Ap Calc Ab Frq 2017

Conquering the 2017 AP Calculus AB Free Response Questions: A practical guide

The 2017 AP Calculus AB Free Response Questions (FRQs) presented a diverse range of challenges, testing students' understanding of fundamental concepts and their ability to apply them to various scenarios. Consider this: mastering these questions can significantly improve your understanding of calculus and boost your exam score. So this in-depth guide will dissect each question, providing detailed solutions, explanations, and valuable insights for current and future AP Calculus AB students. This guide will cover not only the mechanics of solving the problems but also the underlying calculus principles and strategies for tackling similar problems on the exam.

Introduction to the 2017 AP Calculus AB FRQs

The AP Calculus AB exam consists of two sections: multiple choice and free response. The free response section tests your ability to apply your knowledge to solve complex problems, show your work, and justify your reasoning. The 2017 FRQs covered a broad spectrum of topics, including:

• Differentiation: Finding derivatives, applying derivative rules (product, quotient, chain rule), and interpreting derivatives in context (rate of change, slopes of tangent lines).
• Integration: Evaluating definite and indefinite integrals, understanding the fundamental theorem of calculus, and applying integration techniques (substitution).
• Applications of Derivatives: Optimization problems, related rates, analyzing graphs of functions and their derivatives.
• Applications of Integrals: Finding areas, volumes, and accumulating quantities.

Let's walk through each of the six FRQs individually, providing a comprehensive solution and explanation for each part.

Question 1: Particle Motion and Accumulation

Problem: A particle moves along the x-axis with velocity given by v(t) = 3t² – 12t + 9 for time t ≥ 0.

(a) At what times t is the particle at rest?

(b) For what values of t is the particle moving to the left?

(c) Find the total distance traveled by the particle from time t = 0 to time t = 3.

(Solution and Explanation):

(a) The particle is at rest when its velocity is zero. We set v(t) = 0 and solve for t:

3t² – 12t + 9 = 0 t² – 4t + 3 = 0 (t – 1)(t – 3) = 0 t = 1, t = 3

That's why, the particle is at rest at t = 1 and t = 3.

(b) The particle is moving to the left when its velocity is negative. We analyze the sign of v(t):

• For 0 ≤ t < 1, v(t) > 0 (positive)
• For 1 < t < 3, v(t) < 0 (negative)
• For t > 3, v(t) > 0 (positive)

Thus, the particle is moving to the left for 1 < t < 3.

(c) Total distance is the integral of the absolute value of velocity. We need to consider the intervals where the velocity is positive and negative separately:

Total distance = ∫₀¹ |v(t)| dt + ∫₁³ |v(t)| dt + ∫₃³ |v(t)| dt

= ∫₀¹ (3t² – 12t + 9) dt + ∫₁³ -(3t² – 12t + 9) dt

Evaluating these integrals (using the power rule for integration):

= [t³ – 6t² + 9t]₀¹ + [-t³ + 6t² – 9t]₁³ = 4 + 4 = 8

The total distance traveled from t = 0 to t = 3 is 8 units.

Question 2: Area and Volume

Problem: Let R be the region bounded by the graphs of y = eˣ, y = 0, x = 0, and x = 1.

(a) Find the area of region R.

(b) Find the volume of the solid generated when R is rotated about the x-axis.

(Solution and Explanation):

(a) The area of region R is given by the definite integral:

Area = ∫₀¹ eˣ dx = [eˣ]₀¹ = e¹ – e⁰ = e – 1

(b) The volume of the solid generated when R is rotated about the x-axis is given by the disk method:

Volume = π ∫₀¹ (eˣ)² dx = π ∫₀¹ e²ˣ dx

Using u-substitution (let u = 2x, du = 2dx), we get:

Volume = (π/2) ∫₀² eᵘ du = (π/2) [eᵘ]₀² = (π/2)(e² – 1)

Question 3: Related Rates

Problem: A container has the shape of an inverted right circular cone with radius 5 cm and height 10 cm. Water is poured into the container at a constant rate of 10 cm³/sec. Find the rate at which the water level is rising when the height of the water is 4 cm.

(Solution and Explanation):

This problem involves related rates. We need to relate the volume of the cone (V) to its height (h) and radius (r). The volume of a cone is given by:

V = (1/3)πr²h

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Since the cone is inverted, the ratio of radius to height is constant: r/h = 5/10 = 1/2, so r = h/2. Substituting this into the volume equation:

V = (1/3)π(h/2)²h = (1/12)πh³

Now we differentiate both sides with respect to time (t):

dV/dt = (1/4)πh²(dh/dt)

We are given dV/dt = 10 cm³/sec and h = 4 cm. We need to find dh/dt:

10 = (1/4)π(4)²(dh/dt) 10 = 4π(dh/dt) dh/dt = 10/(4π) = 5/(2π) cm/sec

Question 4: Analyzing a Graph and its Derivative

(Problem): The graph of f', the derivative of f, is shown below. The graph of f' has horizontal tangent lines at x = -1, x = 1, and x = 3.

(a) For what values of x in the interval (-2, 4) is f increasing? Justify your answer.

(b) For what values of x in the interval (-2, 4) does f have a relative minimum? Justify your answer.

(c) For what values of x is the graph of f concave up? Justify your answer.

(Solution and Explanation):

(a) f is increasing when f'(x) > 0. Observing the graph, f'(x) > 0 for (-2, -1) and (1, 3).

(b) f has a relative minimum when f'(x) changes from negative to positive. This occurs at x = 1.

(c) f is concave up when f''(x) > 0. Since f''(x) represents the slope of f'(x), we look for where the slope of f'(x) is positive. This occurs in the intervals (-1, 1) and (3, 4).

Question 5: Differential Equation and Slope Fields

(Problem): Consider the differential equation dy/dx = xy.

(a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated.

(b) Let y = f(x) be the particular solution to the given differential equation with the initial condition f(1) = 2. Write an equation for the line tangent to the graph of y = f(x) at x = 1.

(c) Find the particular solution y = f(x) to the given differential equation with the initial condition f(1) = 2.

(Solution and Explanation):

(a) This involves sketching short line segments with slopes determined by xy at each of the twelve indicated points on the graph provided in the original exam.

(b) At x = 1, y = 2. The slope is dy/dx = (1)(2) = 2. The equation of the tangent line is: y - 2 = 2(x - 1) or y = 2x.

(c) This is a separable differential equation. We separate the variables and integrate:

(1/y) dy = x dx

∫(1/y) dy = ∫x dx

ln|y| = (1/2)x² + C

Solving for y:

y = Ae^(x²/2) where A = ±e^C

Using the initial condition f(1) = 2:

2 = Ae^(1/2) A = 2e^(-1/2)

So, the particular solution is y = 2e^((x²-1)/2)

Question 6: Accumulation and the Fundamental Theorem of Calculus

(Problem): Let f be the function defined by f(x) = ∫₀ˣ g(t) dt, where g is a continuous function. The graph of g is shown below.

(a) Find f(3).

(b) Find f'(3).

(c) Find f''(3).

(d) On what intervals is f increasing? Justify your answer.

(Solution and Explanation):

(a) f(3) = ∫₀³ g(t) dt. This represents the area under the curve g(t) from t = 0 to t = 3. By calculating the areas of the triangles and rectangle, we find this area to be 2.5. Because of this, f(3) = 2.5.

(b) By the Fundamental Theorem of Calculus, f'(x) = g(x). That's why, f'(3) = g(3) = 0.5

(c) f''(x) = g'(x). The slope of g at x=3 is found visually on the graph as -1, so f''(3) = g'(3) = -1

(d) f is increasing when f'(x) > 0, which means g(x) > 0. Observing the graph of g, this occurs on the interval (0, 3).

Conclusion: Mastering the 2017 AP Calculus AB FRQs

By thoroughly understanding the solutions and explanations provided for each question in the 2017 AP Calculus AB FRQs, you can significantly enhance your preparation for the exam. Remember to practice regularly, focusing on developing a strong understanding of the underlying concepts and applying various problem-solving strategies. This detailed breakdown provides a solid foundation for success. Think about it: remember to always show your work clearly and justify your answers to maximize your score. Consistent practice and a focused approach will lead to mastery of the material. Good luck!