Calc-OkQuestion 1
Most-appropriate topic codes (CED):
• TOPIC 5.3: Determining Intervals on Which a Function Is Increasing or Decreasing — part (b)
• TOPIC 5.5: Using the Candidates Test to Determine Absolute (Global) Extrema — part (c)
▶️ Answer/Explanation
(a)
Amount that flows in on \([0,8]\): \(\displaystyle \int_{0}^{8} R(t)\,dt\).
\(\displaystyle \int_{0}^{8} 20\sin\!\left(\tfrac{t^{2}}{35}\right)\,dt \approx \boxed{76.570}\ \text{ft}^3.\)
(b)
Let \(A(t)\) be the amount of water. Then \(A'(t)=R(t)-D(t)\) (net rate in minus out).
\(R(3)=20\sin\!\left(\tfrac{9}{35}\right)\approx 5.086,\quad D(3)=-0.04(27)+0.4(9)+0.96(3)=5.400.\)
\(A'(3)=R(3)-D(3)\approx -0.314<0\), so the amount is \(\boxed{\text{decreasing at }t=3\text{ hr}}\).
(c)
Critical times solve \(A'(t)=R(t)-D(t)=0\) on \([0,8]\).
\(R(t)-D(t)=0\) at \(t=0\) and \(t\approx \boxed{3.272}\ \text{hr}.\)
Use the candidates test on \(\{0,\,3.272,\,8\}\) with \(A(t)=30+\displaystyle\int_{0}^{t}[R(x)-D(x)]\,dx\):
\(A(0)=30.000,\quad A(3.272)\approx 27.965,\quad A(8)\approx 48.544.\)
Minimum occurs at \(\boxed{t\approx 3.272\ \text{hr}}\) (smallest value of \(A\)).
(d)
Overflow when \(A(w)=50\) with \(w>8\). An appropriate equation is
\(\displaystyle 30+\int_{0}^{8}[R(t)-D(t)]\,dt+\int_{8}^{w}[R(t)-D(t)]\,dt=50.\)
(Equivalently, since the rates continue for \(t>8\): \(\displaystyle 30+\int_{0}^{w}[R(t)-D(t)]\,dt=50\), \(w>8\).)
Calc-Ok Question 2
Most-appropriate topic codes (CED):
• TOPIC 9.4: Defining and Differentiating Vector-Valued Functions — part (b)
• TOPIC 9.6: Solving Motion Problems Using Parametric and Vector-Valued Functions — parts (c), (d)
▶️ Answer/Explanation
\(x'(t)=\cos(t^2)\), \(x(1)=3\).
\(x(2)=x(1)+\displaystyle\int_{1}^{2}\cos(t^2)\,dt=3+\int_{1}^{2}\cos(t^2)\,dt\).
Numerical value \(\;\Rightarrow\;\boxed{x(2)\approx 2.557}\).
(b)
Slope \(\displaystyle \frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{e^{0.5t}}{\cos(t^2)}\).
Set \(\frac{dy}{dx}=2\): \(e^{0.5t}-2\cos(t^2)=0\).
Solve on \((0,1)\) (calculator) \(\Rightarrow\;\boxed{t\approx 0.840}\).
(c)
Speed \(=\sqrt{(x'(t))^{2}+(y'(t))^{2}}=\sqrt{\cos^{2}(t^{2})+e^{t}}\).
Solve \(\sqrt{\cos^{2}(t^{2})+e^{t}}=3\ \Rightarrow\ \cos^{2}(t^{2})+e^{t}-9=0\).
Numerical solution \(\Rightarrow\;\boxed{t\approx 2.196}\).
(d)
Total distance on \([0,1]\): \(\displaystyle \int_{0}^{1}\sqrt{\cos^{2}(t^{2})+e^{t}}\,dt\).
Numerical value \(\Rightarrow\;\boxed{\text{distance}\approx 1.595}\).
No-Calc Question 3
| t (minutes) | 0 | 12 | 20 | 24 | 40 |
| v(t) (meters per minute) | 0 | 200 | 240 | -220 | 150 |
Approximate the value of \(\displaystyle \int_{0}^{40}\!|v(t)|\,dt\) using a right Riemann sum with the four subintervals indicated in the table.
Find Bob’s acceleration at time \(t=5\).
Most-appropriate topic codes (CED):
• TOPIC 6.3: Riemann Sums, Summation Notation, and Definite Integral Notation — part (b)
• TOPIC 4.2: Straight-Line Motion: Connecting Position, Velocity, and Acceleration — part (c)
• TOPIC 8.1: Finding the Average Value of a Function on an Interval — part (d)
▶️ Answer/Explanation
(a)
Symmetric difference about \(t=16\) using \(t=12\) and \(t=20\):
\(\displaystyle v'(16)\approx \frac{v(20)-v(12)}{20-12}=\frac{240-200}{8}= \boxed{5\ \text{m/min}^2}.\)
(b)
Meaning: \(\displaystyle \int_{0}^{40}\!|v(t)|\,dt\) is the total distance Johanna travels (meters) for \(0\le t\le 40\).
Right Riemann sum on \([0,12],[12,20],[20,24],[24,40]\):
\(\displaystyle \int_{0}^{40}\!|v(t)|\,dt \approx 12\times|v(12)|+8\times|v(20)|+4\times|v(24)|+16\times|v(40)|\)
\(=12\times200+8\times240+4\times220+16\times150\)
\(=2400+1920+880+2400=\boxed{7600\ \text{meters}}.\)
(c)
\(B(t)=t^{3}-6t^{2}+300\ \Rightarrow\ B'(t)=3t^{2}-12t\) (acceleration, m/min\(^2\)).
\(B'(5)=3\cdot25-12\cdot5=75-60=\boxed{15\ \text{m/min}^{2}}.\)
(d)
Average velocity on \([0,10]\): \(\displaystyle \bar v=\frac{1}{10}\int_{0}^{10}B(t)\,dt\).
\(\displaystyle \int B(t)\,dt=\frac{t^{4}}{4}-2t^{3}+300t\). Evaluate \(0\to 10\):
\(\displaystyle \bar v=\frac{1}{10}\Big[\frac{10^{4}}{4}-2\cdot10^{3}+300\cdot10\Big]\) \(=\frac{1}{10}(2500-2000+3000)=\boxed{350\ \text{m/min}}.\)
No-Calc Question 4
Most-appropriate topic codes (CED):
• TOPIC 7.1: Modeling Situations with Differential Equations — part (b)
• TOPIC 5.4: Using the First Derivative Test to Determine Relative (Local) Extrema — part (c)
• TOPIC 2.2: Defining the Derivative of a Function and Using Derivative Notation — part (d)
▶️ Answer/Explanation
(a)
Sample slope field at the six indicated points:
(b)
\(\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(2x-y)=2-\frac{dy}{dx}=2-(2x-y)=2-2x+y.\)
In Quadrant II, \(x<0\) and \(y>0\). Then \(2-2x+y>0\), so \(\displaystyle \frac{d^{2}y}{dx^{2}}>0\).
Therefore, all solution curves are \(\boxed{\text{concave up in Quadrant II}}\).
(c)
\(f'(x)=2x-f(x)\). At \((x,y)=(2,3)\): \(f'(2)=2(2)-3=1\ne0\).
Since the derivative is not zero, \(f\) has \(\boxed{\text{neither a relative minimum nor a relative maximum at }x=2}\).
(d)
Suppose \(y=mx+b\). Then \(\dfrac{dy}{dx}=m\). Substitute into \(y’=2x-y\):
\(m=2x-(mx+b)\Rightarrow (2-m)x-(m+b)=0\ \text{for all }x.\)
Hence \(2-m=0\Rightarrow m=2\) and \(-(m+b)=0\Rightarrow b=-m=-2\).
\(\boxed{m=2,\; b=-2}\).
No-Calc Question 5
Most-appropriate topic codes (CED):
• TOPIC 5.4: Using the First Derivative Test to Determine Relative (Local) Extrema — part (b)
• TOPIC 5.2: EVT, Global vs. Local Extrema, and Critical Points — part (c)
• TOPIC 6.12: Using Linear Partial Fractions — part (d)
▶️ Answer/Explanation
\(f(4)=\dfrac{1}{16-12}=\dfrac{1}{4}\).
\(f'(4)=\dfrac{3-2\cdot 4}{(16-12)^{2}}=\dfrac{-5}{16}\).
Tangent line at \(x=4\): \(\;y-f(4)=f'(4)(x-4)\Rightarrow\boxed{\,y-\tfrac14=-\tfrac{5}{16}(x-4)\,}\).
(b)
For \(k=4\): \(\displaystyle f'(x)=\frac{4-2x}{(x^{2}-4x)^{2}}\). The denominator is positive where \(f\) is defined (except at \(x=0,4\)).
\(\operatorname{sgn}f'(x)=\operatorname{sgn}(4-2x)\): positive for \(x<2\), negative for \(x>2\).
Thus \(f\) increases then decreases \(\Rightarrow\) \(\boxed{\text{relative maximum at }x=2}\).
(c)
A critical point occurs where \(f'(x)=0\) (and \(f\) defined) or where \(f’\) is undefined while \(f\) is defined.
At \(x=-5\), \(f\) is defined provided \(x\neq 0\) and \(x\neq k\) (so \(k\neq -5\)).
Set \(f'(-5)=0:\; \dfrac{k-2(-5)}{((-5)^{2}-k(-5))^{2}}=0 \Rightarrow k+10=0\Rightarrow\boxed{k=-10}.\)
(d)
For \(k=6\): \(f(x)=\dfrac{1}{x^{2}-6x}=\dfrac{1}{x(x-6)}\).
Partial fractions: \(\displaystyle \frac{1}{x(x-6)}=\frac{A}{x}+\frac{B}{x-6}\).
\(1=A(x-6)+Bx \Rightarrow \begin{cases}A+B=0\\ -6A=1\end{cases}\Rightarrow A=-\tfrac16,\;B=\tfrac16.\)
So \(\displaystyle f(x)=\boxed{-\frac{1}{6}\cdot\frac{1}{x}+\frac{1}{6}\cdot\frac{1}{x-6}}\).
Then \[ \int f(x)\,dx=\int\!\left(-\frac{1}{6}\frac{1}{x}+\frac{1}{6}\frac{1}{x-6}\right)dx = -\frac{1}{6}\ln|x|+\frac{1}{6}\ln|x-6|+C = \boxed{\frac{1}{6}\ln\!\left|\frac{x-6}{x}\right|+C }. \]
No-Calc Question 6
\[ \sum_{n=1}^{\infty}\frac{(-3)^{\,n-1}}{n}\,x^{n} = x – \frac{3}{2}x^{2} + 3x^{3} – \dots + \frac{(-3)^{\,n-1}}{n}x^{n} + \dots \]
and converges to \(f(x)\) for \(|x|<R\), where \(R\) is the radius of convergence of the Maclaurin series.
Express \(f’\) as a rational function for \(|x|<R\).
Use the Maclaurin series for \(e^{x}\) to write the third-degree Taylor polynomial for \(g(x)=e^{x}f(x)\) about \(x=0\).
Most-appropriate topic codes (CED):
• TOPIC 10.14: Finding Taylor or Maclaurin Series for a Function — parts (b), (c)
▶️ Answer/Explanation
Let \(a_n=\dfrac{(-3)^{\,n-1}}{n}\,x^{n}\).
\(\displaystyle \left|\frac{a_{n+1}}{a_n}\right| =\left|\frac{(-3)^{n}}{n+1}x^{\,n+1}\times\frac{n}{(-3)^{n-1}x^{\,n}}\right| =\frac{n}{n+1}\times 3|x|\to 3|x|.\)
Converges when \(3|x|<1 \Rightarrow |x|<\tfrac{1}{3}\).
\(\boxed{R=\tfrac{1}{3}}\).
(b)
Differentiate term-by-term (valid for \(|x|<R\)):
\(f'(x)=1-3x+9x^{2}-27x^{3}+\cdots\).
Recognize geometric: first term \(1\), ratio \(-3x\).
Sum (for \(|-3x|<1\)): \(\displaystyle \boxed{f'(x)=\frac{1}{1+3x}}\).
(c)
Maclaurin for \(e^{x}\): \(1+x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{6}+\cdots\).
Given \(f(x)=x-\dfrac{3}{2}x^{2}+3x^{3}+\cdots\).
Multiply to degree \(3\): \(g(x)=e^{x}f(x)\).
Contributions:
• from \(1\times f(x)\): \(x-\dfrac{3}{2}x^{2}+3x^{3}\).
• from \(x\times f(x)\): \(x\times x=x^{2}\) and \(x\times\big(-\dfrac{3}{2}x^{2}\big)=-\dfrac{3}{2}x^{3}\).
• from \(\dfrac{x^{2}}{2}\times f(x)\): \(\dfrac{x^{2}}{2}\times x=\dfrac{1}{2}x^{3}\).
• higher products \(\ge x^{4}\) are ignored for a cubic polynomial.
Combine coefficients:
\(x\) term: \(1\times x = x\).
\(x^{2}\) term: \(-\dfrac{3}{2} + 1 = -\dfrac{1}{2}\).
\(x^{3}\) term: \(3 – \dfrac{3}{2} + \dfrac{1}{2} = 2\).
Therefore the third-degree Taylor polynomial is
\(\boxed{T_{3}(x)=x-\dfrac{1}{2}x^{2}+2x^{3}}.\)