Calc-Ok Question 1
People enter a line for an escalator at a rate modeled by the function \(r\) given by \[ r(t)= \begin{cases} 44\!\left(\dfrac{t}{100}\right)^{3}\!\left(1-\dfrac{t}{300}\right)^{7}, & 0\le t\le 300,\\[8pt] 0, & t>300, \end{cases} \] 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 \(0\le t\le 300\)?
(b) During the time interval \(0\le t\le 300\), 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 \(0\le t\le 300\), 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.
Most-appropriate topic codes (CED):
• TOPIC 8.3: Using Accumulation Functions & Definite Integrals in Applied Contexts — net change via \(\int (\,\text{rate}\,)\,dt\). :contentReference[oaicite:0]{index=0}
• TOPIC 4.1: Interpreting the Meaning of the Derivative in Context — units/rate interpretation \(r(t)-0.7\). :contentReference[oaicite:1]{index=1}
• TOPIC 5.5: Using the Candidates Test to Determine Absolute (Global) Extrema — checking endpoints and where \(N'(t)=0\). :contentReference[oaicite:2]{index=2}
• TOPIC 4.1: Interpreting the Meaning of the Derivative in Context — units/rate interpretation \(r(t)-0.7\). :contentReference[oaicite:1]{index=1}
• TOPIC 5.5: Using the Candidates Test to Determine Absolute (Global) Extrema — checking endpoints and where \(N'(t)=0\). :contentReference[oaicite:2]{index=2}
▶️ Answer/Explanation
(a)
The number entering on \(0\le t\le 300\) is \[ \int_{0}^{300}\! r(t)\,dt =\int_{0}^{300}\! 44\!\left(\frac{t}{100}\right)^{3}\!\left(1-\frac{t}{300}\right)^{7} dt \approx \boxed{270\ \text{people}}. \] (Accumulation via definite integral: Topic 8.3.) :contentReference[oaicite:3]{index=3}
(b)
Let \(N(t)\) be the number in line. Then \[ N(300)=N(0)+\int_{0}^{300}\!\big(r(t)-0.7\big)\,dt. \] With \(N(0)=20\) and part (a): \(\int_{0}^{300}r(t)\,dt=270\). Also \(\displaystyle \int_{0}^{300}0.7\,dt=0.7(300)=210\). Hence \[ N(300)=20+(270-210)=\boxed{80\ \text{people}}. \]
(c)
For \(t>300\), \(r(t)=0\), so people leave at \(0.7\) person/s starting from \(N(300)=80\). Time to empty line \(=\dfrac{80}{0.7}=114.2857\text{ s}\). First time with no one in line: \[ \boxed{t=300+ \frac{80}{0.7}\approx 414.286\ \text{seconds}}. \]
(d) Minimum of \(N(t)\) on \(0\le t\le 300\)
For \(0\le t\le 300\), \[ N(t)=20+\int_{0}^{t}\!\big(r(x)-0.7\big)\,dx, \qquad N'(t)=r(t)-0.7. \] Critical times solve \(r(t)=0.7\). Numerically, there are two solutions in \([0,300]\):
\(\displaystyle t_1\approx 33.013298\ \text{s},\quad t_2\approx 166.574719\ \text{s}.\)
Evaluate candidates and endpoints (candidates test on a closed interval: Topic 5.5). :contentReference[oaicite:4]{index=4}
\[ \begin{array}{c|c} t & \text{People in line} \\ \hline 0 & 20 \\ t_1 & 20+\int_{0}^{t_1}\!\!\big(r-0.7\big)\,dx \approx \mathbf{3.803} \\ t_2 & 20+\int_{0}^{t_2}\!\!\big(r-0.7\big)\,dx \approx 158.070 \\ 300 & 80 \end{array} \] Since \(N'(t)\) changes from negative to positive at \(t_1\) (and \(N(0),N(300)\) are larger), the minimum occurs at \[ \boxed{t\approx 33.013\ \text{s},\quad N_{\min}\approx 4\ \text{people (nearest whole number)}}. \] (Rate meaning and sign analysis: Topics 4.1 & 8.3.) :contentReference[oaicite:5]{index=5} :contentReference[oaicite:6]{index=6}
Calc-Ok Question 2
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 \(p(h)=0.2h^{2}e^{-0.0025h^{2}}\) for \(0\le h\le 30\) and is modeled by \(f(h)\) for \(h\ge 30\). 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 \(0\le f(h)\le u(h)\) for all \(h\ge 30\) and \(\displaystyle \int_{30}^{\infty}u(h)\,dh=105\). 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 \(t\ge 0\), the position of the boat is \((x(t),y(t))\), where \(x'(t)=662\sin(5t)\) and \(y'(t)=880\cos(6t)\). 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 \(0\le t\le 1\).
Most-appropriate topic codes (CED):
• TOPIC 2.7–2.8: Product & Chain Rules — computing \(p'(h)\) (part (a)).
• TOPIC 8.3: Using Accumulation Functions & Definite Integrals in Applied Contexts — total amount from a density (parts (b), (c)).
• TOPIC 6.13 (BC): Evaluating Improper Integrals — bounding the tail with a comparison (part (c)).
• TOPIC 9.4 (BC): Vector-Valued Functions — speed \(=\sqrt{(x’)^{2}+(y’)^{2}}\) and distance \(\int \text{speed}\,dt\) (part (d)).
• TOPIC 8.3: Using Accumulation Functions & Definite Integrals in Applied Contexts — total amount from a density (parts (b), (c)).
• TOPIC 6.13 (BC): Evaluating Improper Integrals — bounding the tail with a comparison (part (c)).
• TOPIC 9.4 (BC): Vector-Valued Functions — speed \(=\sqrt{(x’)^{2}+(y’)^{2}}\) and distance \(\int \text{speed}\,dt\) (part (d)).
▶️ Answer/Explanation
(a) Compute \(p'(25)\) and interpret
\(p(h)=0.2h^{2}e^{-0.0025h^{2}}\). Using product & chain rules, \[ p'(h)=e^{-0.0025h^{2}}\big(0.4h\big)+0.2h^{2}\cdot e^{-0.0025h^{2}}\cdot(-0.005h) =e^{-0.0025h^{2}}\big(0.4h-0.001h^{3}\big). \] Evaluate at \(h=25\): \(0.4h-0.001h^{3}=10-15.625=-5.625\); \(e^{-0.0025(25^{2})}=e^{-1.5625}\). \[ p'(25)=(-5.625)\,e^{-1.5625}\approx \boxed{-1.179}. \] Units/meaning: At a depth of \(25\) meters, the plankton density is decreasing at a rate of about \(\boxed{1.179\ \text{million cells per m}^3\ \text{per meter of depth}}\).
(b) Total plankton (millions) in a \(3\ \text{m}^2\) column from \(h=0\) to \(30\)
Amount \(=\displaystyle \int_{0}^{30}\!\big(\text{area}\times \text{density}\big)\,dh=\int_{0}^{30}\!3\,p(h)\,dh\). Numerically, \[ \int_{0}^{30}\!3\,p(h)\,dh\approx \boxed{1675.414936}. \] To the nearest million: \(\boxed{1675\ \text{million}}\).
(c) Whole column to depth \(K>30\) and the \(2000\) bound
Number of cells (millions) in entire column: \[ \boxed{\;\int_{0}^{30}\!3\,p(h)\,dh\;+\;\int_{30}^{K}\!3\,f(h)\,dh\;} \] Since \(0\le f(h)\le u(h)\) for \(h\ge 30\), \[ 0\le \int_{30}^{K}\!3\,f(h)\,dh \le \int_{30}^{K}\!3\,u(h)\,dh \le 3\!\int_{30}^{\infty}\!u(h)\,dh=3\cdot 105=315. \] Therefore the total is \(\le 1675.415+315=1990.415\le \boxed{2000\ \text{million}}\).
(d) Distance traveled for \(0\le t\le 1\)
Speed \(=\sqrt{(x'(t))^{2}+(y'(t))^{2}}=\sqrt{(662\sin 5t)^{2}+(880\cos 6t)^{2}}\). Total distance \[ \boxed{\;\int_{0}^{1}\!\sqrt{(x'(t))^{2}+(y'(t))^{2}}\,dt\;} \approx \boxed{757.455862\ \text{meters}}\ \ (\text{about }757.456\ \text{m})\]
\(p(h)=0.2h^{2}e^{-0.0025h^{2}}\). Using product & chain rules, \[ p'(h)=e^{-0.0025h^{2}}\big(0.4h\big)+0.2h^{2}\cdot e^{-0.0025h^{2}}\cdot(-0.005h) =e^{-0.0025h^{2}}\big(0.4h-0.001h^{3}\big). \] Evaluate at \(h=25\): \(0.4h-0.001h^{3}=10-15.625=-5.625\); \(e^{-0.0025(25^{2})}=e^{-1.5625}\). \[ p'(25)=(-5.625)\,e^{-1.5625}\approx \boxed{-1.179}. \] Units/meaning: At a depth of \(25\) meters, the plankton density is decreasing at a rate of about \(\boxed{1.179\ \text{million cells per m}^3\ \text{per meter of depth}}\).
(b) Total plankton (millions) in a \(3\ \text{m}^2\) column from \(h=0\) to \(30\)
Amount \(=\displaystyle \int_{0}^{30}\!\big(\text{area}\times \text{density}\big)\,dh=\int_{0}^{30}\!3\,p(h)\,dh\). Numerically, \[ \int_{0}^{30}\!3\,p(h)\,dh\approx \boxed{1675.414936}. \] To the nearest million: \(\boxed{1675\ \text{million}}\).
(c) Whole column to depth \(K>30\) and the \(2000\) bound
Number of cells (millions) in entire column: \[ \boxed{\;\int_{0}^{30}\!3\,p(h)\,dh\;+\;\int_{30}^{K}\!3\,f(h)\,dh\;} \] Since \(0\le f(h)\le u(h)\) for \(h\ge 30\), \[ 0\le \int_{30}^{K}\!3\,f(h)\,dh \le \int_{30}^{K}\!3\,u(h)\,dh \le 3\!\int_{30}^{\infty}\!u(h)\,dh=3\cdot 105=315. \] Therefore the total is \(\le 1675.415+315=1990.415\le \boxed{2000\ \text{million}}\).
(d) Distance traveled for \(0\le t\le 1\)
Speed \(=\sqrt{(x'(t))^{2}+(y'(t))^{2}}=\sqrt{(662\sin 5t)^{2}+(880\cos 6t)^{2}}\). Total distance \[ \boxed{\;\int_{0}^{1}\!\sqrt{(x'(t))^{2}+(y'(t))^{2}}\,dt\;} \approx \boxed{757.455862\ \text{meters}}\ \ (\text{about }757.456\ \text{m})\]
No-Calc Question
The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for \(-5 \le x < 3\), and \(g(x)=2(x-4)^2\) for \(3 \le x \le 6\).
(a) If \(f(1)=3\), what is the value of \(f(-5)\)?
(b) Evaluate \(\displaystyle \int_{1}^{6} g(x)\,dx.\)
(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.
Most-appropriate topic codes (CED):
• TOPIC 8.3: Using accumulation functions & definite integrals in context — parts (a), (b)
• TOPIC 5.3: Determining intervals of increase/decrease from \(f'(x)\) — part (c)
• TOPIC 5.4: Determining concavity and points of inflection using \(f”(x)\) or \(g'(x)\) — parts (c), (d)
• TOPIC 5.3: Determining intervals of increase/decrease from \(f'(x)\) — part (c)
• TOPIC 5.4: Determining concavity and points of inflection using \(f”(x)\) or \(g'(x)\) — parts (c), (d)
▶️ Answer/Explanation
(a) Find \(f(-5)\)
\(f’ = g\), so \(f(b)-f(a)=\displaystyle\int_a^b g(x)\,dx\). \(f(-5)=f(1)+\int_{1}^{-5} g(x)\,dx= f(1) – \int_{-5}^{1} g(x)\,dx.\)
From the graph on \([-5,1]\):
• \([-5,-3]\): rectangle of height \(-3\), width \(2\) → area \(-6\).
• \([-3,0]\): triangle below axis, base \(3\), height \(3\) → \(-\tfrac{9}{2}\).
• \([0,1]\): triangle above axis, base \(1\), height \(2\) → \(+1\).
\(\int_{-5}^{1} g(x)\,dx = -6 – \tfrac{9}{2} + 1 = -\tfrac{19}{2}\).
\(f(-5)=3 – \big(-\tfrac{19}{2}\big)=\boxed{\tfrac{25}{2}}.\)
(b) Evaluate \(\int_{1}^{6} g(x)\,dx\)
Split at \(x=3\): \(\int_{1}^{6} g=\int_{1}^{3} g + \int_{3}^{6} g.\)
• On \([1,3]\), \(g=2\) → area \(2\cdot2=4\).
• On \([3,6]\), \(g(x)=2(x-4)^2\). \(\int_{3}^{6}2(x-4)^2dx=\left[\tfrac{2}{3}(x-4)^3\right]_{3}^{6}= \tfrac{16}{3}-(-\tfrac{2}{3})=6.\)
Total \(=4+6=\boxed{10}.\)
(c) Intervals where \(f\) is increasing and concave up
\(f\) is increasing when \(f'(x)=g(x)>0\). \(f\) is concave up when \(f”(x)=g'(x)>0\). On \(0<x<1\): \(g>0\) and slope of \(g\) is positive. On \(4<x<6\): \(g(x)=2(x-4)^2>0\), and \(g'(x)=4(x-4)>0\). Therefore the intervals are \(\boxed{(0,1)\ \text{and}\ (4,6)}\).
(d) Points of inflection of \(f\)
Points of inflection occur where \(f”(x)=g'(x)\) changes sign. From the graph, \(g\) changes from decreasing to increasing at \(x=4\) (local minimum). Thus, the point of inflection is at \(\boxed{x=4}\).
\(f’ = g\), so \(f(b)-f(a)=\displaystyle\int_a^b g(x)\,dx\). \(f(-5)=f(1)+\int_{1}^{-5} g(x)\,dx= f(1) – \int_{-5}^{1} g(x)\,dx.\)
From the graph on \([-5,1]\):
• \([-5,-3]\): rectangle of height \(-3\), width \(2\) → area \(-6\).
• \([-3,0]\): triangle below axis, base \(3\), height \(3\) → \(-\tfrac{9}{2}\).
• \([0,1]\): triangle above axis, base \(1\), height \(2\) → \(+1\).
\(\int_{-5}^{1} g(x)\,dx = -6 – \tfrac{9}{2} + 1 = -\tfrac{19}{2}\).
\(f(-5)=3 – \big(-\tfrac{19}{2}\big)=\boxed{\tfrac{25}{2}}.\)
(b) Evaluate \(\int_{1}^{6} g(x)\,dx\)
Split at \(x=3\): \(\int_{1}^{6} g=\int_{1}^{3} g + \int_{3}^{6} g.\)
• On \([1,3]\), \(g=2\) → area \(2\cdot2=4\).
• On \([3,6]\), \(g(x)=2(x-4)^2\). \(\int_{3}^{6}2(x-4)^2dx=\left[\tfrac{2}{3}(x-4)^3\right]_{3}^{6}= \tfrac{16}{3}-(-\tfrac{2}{3})=6.\)
Total \(=4+6=\boxed{10}.\)
(c) Intervals where \(f\) is increasing and concave up
\(f\) is increasing when \(f'(x)=g(x)>0\). \(f\) is concave up when \(f”(x)=g'(x)>0\). On \(0<x<1\): \(g>0\) and slope of \(g\) is positive. On \(4<x<6\): \(g(x)=2(x-4)^2>0\), and \(g'(x)=4(x-4)>0\). Therefore the intervals are \(\boxed{(0,1)\ \text{and}\ (4,6)}\).
(d) Points of inflection of \(f\)
Points of inflection occur where \(f”(x)=g'(x)\) changes sign. From the graph, \(g\) changes from decreasing to increasing at \(x=4\) (local minimum). Thus, the point of inflection is at \(\boxed{x=4}\).
No-Calc Question 4
| \(t\) (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 \(2\le t\le 10\).
(d) The height of the tree, in meters, can also be modeled by the function \(G\), given by \(G(x)=\dfrac{100x}{1+x}\), 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?
Most-appropriate topic codes (CED):
• TOPIC 4.3: Rates of Change in Applied Contexts Other Than Motion — parts (a), (b)
• TOPIC 6.3: Riemann Sums & Trapezoidal Rule — part (c)
• TOPIC 4.4: Introduction to Related Rates — part (d) setup
• TOPIC 4.5: Solving Related Rates Problems (Chain Rule) — part (d) computation
• TOPIC 6.3: Riemann Sums & Trapezoidal Rule — part (c)
• TOPIC 4.4: Introduction to Related Rates — part (d) setup
• TOPIC 4.5: Solving Related Rates Problems (Chain Rule) — part (d) computation
▶️ Answer/Explanation
(a) Estimate \(H'(6)\) and interpret
Use a symmetric secant around \(t=6\): points \(t=5\) and \(t=7\).
\[ H'(6)\approx\frac{H(7)-H(5)}{7-5}=\frac{11-6}{2}=\boxed{\tfrac{5}{2}}. \]
Interpretation: At \(t=6\) years, the tree’s height is increasing at about \(\boxed{2.5\ \text{meters per year}}\).
(b) Show a time \(t\) with \(H'(t)=2\) for \(2<t<10\)
On \([3,5]\), the average rate of change is \(\dfrac{H(5)-H(3)}{5-3}=\dfrac{6-2}{2}=2\).
\(H\) is continuous on \([3,5]\) and differentiable on \((3,5)\).
By the Mean Value Theorem, there exists \(c\in(3,5)\) such that \(\boxed{H'(c)=2}\).
(c) Average height on \(2\le t\le 10\) via trapezoids
Subintervals: \([2,3], [3,5], [5,7], [7,10]\) with widths \(1,2,2,3\).
Trapezoidal estimate for the integral: \[ \int_{2}^{10} H(t)\,dt \approx 1\cdot\frac{1.5+2}{2} +2\cdot\frac{2+6}{2} +2\cdot\frac{6+11}{2} +3\cdot\frac{11+15}{2} = 65.75=\frac{263}{4}. \]
Average height \(=\dfrac{1}{10-2}\int_{2}^{10}H(t)\,dt =\dfrac{1}{8}\cdot\frac{263}{4} =\boxed{\tfrac{263}{32}\ \text{meters}}\;(\approx 8.219).\)
(d) Rate of change of height when the tree is 50 m tall
Model: \(G(x)=\dfrac{100x}{1+x}\), with \(x’=\dfrac{dx}{dt}=0.03\ \text{m/yr}\).
First find \(x\) when \(G(x)=50\): \[ \frac{100x}{1+x}=50 \;\Rightarrow\; 100x=50(1+x)\;\Rightarrow\; x=\boxed{1}. \]
Differentiate \(G\): \(G'(x)=\dfrac{100(1+x)-100x}{(1+x)^2}=\dfrac{100}{(1+x)^2}\).
Chain rule: \[ \frac{dG}{dt}=G'(x)\,x’ =\frac{100}{(1+1)^2}\cdot 0.03 =25\cdot 0.03 =\boxed{\tfrac{3}{4}\ \text{m/yr}}. \]
So when the tree is \(50\) m tall, its height is increasing at \(\tfrac{3}{4}\) meter per year.
Use a symmetric secant around \(t=6\): points \(t=5\) and \(t=7\).
\[ H'(6)\approx\frac{H(7)-H(5)}{7-5}=\frac{11-6}{2}=\boxed{\tfrac{5}{2}}. \]
Interpretation: At \(t=6\) years, the tree’s height is increasing at about \(\boxed{2.5\ \text{meters per year}}\).
(b) Show a time \(t\) with \(H'(t)=2\) for \(2<t<10\)
On \([3,5]\), the average rate of change is \(\dfrac{H(5)-H(3)}{5-3}=\dfrac{6-2}{2}=2\).
\(H\) is continuous on \([3,5]\) and differentiable on \((3,5)\).
By the Mean Value Theorem, there exists \(c\in(3,5)\) such that \(\boxed{H'(c)=2}\).
(c) Average height on \(2\le t\le 10\) via trapezoids
Subintervals: \([2,3], [3,5], [5,7], [7,10]\) with widths \(1,2,2,3\).
Trapezoidal estimate for the integral: \[ \int_{2}^{10} H(t)\,dt \approx 1\cdot\frac{1.5+2}{2} +2\cdot\frac{2+6}{2} +2\cdot\frac{6+11}{2} +3\cdot\frac{11+15}{2} = 65.75=\frac{263}{4}. \]
Average height \(=\dfrac{1}{10-2}\int_{2}^{10}H(t)\,dt =\dfrac{1}{8}\cdot\frac{263}{4} =\boxed{\tfrac{263}{32}\ \text{meters}}\;(\approx 8.219).\)
(d) Rate of change of height when the tree is 50 m tall
Model: \(G(x)=\dfrac{100x}{1+x}\), with \(x’=\dfrac{dx}{dt}=0.03\ \text{m/yr}\).
First find \(x\) when \(G(x)=50\): \[ \frac{100x}{1+x}=50 \;\Rightarrow\; 100x=50(1+x)\;\Rightarrow\; x=\boxed{1}. \]
Differentiate \(G\): \(G'(x)=\dfrac{100(1+x)-100x}{(1+x)^2}=\dfrac{100}{(1+x)^2}\).
Chain rule: \[ \frac{dG}{dt}=G'(x)\,x’ =\frac{100}{(1+1)^2}\cdot 0.03 =25\cdot 0.03 =\boxed{\tfrac{3}{4}\ \text{m/yr}}. \]
So when the tree is \(50\) m tall, its height is increasing at \(\tfrac{3}{4}\) meter per year.
No-Calc Question
The graphs of the polar curves \(r=4\) and \(r=3+2\cos\theta\) are shown in the figure above. The curves intersect at \(\theta=\dfrac{\pi}{3}\) and \(\theta=\dfrac{5\pi}{3}\).
(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 \(\theta=\dfrac{\pi}{2}\).
(c) A particle moves along the portion of the curve \(r=3+2\cos\theta\) for \(0<\theta<\dfrac{\pi}{2}\). 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 \(\theta=\dfrac{\pi}{3}\). Indicate units of measure.
Most-appropriate topic codes (CED):
• TOPIC 9.8 (BC): Area of a Polar Region — part (a)
• TOPIC 9.7 Defining Polar Coordinates and Differentiating in Polar Form
• TOPIC 4.5:Solving Related Rates Problems (Chain Rule) — part (c)
• TOPIC 9.7 Defining Polar Coordinates and Differentiating in Polar Form
• TOPIC 4.5:Solving Related Rates Problems (Chain Rule) — part (c)
▶️ Answer/Explanation
(a) Area of \(R\)
Inside the circle \(r=4\) and outside \(r=3+2\cos\theta\) between the intersection angles \(\theta=\dfrac{\pi}{3}\) and \(\theta=\dfrac{5\pi}{3}\):
\[ \boxed{\text{Area}(R)=\frac12\int_{\pi/3}^{5\pi/3}\!\Big(4^{2}-\big(3+2\cos\theta\big)^{2}\Big)\,d\theta }. \]
(b) Slope \(\dfrac{dy}{dx}\) at \(\theta=\dfrac{\pi}{2}\)
For polar \(x=r\cos\theta,\ y=r\sin\theta\): \(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} =\frac{r’\sin\theta+r\cos\theta}{r’\cos\theta-r\sin\theta}\), with \(r’=dr/d\theta\).
Here \(r=3+2\cos\theta\Rightarrow r’=-2\sin\theta\). At \(\theta=\dfrac{\pi}{2}\): \(\sin=\!1,\ \cos=\!0,\ r=3\).
Numerator \(=r’\sin\theta+r\cos\theta=(-2)(1)+3(0)=-2\).
Denominator \(=r’\cos\theta-r\sin\theta=(-2)(0)-3(1)=-3\).
\[ \boxed{\left.\frac{dy}{dx}\right|_{\theta=\pi/2}=\frac{-2}{-3}=\frac{2}{3}}. \]
(c) Angular rate \(\dfrac{d\theta}{dt}\) at \(\theta=\dfrac{\pi}{3}\)
Given the radial distance increases at \(3\) units/s: \(\displaystyle \frac{dr}{dt}=3\). Chain rule: \(\displaystyle \frac{dr}{dt}=\frac{dr}{d\theta}\cdot\frac{d\theta}{dt}\). Since \(r’=dr/d\theta=-2\sin\theta\), \[ \frac{d\theta}{dt}=\frac{\frac{dr}{dt}}{\frac{dr}{d\theta}} =\frac{3}{-2\sin\theta}. \] Evaluate at \(\theta=\dfrac{\pi}{3}\) (\(\sin\frac{\pi}{3}=\frac{\sqrt3}{2}\)): \[ \boxed{\frac{d\theta}{dt}=\frac{3}{-2\cdot(\sqrt3/2)}=-\frac{3}{\sqrt3}=-\sqrt3\ \text{radians/second}}. \] (Negative sign indicates \(\theta\) is decreasing at that instant.)
Inside the circle \(r=4\) and outside \(r=3+2\cos\theta\) between the intersection angles \(\theta=\dfrac{\pi}{3}\) and \(\theta=\dfrac{5\pi}{3}\):
\[ \boxed{\text{Area}(R)=\frac12\int_{\pi/3}^{5\pi/3}\!\Big(4^{2}-\big(3+2\cos\theta\big)^{2}\Big)\,d\theta }. \]
(b) Slope \(\dfrac{dy}{dx}\) at \(\theta=\dfrac{\pi}{2}\)
For polar \(x=r\cos\theta,\ y=r\sin\theta\): \(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} =\frac{r’\sin\theta+r\cos\theta}{r’\cos\theta-r\sin\theta}\), with \(r’=dr/d\theta\).
Here \(r=3+2\cos\theta\Rightarrow r’=-2\sin\theta\). At \(\theta=\dfrac{\pi}{2}\): \(\sin=\!1,\ \cos=\!0,\ r=3\).
Numerator \(=r’\sin\theta+r\cos\theta=(-2)(1)+3(0)=-2\).
Denominator \(=r’\cos\theta-r\sin\theta=(-2)(0)-3(1)=-3\).
\[ \boxed{\left.\frac{dy}{dx}\right|_{\theta=\pi/2}=\frac{-2}{-3}=\frac{2}{3}}. \]
(c) Angular rate \(\dfrac{d\theta}{dt}\) at \(\theta=\dfrac{\pi}{3}\)
Given the radial distance increases at \(3\) units/s: \(\displaystyle \frac{dr}{dt}=3\). Chain rule: \(\displaystyle \frac{dr}{dt}=\frac{dr}{d\theta}\cdot\frac{d\theta}{dt}\). Since \(r’=dr/d\theta=-2\sin\theta\), \[ \frac{d\theta}{dt}=\frac{\frac{dr}{dt}}{\frac{dr}{d\theta}} =\frac{3}{-2\sin\theta}. \] Evaluate at \(\theta=\dfrac{\pi}{3}\) (\(\sin\frac{\pi}{3}=\frac{\sqrt3}{2}\)): \[ \boxed{\frac{d\theta}{dt}=\frac{3}{-2\cdot(\sqrt3/2)}=-\frac{3}{\sqrt3}=-\sqrt3\ \text{radians/second}}. \] (Negative sign indicates \(\theta\) is decreasing at that instant.)
No-Calc Question
The Maclaurin series for \(\ln(1+x)\) is given by \[ x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots+(-1)^{\,n+1}\frac{x^{n}}{n}+\cdots. \] On its interval of convergence, this series converges to \(\ln(1+x)\). Let \(f\) be the function defined by \(f(x)=x\ln\!\left(1+\dfrac{x}{3}\right)\).
(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 \(\big|P_4(2)-f(2)\big|\).
Most-appropriate topic codes (CED):
• TOPIC 10.14: Finding Taylor/Maclaurin Series (build from known series; substitution and multiplication) — parts (a), (b)
• TOPIC 10.13: Radius & Interval of Convergence of Power Series (ratio test; endpoint analysis) — part (b)
• TOPIC 10.10: Alternating Series Test & Error Bound — part (c)
• TOPIC 10.13: Radius & Interval of Convergence of Power Series (ratio test; endpoint analysis) — part (b)
• TOPIC 10.10: Alternating Series Test & Error Bound — part (c)
▶️ Answer/Explanation
(a) First four nonzero terms & general term
Start with \(\ln(1+u)=\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{u^{n}}{n}\) for \(|u|<1\). Substitute \(u=\dfrac{x}{3}\) and multiply by \(x\): \[ f(x)=x\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x/3)^{n}}{n} =\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{\,n+1}}{n\,3^{n}}. \] First four nonzero terms (for \(n=1,2,3,4\)): \[ \boxed{\frac{x^{2}}{3}\;-\;\frac{x^{3}}{2\cdot 3^{2}}\;+\;\frac{x^{4}}{3\cdot 3^{3}}\;-\;\frac{x^{5}}{4\cdot 3^{4}}\;+\;\cdots} \] General term: \[ \boxed{a_n(x)=(-1)^{n+1}\,\frac{x^{\,n+1}}{n\,3^{n}}\qquad(n\ge 1)}. \]
(b) Interval of convergence
The parent series for \(\ln(1+u)\) has radius \(1\). With \(u=\dfrac{x}{3}\), the radius becomes \(3\): \(\;|x|<3\). Check endpoints:
• \(x=3\): series becomes \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{3}{n}\), an alternating harmonic (scaled) → converges (conditionally).
• \(x=-3\): series becomes \(\displaystyle\sum_{n=1}^{\infty}\frac{3}{n}\) → harmonic → diverges.
Therefore the interval of convergence is \[ \boxed{(-3,\,3]}\quad\text{with radius }R=3. \]
(c) Error bound for \(P_4(2)\)
\(P_4\) keeps terms up to degree 4 (i.e., \(n=1,2,3\)). The next omitted term is the degree-5 term (\(n=4\)): \[ T_{5}(x)=\;-\;\frac{x^{5}}{4\cdot 3^{4}}. \] By the alternating series error bound, at \(x=2\) \[ \big|P_4(2)-f(2)\big| \le \Big|\frac{2^{5}}{4\cdot 3^{4}}\Big| =\frac{32}{4\cdot 81} =\boxed{\frac{8}{81}}. \]
Start with \(\ln(1+u)=\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{u^{n}}{n}\) for \(|u|<1\). Substitute \(u=\dfrac{x}{3}\) and multiply by \(x\): \[ f(x)=x\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x/3)^{n}}{n} =\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{\,n+1}}{n\,3^{n}}. \] First four nonzero terms (for \(n=1,2,3,4\)): \[ \boxed{\frac{x^{2}}{3}\;-\;\frac{x^{3}}{2\cdot 3^{2}}\;+\;\frac{x^{4}}{3\cdot 3^{3}}\;-\;\frac{x^{5}}{4\cdot 3^{4}}\;+\;\cdots} \] General term: \[ \boxed{a_n(x)=(-1)^{n+1}\,\frac{x^{\,n+1}}{n\,3^{n}}\qquad(n\ge 1)}. \]
(b) Interval of convergence
The parent series for \(\ln(1+u)\) has radius \(1\). With \(u=\dfrac{x}{3}\), the radius becomes \(3\): \(\;|x|<3\). Check endpoints:
• \(x=3\): series becomes \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{3}{n}\), an alternating harmonic (scaled) → converges (conditionally).
• \(x=-3\): series becomes \(\displaystyle\sum_{n=1}^{\infty}\frac{3}{n}\) → harmonic → diverges.
Therefore the interval of convergence is \[ \boxed{(-3,\,3]}\quad\text{with radius }R=3. \]
(c) Error bound for \(P_4(2)\)
\(P_4\) keeps terms up to degree 4 (i.e., \(n=1,2,3\)). The next omitted term is the degree-5 term (\(n=4\)): \[ T_{5}(x)=\;-\;\frac{x^{5}}{4\cdot 3^{4}}. \] By the alternating series error bound, at \(x=2\) \[ \big|P_4(2)-f(2)\big| \le \Big|\frac{2^{5}}{4\cdot 3^{4}}\Big| =\frac{32}{4\cdot 81} =\boxed{\frac{8}{81}}. \]