The AP Calculus BC exam is composed of 45 multiple choice questions (MCQs) and 6 free response questions (FRQs), each section weighing 50% of the total grade for a possible score from 1 to 5. In this guide, we will go over the strategies for the multiple choice questions and then work through two hard multiple choice questions.
It’s important to note that the BC exam score also contains an AB subscore, which is a grade you receive for solving the questions that cover AB material. Most colleges treat your AB subscore as a valid substitute for the official AB exam score.
Basic Strategies
The multiple choice section of the AP Calculus BC exam is divided into two parts, one with 30 no calculator questions followed by 15 calculator questions. The free response section is split between 3 no calculator and 3 calculator questions.
1. Prepare Well
Before taking the AP Calculus BC exam, it is important to practice. The BC exam is very different from other quizzes, tests, or homework you may have done at school. It is long and comprehensive, with difficult multi-step problems. You should take at least one full-length test to see what it really feels like. You can take official free response practice tests and IAC mock multiple choice test. There is also one full practice test available from 2012 administration of the exam. You can also download I Aced Calculus Flashcards App, which contains all BC material you need to know with video explanations, along with calculator tips, and over 900 MCQ practice problems.
2. Read Each Problem Carefully
It may surprise you, but many students answer questions incorrectly or waste valuable time because they misread the problems. Take your time and read the problem to make sure you fully understand what’s being asked. We cannot emphasize enough how important this step is.
3. Think About the Problem
Sometimes it’s best to not jump right into a problem. Often there are also multiple ways to solve one problem. Before you start solving, consider your options and make a plan. This will help you save time and be more accurate.
4. Keep Moving
You have only a few minutes to work on each problem. In the MCQ section, there’s a huge time crunch. If you spend too much time on a single problem, you might not get to others. Remember, you only need to get about 60%-70% of the problems correct to get a perfect score. If you can’t figure out a multiple choice question, make your best guess and keep moving. Keep in mind that there are no points subtracted for incorrect answers, so make sure to pick an answer for every question.
5. Check Your Work
Set aside some time to check your work for each question. It’s always important to check your work, even briefly, after you answer a question. In the MCQ section, there are answer choices specifically designed to catch students making the most common mistakes. Make sure you don’t fall for those distractor choices.
6. Eliminate Choices
Sometimes you can substitute an answer into the problem to eliminate answer choices. Other times your answer is an expression that involves variables, so you can pick specific values for those variables. For example, if a problem gives a particle’s position as \(at^2 + bt + c\) and asks for the value of \(t\) when the particle’s velocity is \(0\), the answer will be in terms of \(a\), \(b\), and \(c\). However, you may let \(a = 1\), \(b = 2\), \(c = 3\) and the particle’s position will be \(t^2 + 2t + 3\) and its velocity will be \(2t+2\), which can be used to eliminate answer choices. There is another example of selecting specific functions in the second problem later on.
Practice Problems
The most important things on the AP Calculus BC exam are your timing and accuracy – as long as you are confident in your answers and making progress on a problem at all times, only minor adjustments to your strategy may be necessary. Below are two problems from the first section of BC exams.
Sample AP Calculus BC Problem I
Let \(y=f(x)\) be the solution to the differential equation \(\dfrac{dy}{dx} = 1+2y\) with the point \(f(0) = 1\). What is the approximation for \(f(1)\) if Euler’s method is used, starting at \(f(0)\) with a step size of \(0.5\)?
- \(2.5\)
- \(3.5\)
- \(4.0\)
- \(5.5\)
In this question, it is unlikely we will be able to eliminate answer choices in any meaningful way. To begin the problem, let’s remember the formula for Euler’s method:
\(y_n = y_{n-1}+\Delta x \cdot f’(x_{n-1},y_{n-1})\)
In order to use it, however, we need to understand what it means. We start by choosing a starting value, or initial point \(y_{n-1}\). Sometimes the formula is written using \(y_n\) and \(y_{n+1}\) instead of \(y_{n-1}\) and \(y_n\), but these are preferential shifts of the index and don’t affect the actual computation. With this notation, we will first choose a point \(y_{n-1}\) and use the formula to predict the next point \(y_n\) on \(f(x)\). Setting this point to the initial value given in the problem statement, we still need to find a couple more values to complete the Euler step. Namely, we need to find the values \(\Delta x\) and \(f’(x_{n-1},y_{n-1})\). The former value is the step size, given here to be \(0.5\). This means that we will increase our argument value by increments of \(0.5\), stepping from \(f(0)\) to \(f(0.5)\) and finally to \(f(1)\). To find \(f’(x_{n-1},y_{n-1})\), we’ll need the derivative of \(f(x)\)—since we’re told \(f(x)\) is the solution to the given differential equation, that very equation is our derivative. The \(x_i, y_i\) values used as the argument are the coordinates of the initial point \(f(0)\). Thus, we can carry out the Euler scheme as follows:
\[ y_1 = y_0 + \Delta x \cdot f'(x_0, y_0) \]
\[ = 1 + 0.5(1 + 2(1)) \]
\[ = 2.5 \]
\[ y_2 = y_1 + \Delta x \cdot f'(x_1, y_1) \]
\[ = 2.5 + 0.5(1 + 2(2.5)) \]
\[ = 5.5 \]
By Euler’s method, \(y_2 = f(1) = 5.5\), which is answer choice (D). Since there are many computations involved in this question, one of the best ways to check if our solution is indeed correct is to make sure there are no arithmetic errors throughout the process. In order to do this, it is important that you write down your full solution and work in case there is a hidden error.
Sample AP Calculus BC Problem II
The position of a particle moving in the \(xy\)-plane is given by the parametric equations
\(x(t) = \frac{3t}{t+1}, \quad y(t) = \frac{-4}{t^2-1}\)
What is the slope of the line tangent to the path of the particle at the point where \(t=3\)?
- \(2\)
- \(\dfrac{3}{8}\)
- \(\dfrac{1}{5}\)
- \(5\)
To keep moving and begin progress on this question, we first need to understand what is being asked of us. Since the solution will be a number—the slope of the particle’s tangent line—what we first need is a derivative equation. Even though we are dealing with parametric equations, we desire the derivative \(\dfrac{dy}{dx}\), an equation that will give us the slope of the line tangent to a point on the particle’s path. Then, we will substitute \(t=3\) into that equation to find the correct solution. For this particular question, the steps toward the solution aren’t as difficult to come up with as the desired result—the hard part is computing the derivative \(\dfrac{dy}{dx}\). We’re provided with parametric equations of the form \(x(t)\) and \(y(t)\), each independent function of the parameter \(t\). If it’s difficult to remember what to do next to find the desired derivative, we can continue making progress on the question by writing down and evaluating useful equations. If we’re looking for a derivative of some sort, differentiation is necessary. Taking the derivative of \(x(t)\) will give us \(x’(t) = \dfrac{dx}{dt}\), and similarly, taking the derivative of \(y(t)\) will give us \(y’(t) = \dfrac{dy}{dt}\). Hopefully, at this point, you will be comfortable enough with derivative operations to notice, if not earlier, that \(\dfrac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y’(t)}{x’(t)}\). Now, we just need to carry out the computations using the quotient rule:
\[x'(t) = \dfrac{d}{dt}\left(\frac{3t}{t+1}\right)\]
\[= \frac{3(t+1)-3t}{(t+1)^2}\]
\[= \frac{3}{(t+1)^2}\]
and
\[y'(t) = \dfrac{d}{dt}\left(\frac{-4}{t^2-1}\right)\]
\[= \frac{0-(-4)(2t)}{(t^2-1)^2}\]
\[= \frac{8t}{(t^2-1)^2}\]
Since we need to evaluate \(\dfrac{dy}{dx}\) at \(t=3\), we can use the following equivalence to simplify the process:
\[ \left.\dfrac{dy}{dx}\right|_{t=3} = \left.\frac{y'(t)}{x'(t)}\right|_{t=3} = \frac{y'(3)}{x'(3)} \]
Evaluating each of the derivatives individually, we get that \(y’(3) = \dfrac{3}{8}\) and \(x’(3) = \dfrac{3}{16}\). Dividing the former by the latter, we have a choice (A) \(2\) as the correct answer. Here, we couldn’t eliminate answer choices because it is not possible to tell which choices are reasonable candidates until the end of the problem. To check our work, however, we can review the derivative expressions and make sure they were computed accurately.
Conclusion
This article contains general advice and strategies and you may find some work for you, while others don’t. These general strategies apply to most multiple choice exams, including the SATs. You can find a sample multiple choice section for the AP Calculus AB exam with solutions here. You can find a complete list of AP Calculus AB free response questions with solutions here. We hope that you learned something from this article and we wish you get a 5 on the exam. We believe that having strategies is always better than going without them. Good luck!
