Question 1
(a) Topic-6.7 The Fundamental Theorem of Calculus and Definite Integrals
(b) Topic-8.1 Finding the Average Value of a Function on an Interval
(c) Topic-5.3 Determining Intervals on Which a Function is Increasing or Decreasing
(d) Topic-8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts
From \( 5 \, \text{A.M.} \) to \( 10 \, \text{A.M.} \), the rate at which vehicles arrive at a certain toll plaza is given by
\(
A(t) = 450 \sqrt{\sin(0.6t)},
\)
where \( t \) is the number of hours after \( 5 \, \text{A.M.} \) and \( A(t) \) is measured in vehicles per hour. Traffic is flowing smoothly at \( 5 \, \text{A.M.} \) with no vehicles waiting in line.
(a) Write, but do not evaluate, an integral expression that gives the total number of vehicles that arrive at the toll plaza from \( 6 \, \text{A.M.} \) (\( t = 1 \)) to \( 10 \, \text{A.M.} \) (\( t = 5 \)).
(b) Find the average value of the rate, in vehicles per hour, at which vehicles arrive at the toll plaza from \( 6 \, \text{A.M.} \) (\( t = 1 \)) to \( 10 \, \text{A.M.} \) (\( t = 5 \)).
(c) Is the rate at which vehicles arrive at the toll plaza at \( 6 \, \text{A.M.} \) (\( t = 1 \)) increasing or decreasing? Give a reason for your answer.
(d) A line forms whenever \( A(t) \geq 400 \). The number of vehicles in line at time \( t \), for \( a \leq t \leq 4 \), is given by
\(
N(t) = \int_a^t (A(x) – 400) \, dx,
\)
where \( a \) is the time when a line first begins to form. To the nearest whole number, find the greatest number of vehicles in line at the toll plaza in the time interval \( a \leq t \leq 4 \). Justify your answer.
▶️Answer/Explanation
1(a) \( A(t) = 450 \sqrt{\sin(0.6t)} \) is the rate at which vehicles arrive at the toll plaza
in vehicles per hour from 5 a.m. until 10 a.m. The number of vehicles that arrive
at the toll plaza from \( 6 \, \text{a.m.} \) (\( t = 1 \)) until \( 10 \, \text{a.m.} \) (\( t = 5 \)) is
\(
\int_1^5 A(t) \, dt.
\)
1(b) The average value of the rate at which the vehicles arrive at the toll plaza
from time \( t = 1 \) to time \( t = 5 \) is
\(
\frac{1}{5 – 1} \int_1^5 A(t) \, dt = \frac{1}{4} \int_1^5 A(t) \, dt = 375.5369662 \quad \text{or} \quad 375.536 \, \text{or} \, 375.537 \, \text{vehicles per hour.}
\)
1(c) \( A'(1) = 148.9472908 > 0 \)
The rate at which vehicles arrive at the toll plaza at 6 a.m. is
\textbf{increasing} because \( A'(t) > 0 \) when \( t = 1 \).
1(d) When \( A(t) \geq 400 \), \( N(t) = \int_a^t (A(x) – 400) \, dx \) for \( a \leq t \leq 4 \).
We want the absolute maximum value of \( N(t) \) for \( a \leq t \leq 4 \).
This will occur when \( t = a \), or when \( t = 4 \), or when \( N'(t) = 0 \).
\( N'(t) = A(t) – 400 = 0 \implies A(t) = 400 \implies t = 1.4693716 = t_1 \quad \text{and} \quad t = 3.5977133 = t_2 \)
(we knew \( t_1 \) was going to be \( a \) since we know the line started forming at \( t = a \))
Comparing the values of \( N \) at the candidates:
\(
N(a) = \int_a^a (A(t) – 400) \, dt = 0
\)
\(
N(t_1) = \int_a^{t_1} (A(t) – 400) \, dt = 0
\)
\(
N(t_2) = \int_a^{t_2} (A(t) – 400) \, dt = 71.254
\)
\(
N(4) = \int_a^4 (A(t) – 400) \, dt = 62.338
\)
So the greatest number of vehicles in line at the toll plaza in the time
interval \( a \leq t \leq 4 \) is \(\textbf{71}\), and this occurs at \( t = 3.5977133 \).
Question 2
(a) Topic-4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration
(b) Topic-4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration
(c) Topic-6.7 The Fundamental Theorem of Calculus and Definite Integrals
(d) Topic-8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled
A particle moving along a curve in the $xy$-plane is at position $(x(t), y(t))$ at time $t > 0$.
The particle moves in such a way that
\(
\frac{dx}{dt} = \sqrt{1 + t^2} \quad \text{and} \quad \frac{dy}{dt} = \ln(2 + t^2).
\)
At time $t = 4$, the particle is at the point $(1, 5)$.
a) Find the slope of the line tangent to the path of the particle at time $t = 4$.
b)Find the speed of the particle at time $t = 4$, and find the acceleration vector of the particle at time $t = 4$.
c)Find the $y$-coordinate of the particle’s position at time $t = 6$.
d) Find the total distance the particle travels along the curve from time $t = 4$ to time $t = 6$.
▶️Answer/Explanation
\(\textbf{2(a)}\)
\(
m_{t=4} = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \bigg|_{t=4} = 0.701 \quad \text{(using your calculator)}
\)
or
\(
m = \frac{\ln(2 + t^2)}{\sqrt{1 + t^2}} \implies m_{t=4} = \frac{\ln(18)}{\sqrt{17}} \quad \text{(calculator not needed)}
\)
\(\textbf{2(b)}\)
\(
\text{speed}_{t=4} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \bigg|_{t=4} = 5.0353
\)
or
\(
\text{speed}_{t=4} = \sqrt{(\sqrt{17})^2 + (\ln(18))^2}
\)
\(
\text{acceleration vector}_{t=4} = \langle x^{”}(4), y^{”}(4) \rangle = \{0.970, 0.444\}
\)
\(\textbf{2(c)}\)
\(
\int_{4}^{6} \frac{dy}{dt} \, dt = y(6) – y(4) = y(6) – 5
\)
\(
y(6) = 5 + \int_{4}^{6} \frac{dy}{dt} \, dt = 11.5705 \quad \text{or} \quad 11.570 \quad \text{or} \quad 11.571
\)
\(\textbf{2(d)}\)
\(
\text{total distance} = \int_{4}^{6} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt = 12.136
\)
Question 3
(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-5.3 Determining Intervals on Which a Function is Increasing or Decreasing
(d) Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema
Let $f$ be a differentiable function with $f(4) = 3$. On the interval $0 \leq x \leq 7$, the graph of $f’$,
the derivative of $f$, consists of a semicircle and two line segments, as shown in the figure above.
a) Find $f(0)$ and $f(5)$.
b) Find the $x$-coordinates of all points of inflection of the graph of $f$ for $0 < x < 7$.Justify your answer.
c) Let $g$ be the function defined by $g(x) = f(x) – x$. On what intervals, if any, is $g$ decreasing for $0 \leq x \leq 7$?
Show the analysis that leads to your answer.
d) For the function $g$ defined in part (c), find the absolute minimum value on the interval $0 \leq x \leq 7$.
Justify your answer.
▶️Answer/Explanation
\(\textbf{3(a)}\)
\(
f(4) = 3, \quad \int_{0}^{4} f'(x) \, dx = f(4) – f(0) \implies f(0) = f(4) – \int_{0}^{4} f'(x) \, dx
\)
\(
f(0) = 3 – \left(-\frac{12\pi}{2}\right) \quad \text{or} \quad f(0) = 3 + 2\pi
\)
\(
\int_{4}^{5} f'(x) \, dx = f(5) – f(4) \implies f(5) = f(4) + \int_{4}^{5} f'(x) \, dx
\)
\(
f(5) = 3 + 12(1)(1) \quad \text{or} \quad f(5) = 3.5
\)
\(\textbf{3(b)}\)
\(
f \text{ is continuous because } f \text{ is differentiable.}
\)
\(
f^{”}(x) \text{ is undefined at } x = 4 \text{ and } x = 6. \quad f^{”}(x) = 0 \text{ at } x = 2.
\)
\(
f \text{ has inflection points at } x = 2 \text{ and } x = 6 \text{ because }
f’ \text{ changes from decreasing to increasing or vice versa.}
\)
\(\textbf{3(c)}\)
\(
g(x) = f(x) – x \implies g'(x) = f'(x) – 1
\)
\(
g'(x) < 0 \implies f'(x) – 1 < 0 \implies f'(x) < 1 \text{ when } 0 < x < 5
\)
\(
\text{So, } g \text{ is decreasing on } [0, 5] \text{ because } f'(x) \leq 1 \text{ there.}
\)
\(\textbf{3(d)}\)
\(\textbf{Method 1:}\)
\(
\text{The absolute minimum value of } g \text{ on } [0, 7] \text{ occurs at } x = 0, x = 5, \text{ or } x = 7.
\)
\(
g'(x) = 0 \text{ at } x = 5 \text{ and } g(0) = f(0) – 0 = 3 + 2\pi
\)
\(
g(5) = f(5) – 5 = 3.5 – 5 = -1.5
\)
\(
g(7) = f(7) – 7 = 3 + \int_{4}^{7} f'(x) \, dx – 7 = -0.5
\)
\(
\text{The minimum value of } g \text{ on the interval is } -1.5 \text{ at } x = 5.
\)
\(\textbf{Method 2:}\)
\(
\text{There is only one critical point on } (0, 7) \text{ at } x = 5.
\)
\(
g'(x) < 0 \text{ on } [0, 5) \text{ and } g'(x) \geq 0 \text{ on } [5, 7].
\)
\(
\text{The absolute minimum is at } x = 5 \text{ with } g(5) = f(5) – 5 = 3.5 – 5 = -1.5.
\)
Question 4
(a) Topic-3.6 Calculating Higher-Order Derivatives
(b) Topic-2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist
(c) Topic-6.3 Riemann Sums, Summation Notation, and Definite Integral Notation
(d) Topic-4.5 Solving Related Rates Problems
An ice sculpture melts in such a way that it can be modeled as a cone that maintains a conical shape as it decreases in size. The radius of the base of the cone is given by a twice-differentiable function $r$, where $r(t)$ is measured in centimeters and $t$ is measured in days. The table above gives selected values of $r'(t)$, the rate of change of the radius, over the time interval $0 \leq t \leq 12$.
\(\text{(a)}\) Approximate $r”(8.5)$ using the average rate of change of $r’$ over the interval $7 \leq t \leq 10$. Show the computations that lead to your answer, and indicate units of measure.
\(\text{(b)}\) Is there a time $t$, $0 \leq t \leq 3$, for which $r'(t) = -6$? Justify your answer.
\(\text{(c)}\) Use a right Riemann sum with the four subintervals indicated in the table to approximate the value of
\(
\int_{0}^{12} r'(t) \, dt.
\)
\(\text{(d)}\) The height of the cone decreases at a rate of $2$ centimeters per day. At time $t = 3$ days, the radius is $100$ centimeters and the height is $50$centimeters. Find the rate of change of the volume of the cone with respect to time, in cubic centimeters per day, at time $t = 3$ days. (The volume $V$ of a cone with radius $r$ and height $h$ is $V = \frac{1}{3}\pi r^2 h$.)
▶️Answer/Explanation
\(\textbf{4(a)}\)
\(
r^{”}(8.5) \approx \frac{r'(10) – r'(7)}{10 – 7} = \frac{-3.8 – (-4.4)}{10 – 7} = \frac{0.6}{3} \, \frac{\text{cm}}{\text{day}} = 0.2 \, \frac{\text{cm}}{\text{day}^2}
\)
\(\textbf{4(b)}\)
Since $r$ is twice differentiable, then $r’$ is differentiable and, hence, $r’$ is also continuous.
So, by the Intermediate Value Theorem,
$r’$ takes on all values between $r'(0) = -6.1$ and $r'(3) = -5.0$.
Hence, since $-6.1 = r'(0) < -6 < -5 = r'(3)$, there is a time on $[0, 3]$ for which $r'(t) = -6$.
\(\textbf{4(c)}\)
Right Riemann sum with 4 subintervals:
\(
\int_{0}^{12} r'(t) \, dt \approx (3 – 0)r'(3) + (7 – 3)r'(7) + (10 – 7)r'(10) + (12 – 10)r'(12)
\)
\(
= 3(-5) + 4(-4.4) + 3(-3.8) + 2(-3.5)
\)
\(
= -51 \, \text{cm}
\)
\(\textbf{4(d)}\)
\(
\frac{dh}{dt} = -2, \quad \text{when } t = 3, \, r = 100, \, h = 50. \quad \text{Find } \frac{dV}{dt} \text{ when } t = 3.
\)
\(
V = \frac{1}{3}\pi r^2 h
\)
\(
\frac{dV}{dt} = \frac{1}{3}\pi \left( r^2 \frac{dh}{dt} + h 2r \frac{dr}{dt} \right) \quad \text{(using the product rule).}
\)
\(
\text{Note: } \frac{dr}{dt} \big|_{t=3} = -5 \text{ (from the table).}
\)
\(
\frac{dV}{dt} = \frac{1}{3}\pi \left( (100)^2(-2) + (50)(2)(100)(-5) \right)
\)
\(
= \frac{1}{3}\pi \left( -20000 – 50000 \right) = \frac{-70000\pi}{3}
\)
Question 5
(a) Topic-8.4 Finding the Area Between Curves Expressed as Functions of x
(b) Topic-8.7 Volumes with Cross Sections: Squares and Rectangles
(c) Topic-8.9 Volume with Disc Method: Revolving Around the x- or y-Axis
Figures 1 and 2, shown above, illustrate regions in the first quadrant associated with the graphs of $y = \frac{1}{x}$ and $y = \frac{1}{x^2}$, respectively. In Figure 1, let $R$ be the region bounded by the graph of $y = \frac{1}{x}$, the $x$-axis, and the vertical lines $x = 1$ and $x = 5$. In Figure 2, let $W$ be the unbounded region between the graph of $y = \frac{1}{x^2}$ and the $x$-axis that lies to the right of the vertical line $x = 3$.
(a) Find the area of region $R$.
(b) Region $R$ is the base of a solid. For the solid, at each $x$, the cross section perpendicular to the $x$-axis is a rectangle with area given by $x e^{x/5}$. Find the volume of the solid.
(c) Find the volume of the solid generated when the unbounded region $W$ is revolved about the $x$-axis.
▶️Answer/Explanation
\(\textbf{5(a)}\)
\(
\text{Area}_R = \int_{1}^{5} \frac{1}{x} \, dx
\)
\(
= \ln(x) \Big|_{1}^{5}
\)
\(
= \ln(5) – \ln(1) = \ln(5)
\)
\(\textbf{5(b)}\)
\(
\text{Volume}_R = \int_{1}^{5} \text{(Area of cross section)} \, dx
\)
\(
= \int_{1}^{5} (x e^{x/5}) \, dx
\)
Integrate by parts:
\(
u = x, \quad v = 5e^{x/5}, \quad du = dx, \quad dv = e^{x/5}
\)
\(
= \Big[ 5x e^{x/5} – 5e^{x/5} \Big]_{1}^{5}
\)
\(
= \Big[ 5 \cdot 5e^{5/5} – 25e^{5/5} \Big] – \Big[ 5 \cdot 1e^{1/5} – 25e^{1/5} \Big]
\)
\(
= \Big(5 \cdot e^{5/5} – 25e^{5/5}\Big) – \Big(5 \cdot e^{1/5} – 25e^{1/5}\Big)
\)
\(
= 20e^{1/5}
\)
\(\textbf{5(c)}\)
When region $W$ is rotated about the $x$-axis, the circular disk will have radius $= \frac{1}{x^2}$.
\(
\text{Volume} = \pi \int_{3}^{\infty} \left( \frac{1}{x^2} \right)^2 \, dx
\)
\(
= \lim_{b \to \infty} \pi \int_{3}^{b} x^{-4} \, dx
\)
\(
= \lim_{b \to \infty} \pi \Bigg[ \frac{-1}{3x^3} \Bigg]_{3}^{b}
\)
\(
= \frac{\pi}{3} \lim_{b \to \infty} \Bigg[ \frac{-1}{b^3} – \frac{-1}{3^3} \Bigg]
\)
\(
= \frac{\pi}{3} \Bigg[ 0 – \frac{-1}{27} \Bigg]
\)
\(
= \frac{\pi}{3} \cdot \frac{1}{27} = \frac{\pi}{81}
\)
Question 6
(a) Topic-10.8 Ratio Test for Convergence
(b) Topic-10.10 Alternating Series Error Bound
(c) Topic-10.14 Finding Taylor or Maclaurin Series for a Function
(d) Topic-10.2 Working with Geometric Series
The function $f$ is defined by the power series
\(
f(x) = x – \frac{x^3}{3} + \frac{x^5}{5} – \frac{x^7}{7} + \cdots + \frac{(-1)^n x^{2n+1}}{2n+1} + \cdots
\)
for all real numbers $x$ for which the series converges.
(a) Using the ratio test, find the interval of convergence of the power series for $f$. Justify your answer.
(b) Show that
\(
\left| f\left(\frac{1}{2}\right) – \frac{1}{2} \right| < \frac{1}{10}.
\)
Justify your answer.
(c) Write the first four nonzero terms and the general term for an infinite series that represents $f'(x)$.
(d) Use the result from part (c) to find the value of $f’\left(\frac{1}{6}\right)$.
▶️Answer/Explanation
\(\textbf{6(a)}\)
Using the ratio test, we want to find all $x$ such that
\(
\lim_{n \to \infty} \left| \frac{x^{2(n+1)+1}}{2(n+1)+1} \cdot \frac{2n+1}{x^{2n+1}} \right|
= \lim_{n \to \infty} \left| \frac{x^2 (2n+1)}{2n+3} \right| < 1.
\)
\(
\lim_{n \to \infty} \frac{2n+1}{2n+3} = 1, \quad \text{so } |x|^2 < 1 \implies -1 < x < 1,
\)
and the radius of convergence is $1$.
Testing the endpoints:
When $x = -1$:
\(
\sum_{n=0}^\infty \frac{(-1)^n (-1)^{2n+1}}{2n+1} = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{2n+1}.
\)
When $x = 1$:
\(
\sum_{n=0}^\infty \frac{(-1)^n (1)^{2n+1}}{2n+1} = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}.
\)
Both are alternating series whose terms decrease in absolute value to $0$, so they both converge.
In other words, they are alternating series, $a_{n+1} < a_n$, and $\lim_{n \to \infty} \frac{1}{2n+1} = 0$.
So, the interval of convergence of $f$ is $-1 \leq x \leq 1$.
\(\textbf{6(b)}\)
\(
f\left(\frac{1}{2}\right) \approx \frac{1}{2}, \quad \text{so we can say that this represents } P_1\left(\frac{1}{2}\right),
\)
the first-degree Taylor polynomial for the alternating series $f(x)$ when $x = \frac{1}{2}$.
\(
\left| f(x) – \frac{1}{2} \right| = \left| f(x) – P_1\left(\frac{1}{2}\right) \right| \quad \text{is the error form for the alternating series.}
\)
Hence, the alternating series error bound is the first omitted term:
\(
\left| f(x) – \frac{1}{2} \right| = \left| f(x) – P_1\left(\frac{1}{2}\right) \right| \leq \left| \frac{\left(\frac{1}{2}\right)^3}{3} \right| = \frac{1}{24} < \frac{1}{10}.
\)
\(\textbf{(c)}\)
\(
f'(x) = 1 – \frac{3x^2}{3} + \frac{5x^4}{5} – \frac{7x^6}{7} + \cdots + \frac{(2n+1)(-1)^n x^{2n}}{2n+1} + \cdots
\)
or
\(
f'(x) = 1 – x^2 + x^4 – x^6 + \cdots + (-1)^n x^{2n} + \cdots
\)
\(\textbf{(d)}\)
\(
f’\left(\frac{1}{6}\right) \approx 1 – \left(\frac{1}{6}\right)^2 + \left(\frac{1}{6}\right)^4 – \left(\frac{1}{6}\right)^6 + \cdots
\)
This forms a geometric series where $a_1 = 1$ and $r = \left(\frac{1}{6}\right)^2$.
The sum of the series is given by:
\(
\text{Sum} = \frac{a_1}{1 – r} = \frac{1}{1 – \left(\frac{1}{6}\right)^2}.
\)
Simplify:
\(
\text{Sum} = \frac{1}{1 – \frac{1}{36}} = \frac{1}{\frac{36 – 1}{36}} = \frac{36}{35}.
\)
Thus:
\(
f’\left(\frac{1}{6}\right) = \frac{36}{35}.
\)