the interval on which you are performing the definite integration. x1 is the lower limit, and thus is usually a lower number, and x2 is the upper limit, and thus usually a higher number. This idea will become more clear as we practice a few problems, but in order to do this, we must look at arguably the most important theorem in calculus:The Fundamental Theorem of Calculus
f `(x)dx = f(x2) - f(x1)
∫ 3dx = 3x + c
(3*7 + c) - (3*2 + c)
21 + c - 6 - c
21 - 6 = 15
Note that the c from each equation cancels out because of the distribution of the negative sign to the second equation. This will happen in every problem, so the + c can be ignored in definite integration.
Example: -3x + 4dx
∫ -3x + 4dx = -3x2/2 + 4x + c
(-3*52/2 + 4*5) - (-3*22/2 + 4*2)
-3*25/2 + 20 + 3*4/2 - 8
-75/2 + 18
-75/2 + 36/2 = -39/2 = -19.5
Note that in calculus, you would normally leave the answer as a simplified improper fraction, like -39/2. In physics, we generally use decimal answers, like -19.5. Also, be very careful to distribute the negative to all terms of the second equation. Failing to distribute this negative properly can result in radically incorrect answers.
Now that we have seen how to use definite integration on basic polynomials, we need to understand what it means and when it is applicable to physics. The more relevant explanation of what a definite integral means would be the total change in f(x) from x1 to x2. So for example, if you are given a velocity function, v(t), and took the definite integral from ti to tf, then since the integral of velocity is position, you would be finding the change in position, or displacement. If you did the same to an acceleration function, a(t), then you would be finding the change in velocity over that time interval.
This should be all the calculus you need to know for the AP Physics C exam.Return to Top