Question 1
(a) Topic-6.7-The Fundamental Theorem of Calculus and Definite Integrals
(b) Topic-6.7-The Fundamental Theorem of Calculus and Definite Integrals
(c) Topic-6.7-The Fundamental Theorem of Calculus and Definite Integrals
(d) Topic-6.7-The Fundamental Theorem of Calculus and Definite Integrals
People enter a line for an escalator at a rate modeled by the function r given by
\(r(t)=\left\{\begin{matrix}
44\left ( \frac{t}{100} \right )^{3} \left ( 1-\frac{t}{300} \right )^{7}& for 0\leq t\leq 300 & \\
0&for t>300, &
\end{matrix}\right.\)
where r (t) is measured in people per second and t is measured in seconds. As people get on the escalator, they exit the line at a constant rate of 0.7 person per second. There are 20 people in line at time t = 0.
(a) How many people enter the line for the escalator during the time interval 0 ≤ t ≤ 300 ?
(b) During the time interval 0 ≤ t ≤ 300, there are always people in line for the escalator. How many people are in line at time t = 300 ?
(c) For t > 300, what is the first time t that there are no people in line for the escalator?
(d) For 0 ≤ t ≤ 300, at what time t is the number of people in line a minimum? To the nearest whole number, find the number of people in line at this time. Justify your answer.
▶️Answer/Explanation
\(\textbf{1(a)} \quad \int_{0}^{300} r(t) \, dt = 270\)
\(\text{According to the model, 270 people enter the line for the escalator during the time interval } 0 \leq t \leq 300.\)
\(\textbf{1(b)} \quad 20 + \int_{0}^{300} \big(r(t) – 0.7\big) \, dt = 20 + \int_{0}^{300} r(t) \, dt – 0.7 \cdot 300 = 20 + 270 – 210 = 80\)
\(\text{According to the model, 80 people are in line at time } t = 300.\)
\(\textbf{1(c)} \quad \text{Based on part (b), the number of people in line at time } t = 300 \text{ is 80.}\)
\(\quad \text{The first time } t \text{ that there are no people in line is}\)
\(\quad 300 + \frac{80}{0.7} = 414.286 \text{ (or 414.285) seconds.}\)
\(\textbf{1(d)} \quad \text{The total number of people in line at time } t, \, 0 \leq t \leq 300, \text{ is modeled by}\)
\(\quad 20 + \int_{0}^{t} r(x) \, dx – 0.7t.\)
\(\quad r(t) – 0.7 = 0 \implies t_1 = 33.013298, \, t_2 = 166.574719\)
\(\begin{array}{|c|c|}
\hline
t & \text{People in line for escalator} \\
\hline
0 & 20 \\
t_1 & 3.803 \\
t_2 & 158.070 \\
300 & 80 \\
\hline
\end{array}\)
\(\text{The number of people in line is a minimum at time } t = 33.013 \text{ seconds, when there are 4 people in line.}\)
Question 2
(a) Topic-2.3-Estimating Derivatives of a Function at a Point
(b) Topic-6.2-Approximating Areas with Riemann Sums
(c) Topic-6.6-Applying Properties of Definite Integrals
(d) Topic-8.13-The Arc Length of a Smooth Planar Curve and Distance Traveled
Researchers on a boat are investigating plankton cells in a sea. At a depth of \( h \) meters, the density of plankton cells, in millions of cells per cubic meter, is modeled by
\(
p(h) = 0.2h e^{-0.0025h^2} \quad \text{for } 0 \leq h \leq 30
\)
and is modeled by \( f(h) \) for \( h \geq 30 \). The continuous function \( f \) is not explicitly given.
(a) Find \( p'(25) \). Using correct units, interpret the meaning of \( p'(25) \) in the context of the problem.
(b) Consider a vertical column of water in this sea with horizontal cross sections of constant area \( 3 \) square meters. To the nearest million, how many plankton cells are in this column of water between \( h = 0 \) and \( h = 30 \) meters?
(c) There is a function \( u \) such that \( 0 \leq f(h) \leq u(h) \) for all \( h \geq 30 \) and
\(
\int_{30}^\infty u(h) \, dh = 105.
\)
The column of water in part (b) is \( K \) meters deep, where \( K > 30 \). Write an expression involving one or more integrals that gives the number of plankton cells, in millions, in the entire column. Explain why the number of plankton cells in the column is less than or equal to \( 2000 \) million.
(d) The boat is moving on the surface of the sea. At time \( t \geq 0 \), the position of the boat is \( (x(t), y(t)) \), where
\(
x'(t) = 662 \sin(5t) \quad \text{and} \quad y'(t) = 880 \cos(6t).
\)
Time \( t \) is measured in hours, and \( x(t) \) and \( y(t) \) are measured in meters. Find the total distance traveled by the boat over the time interval \( 0 \leq t \leq 1 \).
▶️Answer/Explanation
\(\textbf{2(a)}\)
Using the calculator and the given expression for \( p(h) \):
\(
p'(25) \approx -1.179060448 \quad \text{or} \quad -1.179.
\)
This means that at a depth of 25 meters, the density of the plankton is decreasing at a rate of \( 1.179060448 \, \frac{\text{millions of cells}}{\text{cubic meter}} \, \text{per meter.} \)
\(\textbf{2(b)}\)
Think of finding the number of cells as a Riemann sum and use units:
\(
\text{millions of cells} = \frac{\text{millions of cells}}{\text{m}^3} \cdot \text{m}^2 \cdot \text{m} = \text{density} \cdot \text{area} \cdot \Delta h.
\)
So, the number of millions of cells between \( h = 0 \) and \( h = 30 \) is:
\(
\int_0^{30} p(h) \cdot 3 \, dh \approx 1675.414936 \, \text{million cells.}
\)
To the nearest million:
\(
\boxed{1675 \, \text{million cells.}}
\)
\(\textbf{2(c)}\)
Millions of plankton cells in the entire column:
It is given that \( 0 \leq f(h) \leq u(h) \) for \( h \geq 30 \) and
\(
\int_{30}^\infty u(h) \, dh = 105.
\)
Millions of cells \(= \int_0^{30} p(h) \cdot 3 \, dh + \int_{30}^K f(h) \cdot 3 \, dh\).
\(
\leq a + \int_{30}^K u(h) \cdot 3 \, dh.
\)
\(
\leq a + \int_{30}^\infty u(h) \cdot 3 \, dh.
\)
\(
\leq a + 3(105).
\)
\(
= 1990.415 \quad \leq \quad 2000.
\)
\(\textbf{2(d)}\)
Total distance traveled over \( [0, 1] \):
Total distance = \(\int_0^1 \sqrt{(x'(t))^2 + (y'(t))^2} \, dt\).
\(
\approx 757.4558623 \, \text{meters} \quad \text{or} \quad 757.456 \, \text{m.}
\)
Question 3
(a) Topic-6.7-The Fundamental Theorem of Calculus and Definite Integrals
(b) Topic-6.6-Applying Properties of Definite Integrals
(c) Topic-5.6-Determining Concavity of Functions over Their Domains
(d) Topic-5.6-Determining Concavity of Functions over Their Domains
The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for − 5 ≤ x ≤ 3 , and g(x) = 2 ( x− 4)2 for 3 ≤ x ≤ 6.
(a) If f (1) = 3, what is the value of f (−5) ?
(b) Evaluate \(\int_{1}^{6}g(x)dx\) .
(c) For − 5 ≤ x ≤ 6, on what open intervals, if any, is the graph of f both increasing and concave up? Give a reason for your answer.
(d) Find the x-coordinate of each point of inflection of the graph of f. Give a reason for your answer.
▶️Answer/Explanation
\(\textbf{3(a)} \quad f(-5) = f(1) + \int_{1}^{-5} g(x) \, dx = f(1) – \int_{-5}^{1} g(x) \, dx\)
\(\quad = 3 – \left(-9 – \frac{3}{2} + 1\right) = 3 – \left(-\frac{19}{2}\right) = \frac{25}{2}\)
\(\textbf{3(b)} \quad \int_{1}^{6} g(x) \, dx = \int_{1}^{3} g(x) \, dx + \int_{3}^{6} g(x) \, dx\)
\(\quad = \int_{1}^{3} 2 \, dx + \int_{3}^{6} 2(x – 4)^2 \, dx\)
\(\quad = 4 + \left[\frac{2}{3}(x – 4)^3\right]_{x=3}^{x=6} = 4 + \frac{16}{3} – \left(-\frac{2}{3}\right) = 10\)
\(\textbf{3(c)} \quad \text{The graph of } f \text{ is increasing and concave up on } 0 < x < 1 \text{ and } 4 < x < 6 \text{ because } f'(x) = g(x) > 0 \text{ and } f^{”}(x) = g(x)\)
\(\text{ is increasing on those intervals.}\)
\(\textbf{3(d)} \quad \text{The graph of } f \text{ has a point of inflection at } x = 4 \text{ because } f'(x) = g(x) \text{ changes from decreasing to increasing at } x = 4.\)
Question 4
(a) Topic-2.3-Estimating Derivatives of a Function at a Point
(b) Topic-5.1-Using the Mean Value Theorem
(c) Topic-8.1-Finding the Average Value of a Function on an Interval
(d) Topic-2.9-The Quotient Rule
t (years) | 2 | 3 | 5 | 7 | 10 |
H(t) (meters) | 1.5 | 2 | 6 | 11 | 15 |
The height of a tree at time t is given by a twice-differentiable function H, where H (t) is measured in meters and t is measured in years. Selected values of H (t) are given in the table above.
(a) Use the data in the table to estimate H'(6). Using correct units, interpret the meaning of H'(6) in the context of the problem.
(b) Explain why there must be at least one time t, for 2 < t <10 , such that H'(t) = 2.
(c) Use a trapezoidal sum with the four subintervals indicated by the data in the table to approximate the average height of the tree over the time interval 2 ≤ t ≤ 10 .
(d) The height of the tree, in meters, can also be modeled by the function G, given by \(G(x)=\frac{100x}{1+x}\) , where x is the diameter of the base of the tree, in meters. When the tree is 50 meters tall, the diameter of the base of the tree is increasing at a rate of 0.03 meter per year. According to this model, what is the rate of change of the height of the tree with respect to time, in meters per year, at the time when the tree is 50 meters tall?
▶️Answer/Explanation
\(\textbf{4(a)} \quad H'(6) \approx \frac{H(7) – H(5)}{7 – 5} = \frac{11 – 6}{2} = \frac{5}{2}\)
\(\quad H'(6) \text{ is the rate at which the height of the tree is changing, in meters per year, at time } t = 6 \text{ years.}\)
\(\textbf{4(b)} \quad \frac{H(5) – H(3)}{5 – 3} = \frac{6 – 2}{2} = 2\)
\(\quad \text{Because } H \text{ is differentiable on } 3 \leq t \leq 5, \, H \text{ is continuous on } 3 \leq t \leq 5.\)
\(\quad \text{By the Mean Value Theorem, there exists a value } c, \, 3 < c < 5, \text{ such that } H'(c) = 2.\)
\(\textbf{4(c)} \quad \text{The average height of the tree over the time interval } 2 \leq t \leq 10 \text{ is given by}\)
\(\quad \frac{1}{10 – 2} \int_{2}^{10} H(t) \, dt.\)
\(\quad \frac{1}{8} \int_{2}^{10} H(t) \, dt \approx \frac{1}{8} \bigg(\frac{1.5 + 2}{2} \cdot 1 + \frac{2 + 6}{2} \cdot 2 + \frac{6 + 11}{2} \cdot 2 + \frac{11 + 15}{2} \cdot 3\bigg)\)
\(\quad = \frac{1}{8} \cdot (65.75) = \frac{263}{32}\)
\(\quad \text{The average height of the tree over the time interval } 2 \leq t \leq 10 \text{ is } \frac{263}{32} \text{ meters.}\)
\(\textbf{4(d)} \quad G(x) = 50 \implies x = 1\)
\(\quad \frac{d}{dt}\big(G(x)\big) = \frac{d}{dx}\big(G(x)\big) \cdot \frac{dx}{dt} = \frac{(1 + x)100 – 100x \cdot 1}{(1 + x)^2} \cdot \frac{dx}{dt}\)
\(\quad \frac{d}{dt}\big(G(x)\big)\bigg|_{x=1} = \frac{100}{(1 + 1)^2} \cdot 0.03 = \frac{3}{4}\)
\(\quad \text{According to the model, the rate of change of the height of the tree with respect to time when the tree is 50 meters tall is } \frac{3}{4} \text{ meter per year.}\)
Question 5
(a) Topic-9.9-Finding the Area of the Region Bounded by Two Polar Curves
(b) Topic-9.4-Defining and Differentiating Vector-Valued Functions
(c) Topic-9.4-Defining and Differentiating Vector-Valued Functions
The graphs of the polar curves \( r = 4 \) and \( r = 3 + 2\cos\theta \) are shown in the figure above. The curves intersect at \( \theta = \frac{\pi}{3} \) and \( \theta = \frac{5\pi}{3} \).
(a) Let \( R \) be the shaded region that is inside the graph of \( r = 4 \) and also outside the graph of \( r = 3 + 2\cos\theta \), as shown in the figure above. Write an expression involving an integral for the area of \( R \).
(b) Find the slope of the line tangent to the graph of \( r = 3 + 2\cos\theta \) at \( \theta = \frac{\pi}{2} \).
(c) A particle moves along the portion of the curve \( r = 3 + 2\cos\theta \) for \( 0 < \theta < \frac{\pi}{2} \). The particle moves in such a way that the distance between the particle and the origin increases at a constant rate of 3 units per second. Find the rate at which the angle \( \theta \) changes with respect to time at the instant when the position of the particle corresponds to \( \theta = \frac{\pi}{3} \). Indicate units of measure.
▶️Answer/Explanation
\(\textbf{5(a)}\)
\(
\text{Area inside of } r = 4 \text{ and outside of } r = 3 + 2\cos\theta =
\frac{1}{2} \int_{\pi/3}^{5\pi/3} \left[ (4)^2 – (3 + 2\cos\theta)^2 \right] \, d\theta.
\)
\(\textbf{5(b)}\)
slope \(m = \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \quad \text{and} \quad r = 3 + 2\cos\theta.
\)
\(
y = r\sin\theta = (3 + 2\cos\theta)\sin\theta = 3\sin\theta + 2\cos\theta\sin\theta = 3\sin\theta + \sin 2\theta.
\)
\(
\frac{dy}{d\theta} = 3\cos\theta + 2\cos 2\theta \implies \frac{dy}{d\theta}\bigg|_{\theta = \pi/2} = 3\cos\frac{\pi}{2} + 2\cos\pi = -2.
\)
\(
x = r\cos\theta = (3 + 2\cos\theta)\cos\theta = 3\cos\theta + 2\cos^2\theta.
\)
\(
\frac{dx}{d\theta} = -3\sin\theta – 4\cos\theta\sin\theta \implies \frac{dx}{d\theta}\bigg|_{\theta = \pi/2} = -3\sin\frac{\pi}{2} – 4\cos\frac{\pi}{2}\sin\frac{\pi}{2} = -3.
\)
\(
\text{slope } m_{\theta = \pi/2} = \frac{dy/d\theta}{dx/d\theta}\bigg|_{\theta = \pi/2} = \frac{-2}{-3} = \frac{2}{3}.
\)
\(\textbf{5(c)}\)
\(
r = 3 + 2\cos\theta \quad \text{and} \quad \frac{dr}{dt} = 3. \text{ Find } \frac{d\theta}{dt} \text{ when } \theta = \frac{\pi}{3}.
\)
\(
\frac{dr}{d\theta} = -2\sin\theta \implies 3 = -2\sin\frac{\pi}{3} \left( \frac{d\theta}{dt} \right).
\)
\(
\implies 3 = -2 \left( \frac{\sqrt{3}}{2} \right) \frac{d\theta}{dt} \implies \frac{d\theta}{dt} = \frac{-3}{\sqrt{3}} \quad \text{or} \quad -\sqrt{3} \, \text{radians per second}.
\)
Question 6
(a) Topic-10.14-Finding Taylor or Maclaurin Series for a Function
(b) Topic-10.5-Harmonic Series and p-Series
(c) Topic-10.10-Alternating Series Error Bound
On its interval of convergence, this series converges to \( \ln(1 + x) \). Let \( f \) be the function defined by
\(
f(x) = x \ln\left(1 + \frac{x}{3}\right).
\)
(a) Write the first four nonzero terms and the general term of the Maclaurin series for \( f \).
(b) Determine the interval of convergence of the Maclaurin series for \( f \). Show the work that leads to your answer.
(c) Let \( P_4(x) \) be the fourth-degree Taylor polynomial for \( f \) about \( x = 0 \). Use the alternating series error bound to find an upper bound for \( |P_4(2) – f(2)| \).
▶️Answer/Explanation
\(\textbf{6(a)}\)
Using the Maclaurin series for \( \ln(1 + x) \) that was given:
\(
f(x) = x \ln\left(1 + \frac{x}{3}\right) = x \left[ \frac{x}{3} – \frac{\left(\frac{x}{3}\right)^2}{2} + \frac{\left(\frac{x}{3}\right)^3}{3} – \frac{\left(\frac{x}{3}\right)^4}{4} + \cdots + (-1)^{n+1} \frac{\left(\frac{x}{3}\right)^n}{n} \right].
\)
\(
= \frac{x^2}{3} – \frac{x^3}{3 \cdot 2^2} + \frac{x^4}{3 \cdot 3^3} – \frac{x^5}{4 \cdot 3^4} + \cdots + (-1)^{n+1} \frac{x^{n+1}}{n \cdot 3^n}.
\)
\(\textbf{6(b)}\)
Using the ratio test to find the interval of convergence for \( f \):
\(
\frac{(-1)^{n+2} \frac{x^{n+2}}{(n+1) \cdot 3^{n+1}}}{(-1)^{n+1} \frac{x^{n+1}}{n \cdot 3^n}} = \lim_{n \to \infty} \frac{(-1)^{n+2} \cdot x^{n+2} \cdot n \cdot 3^n}{(-1)^{n+1} \cdot x^{n+1} \cdot (n+1) \cdot 3^{n+1}}.
\)
\(
= \lim_{n \to \infty} \frac{n}{n+1} \cdot \frac{|x|}{3} = \frac{|x|}{3} < 1 \quad \text{for convergence.}
\)
So, \(-1 < \frac{x}{3} < 1 \implies -3 < x < 3\).
Now, testing the endpoints of the interval:
\(
x = 3: \quad \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n \cdot 3^n} = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}, \quad \text{which is an alternating harmonic series, so it converges.}
\)
\(
x = -3: \quad \sum_{n=1}^\infty \frac{(-1)^{n+1}(-3)^n}{n \cdot 3^n} = \sum_{n=1}^\infty \frac{(-1)^n}{n}, \quad \text{which is a harmonic series, so it diverges.}
\)
\(
\therefore \text{The interval of convergence for } f \text{ is } -3 < x \leq 3.
\)
\(\textbf{6(c)}\)
Since \( 2 \) is within the interval of convergence for \( f \), the alternating series error bound using the 4th-degree Taylor polynomial would be the absolute value of the next term, the 5th-degree term, evaluated at \( x = 2 \).
So \(\left| P_4(2) – f(2) \right| < \frac{2^5}{4 \cdot 3^4}\).