4 Original FRQs Mirroring Exam Style — With Complete Step‑by‑Step Solutions
Master the AP Calculus AB exam by practicing these rigorously designed Free Response Questions. This workshop covers heavily tested topics including Rate & Accumulation, Area & Volume, Graphical Function Analysis, and Particle Motion.
Tip: Pay attention to the calculator-active versus non-calculator sections, and try to write out your mathematical justifications fully before expanding the solutions.
A large storage tank initially contains 80 liters of brine solution. Water flows in at \( R_{in}(t) = 10 + 3\sin\left(\frac{t}{2}\right) \) L/hr and out at \( R_{out}(t) = 12 + 2\cos t \) L/hr, for \( 0 \le t \le 10 \).
a) Rate at t = 4:
\( A'(4) = R_{in}(4) - R_{out}(4) \)
\( A'(4) \approx 12.7279 - 10.6927 = 2.035 \)
Since \( A'(4) > 0 \), the amount of brine is increasing.
b) Accumulation Function A(t):
The rate of change is \( A'(t) = 10 + 3\sin\left(\frac{t}{2}\right) - (12 + 2\cos t) = -2 + 3\sin\left(\frac{t}{2}\right) - 2\cos t \).
Integrating gives: \( A(t) = \int \left( -2 + 3\sin\left(\frac{t}{2}\right) - 2\cos t \right) dt \)
\( A(t) = -2t - 6\cos\left(\frac{t}{2}\right) - 2\sin t + C \)
Using the initial condition \( A(0) = 80 \):
\( 80 = -0 - 6(1) - 0 + C \implies C = 86 \)
So, \( A(t) = 86 - 2t - 6\cos\left(\frac{t}{2}\right) - 2\sin t \).
Evaluating at \( t = 6 \):
\( A(6) \approx 80.5 \) liters → 81 liters.
c) Critical Points and Absolute Minimum:
Set \( A'(t) = 0 \) to find critical points:
\[ -2 + 3\sin\left(\frac{t}{2}\right) - 2\cos t = 0 \]
Using the double angle identity \( \cos t = 1 - 2\sin^2\left(\frac{t}{2}\right) \):
\[ -2 + 3\sin\left(\frac{t}{2}\right) - 2\left(1 - 2\sin^2\left(\frac{t}{2}\right)\right) = 0 \]
Let \( u = \sin\left(\frac{t}{2}\right) \). The equation becomes: \( 4u^2 + 3u - 4 = 0 \).
By the quadratic formula: \( u = \frac{-3 + \sqrt{73}}{8} \approx 0.693 \) (ignoring the negative root).
Solving \( \sin\left(\frac{t}{2}\right) = 0.693 \) yields two critical points in the interval:
1) \( t_1 = 2\arcsin(u) \approx \mathbf{1.530} \) (Relative Minimum, as \( A' \) changes from negative to positive)
2) \( t_2 = 2\pi - 2\arcsin(u) \approx \mathbf{4.753} \) (Relative Maximum, as \( A' \) changes from positive to negative)
To find the absolute minimum, evaluate \( A(t) \) at the endpoints and critical points:
The absolute minimum occurs at \( t = 10 \) with approximately 65.4 liters.
Let \( f(x) = 2e^{-x/3} + 1 \) and \( g(x) = 0.2x^2 - 1.5x + 4 \) for \( x \geq 0 \). The two curves intersect at two points in the first quadrant. Region \( R \) is bounded by \( y = f(x) \) and \( y = g(x) \) for \( x \) between the leftmost and rightmost intersection.
a) Intersection points:
Set \( f(x) = g(x) \) and solve using a graphing calculator:
\[ 2e^{-x/3} + 1 = 0.2x^2 - 1.5x + 4 \]
The intersections are at \( a \approx 1.474 \) and \( b \approx 4.792 \).
b) Area of R:
On the interval \( [1.474, 4.792] \), \( f(x) \geq g(x) \). The area is given by:
\[ \text{Area} = \int_{1.474}^{4.792} \left[ (2e^{-x/3} + 1) - (0.2x^2 - 1.5x + 4) \right] dx \]
\[ \text{Area} = \int_{1.474}^{4.792} \left( 2e^{-x/3} - 0.2x^2 + 1.5x - 3 \right) dx \]
Using numerical integration, the Area is \( \approx 0.973 \).
c) Volume around y = 6:
Using the washer method, the axis of revolution is above the region. The outer radius is \( R(x) = 6 - g(x) \) and the inner radius is \( r(x) = 6 - f(x) \).
\[ V = \pi \int_{1.474}^{4.792} \left[ (6 - g(x))^2 - (6 - f(x))^2 \right] dx \]
Substitute the functions:
\[ \mathbf{V = \pi \int_{1.474}^{4.792} \left[ (2 + 1.5x - 0.2x^2)^2 - (5 - 2e^{-x/3})^2 \right] dx} \]
The figure below shows the graph of \( f' \), the derivative of a differentiable function \( f \), on the closed interval \( [-3, 5] \). The graph of \( f' \) consists of five line segments. It is known that \( f(0) = 4 \).
a) Relative Maxima:
\( f \) has a relative maximum at \( x = -2 \) and \( x = 3 \).
Justification: A relative maximum occurs where \( f'(x) \) changes sign from positive to negative. Looking at the graph, \( f'(x) \) crosses the x-axis from above to below at both \( x = -2 \) and \( x = 3 \).
b) Concavity:
The graph of \( f \) is concave up on the interval \( (-0.5, 2) \).
Justification: The graph of \( f \) is concave up when its second derivative \( f''(x) > 0 \). This occurs precisely when \( f'(x) \) is increasing. Based on the graph, the slopes of the segments are positive only between \( x = -0.5 \) and \( x = 2 \).
c) Average Value & FTC:
Using the average value formula for \( f' \) on \( [0, 4] \):
\[ \frac{1}{4-0} \int_{0}^{4} f'(x) \, dx = -0.5 \]
Multiplying by 4 gives:
\[ \int_{0}^{4} f'(x) \, dx = -2 \]
By the Fundamental Theorem of Calculus:
\[ f(4) - f(0) = -2 \]
Substituting the given initial condition \( f(0) = 4 \):
\[ f(4) - 4 = -2 \implies \mathbf{f(4) = 2} \]
d) Decreasing & Concave Down:
\( f \) is decreasing and concave down on \( (-2, -0.5) \) and \( (3, 5) \).
Justification: \( f \) is decreasing when \( f'(x) < 0 \) (below the x-axis) and concave down when \( f' \) is decreasing (negative slope). The portions of the graph where \( f' \) is both below the x-axis AND going downwards are the intervals \( (-2, -0.5) \) and \( (3, 5) \).
A particle moves along the \( x \)-axis with velocity \( v(t) = t^2 - 6t + 8 \) for \( t \ge 0 \). At time \( t=0 \), the particle is at position \( x(0)=2 \).
a) Acceleration and Speeding Up:
Acceleration is the derivative of velocity: \( a(t) = v'(t) = 2t - 6 \).
The particle is speeding up when \( v(t) \) and \( a(t) \) have the same sign. Let's analyze the signs by factoring velocity: \( v(t) = (t-2)(t-4) \).
The particle is speeding up on \( (2, 3) \cup (4, \infty) \).
b) Rest and Total Distance:
The particle is at rest when \( v(t) = 0 \implies (t-2)(t-4) = 0 \implies \mathbf{t=2} \) and \( \mathbf{t=4} \).
Total distance = \( \int_{0}^{5} |v(t)| \, dt = \int_{0}^{2} v(t) \, dt - \int_{2}^{4} v(t) \, dt + \int_{4}^{5} v(t) \, dt \).
Using the general antiderivative \( F(t) = \frac{1}{3}t^3 - 3t^2 + 8t \):
Distance = \( |F(2) - F(0)| + |F(4) - F(2)| + |F(5) - F(4)| \)
Distance = \( \left|\frac{20}{3}\right| + \left|\frac{16}{3} - \frac{20}{3}\right| + \left|\frac{20}{3} - \frac{16}{3}\right| = \frac{20}{3} + \frac{4}{3} + \frac{4}{3} = \mathbf{\frac{28}{3}} \) units.
c) Position Function and Location:
The position function \( x(t) \) is the integral of velocity:
\[ x(t) = \int (t^2 - 6t + 8) \, dt = \frac{1}{3}t^3 - 3t^2 + 8t + C \]
Using the initial condition \( x(0) = 2 \):
\[ 2 = 0 - 0 + 0 + C \implies C = 2 \]
Position function: \( x(t) = \frac{1}{3}t^3 - 3t^2 + 8t + 2 \).
At \( t = 4 \):
\[ x(4) = \frac{64}{3} - 3(16) + 8(4) + 2 = \frac{64}{3} - 48 + 32 + 2 = \frac{64}{3} - 14 = \mathbf{\frac{22}{3}} \]
d) Moving Toward/Away from Origin:
At \( t = 3 \):
Position: \( x(3) = \frac{27}{3} - 3(9) + 8(3) + 2 = 9 - 27 + 24 + 2 = 8 \).
Velocity: \( v(3) = (3)^2 - 6(3) + 8 = 9 - 18 + 8 = -1 \).
Since the position is positive (to the right of the origin) and the velocity is negative (moving to the left), the particle is moving toward the origin.