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ap15_frq_calculus_bc

Question 1

(a) Topic-6.7-The Fundamental Theorem of Calculus and Definite Integrals

(b) Topic-5.3-Determining Intervals on Which a Function Is Increasing or Decreasing

(c) Topic-6.2-Approximating Areas with Riemann Sums

(d) Topic-6.7-The Fundamental Theorem of Calculus and Definite Integrals

The rate at which rainwater flows into a drainpipe is modeled by the function R, where \(R(t)=20sin\left ( \frac{t^{2}}{35} \right )\) cubic feet per hour, t is measured in hours, and 0 ≤ t ≤ 8. The pipe is partially blocked, allowing water to drain out the other end of the pipe at a rate modeled by D(t) = -0.04t3 + 0.4t2 + 0.96t cubic feet per hour, for 0 ≤ t ≤ 8. There are 30 cubic feet of water in the pipe at time t = 0.
(a) How many cubic feet of rainwater flow into the pipe during the 8-hour time interval 0 ≤ t ≤ 8?
(b) Is the amount of water in the pipe increasing or decreasing at time t = 3 hours? Give a reason for your answer.
(c) At what time t, 0 ≤ t ≤ 8, is the amount of water in the pipe at a minimum? Justify your answer.
(d) The pipe can hold 50 cubic feet of water before overflowing. For t > 8, water continues to flow into and out of the pipe at the given rates until the pipe begins to overflow. Write, but do not solve, an equation involving one or more integrals that gives the time w when the pipe will begin to overflow.

▶️Answer/Explanation

\(\textbf{1(a) } \int_{0}^{8} R(t) \, dt = 76.570\)

\(\textbf{1(b) } R(3) – D(3) = -0.313632 < 0\)

\(\quad \text{Since } R(3) < D(3), \text{ the amount of water in the pipe is decreasing at time } t = 3 \text{ hours.}\)

\(\textbf{1(c)} \text{The amount of water in the pipe at time } t, \, 0 \leq t \leq 8, \text{ is}\)

\(\quad 30 + \int_{0}^{t} \left[ R(x) – D(x) \right] dx.\)

\(\quad R(t) – D(t) = 0 \implies t = 0, \, 3.271658\)

\(
\begin{array}{|c|c|}
\hline
t & \text{Amount of water in the pipe} \\
\hline
0 & 30 \\
3.271658 & 27.964561 \\
8 & 48.543686 \\
\hline
\end{array}
\)

\(\quad \text{The amount of water in the pipe is a minimum at time } t = 3.272 \, \text{(or 3.271) hours.}\)

\(\textbf{1(d) } 30 + \int_{0}^{w} \left[ R(t) – D(t) \right] dt = 50\)

Question 2

(a) Topic-6.7-The Fundamental Theorem of Calculus and Definite Integrals

(b) Topic-9.4-Defining and Differentiating Vector-Valued Functions

(c) Topic-9.6-Solving Motion Problems Using Parametric and Vector-Valued Functions

(d) Topic-9.3-Finding Arc Lengths of Curves Given by Parametric Equations

At time $t \geq 0$, a particle moving along a curve in the $xy$-plane has position $(x(t), y(t))$ with velocity vector
\(
v(t) = \left( \cos(t^2), e^{0.5t} \right).
\)
At $t = 1$, the particle is at the point $(3, 5)$.
(a) Find the $x$-coordinate of the position of the particle at time $t = 2$.
(b) For $0 < t < 1$, there is a point on the curve at which the line tangent to the curve has a slope of $2$. At what time is the object at that point?
(c) Find the time at which the speed of the particle is $3$.
(d) Find the total distance traveled by the particle from time $t = 0$ to time $t = 1$.

▶️Answer/Explanation

\(\text{2(a)}\) To find $x(2)$, integrate $x'(t)$:
\(
x(2) = x(1) + \int_1^2 x'(t) \, dt = 3 + \int_1^2 \cos(t^2) \, dt.
\)
Using a calculator,
\(
x(2) \approx 2.556937225 \approx 2.557.
\)
\(\text{2(b)}\) The slope of the tangent is given by:
\(
\frac{dy}{dx} = \frac{y'(t)}{x'(t)} = \frac{e^{0.5t}}{\cos(t^2)}.
\)
Set the slope equal to $2$:
\(
\frac{e^{0.5t}}{\cos(t^2)} = 2 \quad \Rightarrow \quad e^{0.5t} – 2\cos(t^2) = 0.
\)
Using a calculator, the zero occurs at $t \approx 0.84016447 \approx 0.840$.
\(\text{2(c)}\) The speed of the particle is:
Speed = $\sqrt{(x'(t))^2 + (y'(t))^2}$.
Set the speed equal to $3$:
\(
\sqrt{\cos^2(t^2) + e^{t}} = 3 \quad \Rightarrow \quad \cos^2(t^2) + e^t – 9 = 0.
\)
Solve using a calculator:
\(
t \approx 2.1958952 \approx 2.196.
\)
\(\text{2(d)}\) The total distance traveled is given by:
\(
\int_0^1 \sqrt{(x'(t))^2 + (y'(t))^2} \, dt.
\)
Using a calculator:
Total distance $\approx 1.59460885 \approx 1.595$.

Question 3

(a) Topic-2.3-Estimating Derivatives of a Function at a Point

(b) Topic-6.3-Riemann Sums Summation Notation and Definite Integral Notation

(c) Topic-4.2-Straight-Line Motion Connecting Position Velocity and Acceleration

(d) Topic-8.1-Finding the Average Value of a Function on an Interval

t
(minutes)
012202440
v (t)
(meters per minute)
0200240-220150

Johanna jogs along a straight path. For 0 ≤ t ≤ 40, Johanna’s velocity is given by a differentiable function v.
Selected values of v (t), where t is measured in minutes and v (t) is measured in meters per minute, are given in the table above.
(a) Use the data in the table to estimate the value of v'(16).
(b) Using correct units, explain the meaning of the definite integral \(\int_{0}^{40}|v(t)|dt\) in the context of the problem.
Approximate the value of \(\int_{0}^{40}|v(t)|dt\) using a right Riemann sum with the four subintervals indicated in the table.
(c) Bob is riding his bicycle along the same path. For 0 ≤ t ≤ 10, Bob’s velocity is modeled by B(t) = t3 – 6t + 300, where t is measured in minutes and B (t) is measured in meters per minute.
Find Bob’s acceleration at time t = 5.
(d) Based on the model B from part (c), find Bob’s average velocity during the interval0 ≤ t ≤ 10.

▶️Answer/Explanation

\(\textbf{3(a) } v'(16) \approx \frac{240 – 200}{20 – 12} = 5 \, \text{meters/min}^2\)

\(\textbf{3(b) } \int_{0}^{40} |v(t)| \, dt \text{ is the total distance Johanna jogs, in meters, over the time interval } 0 \leq t \leq 40 \, \text{minutes.}\)

\(
\int_{0}^{40} |v(t)| \, dt \approx 12 \cdot |v(12)| + 8 \cdot |v(20)| + 4 \cdot |v(24)| + 16 \cdot |v(40)|
\)

\(
= 12 \cdot 200 + 8 \cdot 240 + 4 \cdot 220 + 16 \cdot 150
\)

\(
= 2400 + 1920 + 880 + 2400 = 7600 \, \text{meters.}
\)

\(\textbf{3(c)} \text{Bob’s acceleration is } B'(t) = 3t^2 – 12t.\)

\(
B'(5) = 3(25) – 12(5) = 15 \, \text{meters/min}^2.
\)

\(\textbf{3(d)} \text{Avg vel } = \frac{1}{10} \int_{0}^{10} \left(t^3 – 6t^2 + 300t\right) \, dt\)

\(
= \frac{1}{10} \left[\frac{t^4}{4} – 2t^3 + 300t\right]_{0}^{10}
\)

\(
= \frac{1}{10} \left[\frac{10000}{4} – 2000 + 3000\right]
\)

\(
= 350 \, \text{meters/min.}
\)

Question 4

(a) Topic-7.3-Sketching Slope Fields

(b) Topic-7.1-Modeling Situations with Differential Equations

(c) Topic-5.4-Using the First Derivative Test to Determine Relative Local Extrema

(d) Topic-2.2-Defining the Derivative of a Function and Using Derivative Notation

Consider the differential equation \(\frac{dy}{dx}=2x – y.\)

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

(b) Find \(\frac{d^{2}y}{dx^{2}}\) in terms of x and y. Determine the concavity of all solution curves for the given differential equation in Quadrant II. Give a reason for your answer.
(c) Let y = f(x) be the particular solution to the differential equation with the initial condition f (2) = 3.
Does f have a relative minimum, a relative maximum, or neither at x = 2 ? Justify your answer.
(d) Find the values of the constants m and b for which y = mx + b is a solution to the differential equation. 

▶️Answer/Explanation

\(\textbf{4(a)}\)

\(\textbf{4(b) } \frac{d^2y}{dx^2} = 2 – \frac{dy}{dx} = 2 – (2x – y) = 2 – 2x + y\)

\(\text{In Quadrant II, } x < 0 \text{ and } y > 0, \text{ so } 2 – 2x + y > 0.\)

\(\text{Therefore, all solution curves are concave up in Quadrant II.}\)

\(\textbf{4(c) } \frac{dy}{dx}\bigg|_{(x, y) = (2, 3)} = 2(2) – 3 = 1 \neq 0\)

\(\text{Therefore, } f \text{ has neither a relative minimum nor a relative maximum at } x = 2.\)

\(\textbf{4(d) } y = mx + b \implies \frac{dy}{dx} = \frac{d}{dx} (mx + b) = m\)

\(
2x – y = m
\)

\(
2x – (mx + b) = m
\)

\(
(2 – m)x – (m + b) = 0
\)

\(
2 – m = 0 \implies m = 2
\)

\(
b = -m \implies b = -2
\)

\(\text{Therefore, } m = 2 \text{ and } b = -2.\)

Question 5

(a) Topic-2.3-Estimating Derivatives of a Function at a Point

(b) Topic-5.4-Using the First Derivative Test to Determine Relative Local Extrema

(c) Topic-5.2-Extreme Value Theorem Global Versus Local Extrema and Critical Points

(d) Topic-6.12-Using Linear Partial Fractions

Consider the function
\(
f(x) = \frac{1}{x^2 – kx},
\)
where $k$ is a nonzero constant. The derivative of $f$ is given by
\(
f'(x) = \frac{k – 2x}{(x^2 – kx)^2}.
\)
(a) Let $k = 3$, so that $f(x) = \frac{1}{x^2 – 3x}$. Write an equation for the line tangent to the graph of $f$ at the point whose $x$-coordinate is $4$.
(b) Let $k = 4$, so that $f(x) = \frac{1}{x^2 – 4x}$. Determine whether $f$ has a relative minimum, a relative maximum, or neither at $x = 2$. Justify your answer.
(c) Find the value of $k$ for which $f$ has a critical point at $x = -5$.
(d) Let $k = 6$, so that $f(x) = \frac{1}{x^2 – 6x}$. Find the partial fraction decomposition for the function $f$. Find $\int f(x) \, dx$.

▶️Answer/Explanation

The function is \( f(x) = \frac{1}{x^2 – kx} \), where \( k \neq 0 \). The derivative is given by:
\(
f'(x) = \frac{k – 2x}{(x^2 – kx)^2}.
\)
\(\textbf{5(a)}\) Tangent Line Equation for \( k = 3 \)
For \( k = 3 \), \( f(x) = \frac{1}{x^2 – 3x} \). At \( x = 4 \):
\(
f(4) = \frac{1}{16 – 12} = \frac{1}{4},
\)
\(
f'(4) = \frac{3 – 2(4)}{(4^2 – 3(4))^2} = \frac{-5}{16}.
\)
The equation of the tangent line is:
\(
y – f(4) = f'(4)(x – 4) \quad \Rightarrow \quad y – \frac{1}{4} = -\frac{5}{16}(x – 4).
\)
\(\textbf{5(b)}\) Relative Extrema for \( k = 4 \)
For \( k = 4 \), \( f(x) = \frac{1}{x^2 – 4x} \). The derivative is:
\(
f'(x) = \frac{4 – 2x}{(x^2 – 4x)^2}.
\)
At \( x = 2 \), \( f'(2) = 0 \).
The second derivative test shows \( f^{”}(x) < 0 \) at \( x = 2 \), so \( f(x) \) has a relative maximum at \( x = 2 \).
\(\textbf{5(c)}\) Critical Point for \( k \) such that \( x = -5 \)
To find \( k \) for which \( x = -5 \) is a critical point, set \( f'(x) = 0 \):
\(
f'(x) = \frac{k – 2(-5)}{((-5)^2 – k(-5))^2} = 0.
\)
This gives:
\(
k + 10 = 0 \quad \Rightarrow \quad k = -10.
\)
\(\textbf{5(d)}\) Partial Fraction Decomposition and Integral for \( k = 6 \)
For \( k = 6 \), \( f(x) = \frac{1}{x^2 – 6x} \). Factor the denominator:
\(
f(x) = \frac{1}{x(x – 6)}.
\)
Perform partial fraction decomposition:
\(
\frac{1}{x(x – 6)} = \frac{A}{x} + \frac{B}{x – 6}.
\)
Solving for \( A \) and \( B \):
\(
1 = A(x – 6) + Bx \quad \Rightarrow \quad A = \frac{1}{6}, \, B = -\frac{1}{6}.
\)
The integral is:
\(
\int f(x) \, dx = \int \frac{1}{x(x – 6)} \, dx = \frac{1}{6} \ln|x| – \frac{1}{6} \ln|x – 6| + C.
\)

Question 6

(a) Topic-10.8-Ratio Test for Convergence

(b) Topic-10.14-Finding Taylor or Maclaurin Series for a Function

(c) Topic-10.14-Finding Taylor or Maclaurin Series for a Function

The Maclaurin series for a function \( f \) is given by:
\(
\sum_{n=1}^\infty \frac{(-3)^{n-1}}{n} x^n = x – \frac{3}{2}x^2 + 3x^3 – \dots + \frac{(-3)^{n-1}}{n} x^n + \dots
\)
and converges to \( f(x) \) for \( |x| < R \), where \( R \) is the radius of convergence of the Maclaurin series.
(a) Use the ratio test to find \( R \).
(b) Write the first four nonzero terms of the Maclaurin series for \( f’ \), the derivative of \( f \). Express \( f’ \) as a rational function for \( |x| < R \).
(c) Write the first four nonzero terms of the Maclaurin series for \( e^x \). Use the Maclaurin series for \( e^x \) to write the third-degree Taylor polynomial for \( g(x) = e^x f(x) \) about \( x = 0 \).

▶️Answer/Explanation

\(
\sum_{n=1}^\infty \frac{(-3)^{n-1} x^n}{n} = x – \frac{3}{2}x^2 + 3x^3 – \cdots + \frac{(-3)^{n-1}x^n}{n} + \cdots
\)
converges to \( f(x) \) for \( |x| < R \), the radius of convergence.
\(\textbf{6(a)}\)
\(
\left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{\frac{3^{n+1} x^{n+1}}{n+1}}{\frac{3^n x^n}{n}} =
\lim_{n \to \infty} \frac{3^{n+1} x^{n+1} n}{3^n x^n (n+1)} =
\lim_{n \to \infty} \frac{3^n \cdot 3 x^n \cdot x \cdot n}{3^n x^n (n+1)}
\)
\(
= \lim_{n \to \infty} \frac{3x \cdot n}{n+1} = \lim_{n \to \infty} \frac{n}{n+1} \cdot 3|x| = 3|x| < 1 \implies |x| < \frac{1}{3} \implies R = \frac{1}{3}.
\)
\(\textbf{6(b)}\)
\(
f'(x) = 1 – 3x + 9x^2 – 27x^3 + \cdots + (-3x)^n + \cdots \quad \text{or} \quad \sum_{n=0}^\infty (-3x)^n,
\)
which is a geometric series where \( a_1 = 1 \) and \( r = -3x \).
So, \( f'(x) = \frac{1}{1+3x} \).
\(\textbf{6(c)}\)
\(
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots
\)
\(
g(x) = e^x f'(x) = \left( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \right) \left( -\frac{3}{2}x^2 + 3x^3 – \cdots \right)
\)
\(
= \left( -\frac{3}{2}x^2 + 3x^3 – \cdots \right) + x \left( -\frac{3}{2}x^2 + 3x^3 – \cdots \right) + \frac{x^2}{2!} \left( -\frac{3}{2}x^2 + 3x^3 – \cdots \right) + \cdots
\)
\(
\approx -\frac{3}{2}x^2 + \cdots – \frac{3}{2}x^2 \cdot \frac{1}{2!} + \cdots + x – 1 + 2x^3 + \cdots = x – \frac{1}{2}x^2 + 2x^3.
\)

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