Calc-Ok Question
| \(t\) (seconds) | 0 | 60 | 90 | 120 | 135 | 150 |
|---|---|---|---|---|---|---|
| \(f(t)\) (gal/s) | 0 | 0.1 | 0.15 | 0.1 | 0.05 | 0 |
(a) Using correct units, interpret the meaning of \(\displaystyle \int_{60}^{135} f(t)\,dt\) in the context of the problem. Use a right Riemann sum with the three subintervals \([60,90],\,[90,120],\,[120,135]\) to approximate \(\displaystyle \int_{60}^{135} f(t)\,dt\).
(b) Must there exist a value of \(c\), for \(60<c<120\), such that \(f'(c)=0\)? Justify your answer.
(c) The rate of flow of gasoline (gal/s) can also be modeled by \[ g(t)=\frac{t}{500}\cos\!\left(\left(\frac{t}{120}\right)^{\!2}\right), \quad 0\le t\le 150. \] Using this model, find the average rate of flow of gasoline over \(0\le t\le 150\). Show the setup for your calculations.
(d) Using the model \(g\) from part (c), find the value of \(g'(140)\). Interpret the meaning of your answer in the context of the problem.
Most-appropriate topic codes:
TOPIC 6.4: FTC & Accumulation (interpretation of \(\int f\)) — part (a)
TOPIC 5.1: Mean Value Theorem — part (b)
TOPIC 8.1: Average Value of a Function — part (c)
TOPIC 4.1: Interpreting the Derivative in Context — part (d)
▶️ Answer/Explanation
Right Riemann sum (widths \(30, 30, 15\); right endpoints \(90,120,135\)): \[ \int_{60}^{135} f(t)\,dt \approx f(90)(90-60)+f(120)(120-90)+f(135)(135-120). \] \[ = (0.15)(30)+(0.10)(30)+(0.05)(15) = 4.5+3.0+0.75 = \boxed{8.25\ \text{gallons}}. \]
Compute the average rate of change: \[ \frac{f(120)-f(60)}{120-60}=\frac{0.10-0.10}{60}=0. \] By the Mean Value Theorem, there exists \(c\in(60,120)\) such that \[ f'(c)=0. \] Yes, such a \(c\) must exist.
Then \[ \int_{0}^{150}\frac{t}{500}\cos(u)\,dt =\frac{7200}{500}\int_{u=0}^{u=(150/120)^{2}}\cos u\,du =14.4\,\sin u\Big|_{0}^{(150/120)^{2}}. \] \[ =14.4\,\sin\!\left(\frac{25}{16}\right). \] Average value: \[ \boxed{\frac{1}{150}\int_{0}^{150} g(t)\,dt = \frac{14.4}{150}\sin\!\left(\frac{25}{16}\right) = \frac{12}{125}\sin\!\left(\frac{25}{16}\right) \approx 0.096\ \text{gal/s}.} \]
Let \(u(t)=\left(\tfrac{t}{120}\right)^{2}\) so \(u'(t)=\tfrac{t}{7200}\).
Product/chain rule: \[ g'(t)=\frac{1}{500}\cos u(t)\;-\;\frac{t}{500}\sin u(t)\cdot \frac{t}{7200} =\frac{1}{500}\cos u(t)\;-\;\frac{t^{2}}{3{,}600{,}000}\sin u(t). \] At \(t=140\): \(u(140)=\left(\tfrac{140}{120}\right)^{2}=\tfrac{49}{36}\). \[ g'(140)=\frac{1}{500}\cos\!\left(\frac{49}{36}\right) -\frac{49}{9000}\sin\!\left(\frac{49}{36}\right) \approx \boxed{-0.0049\ \text{(gal/s)}^{2}}. \] Interpretation: At \(t=140\) s, the rate at which gasoline is flowing into the tank is decreasing at about \(0.005\) gallon per second per second.
Calc-Ok Question
For \(0\le t\le \pi\), a particle moves along a curve so that its position at time \(t\) is \((x(t),y(t))\).
The \(x(t)\) component is not explicitly given and \(y(t)=2\sin t\). It is known that \(\displaystyle \frac{dx}{dt}=e^{\cos t}\). At \(t=0\) the particle is at \((1,0)\).
(a) Find the acceleration vector of the particle at time \(t=1\). Show the setup for your calculations.
(b) For \(0\le t\le \pi\), find the first time \(t\) at which the speed of the particle is \(1.5\). Show the work that leads to your answer.
(c) Find the slope of the line tangent to the path at time \(t=1\). Find the \(x\)-coordinate of the position of the particle at time \(t=1\). Show the work that leads to your answers.
(d) Find the total distance traveled by the particle over \(0\le t\le \pi\). Show the setup for your calculations.
Most-appropriate topic codes:
• TOPIC 9.6: Solving Motion Problems Using Parametric & Vector-Valued Functions — part (b)
• TOPIC 9.1: Defining & Differentiating Parametric Equations — part (c, slope)
• TOPIC 9.5: Integrating Vector-Valued Functions — part (c, find \(x(1)\) from \(x'(t)\)+IV)
• TOPIC 9.3: Finding Arc Lengths of Curves Given by Parametric Equations — part (d)
▶️ Answer/Explanation
\(y(t)=2\sin t\Rightarrow y'(t)=2\cos t,\; y”(t)=-2\sin t.\)
Evaluate at \(t=1\): \(x”(1)=-e^{\cos 1}\sin 1\approx-1.444\), \(y”(1)=-2\sin 1\approx-1.683\).
\(\displaystyle \boxed{\mathbf a(1)=\langle x”(1),\,y”(1)\rangle\approx\langle-1.444,\,-1.683\rangle}\).
Solve for \(0\le t\le\pi\): \[ \sqrt{e^{2\cos t}+4\cos^{2}t}=1.5. \] Numerically (unique solution in \((0,\tfrac{\pi}{2})\)): \(\boxed{t\approx 1.254}\) (radians).
To find \(x(1)\), use the initial value \(x(0)=1\) and accumulate change: \[ x(1)=x(0)+\int_{0}^{1}x'(t)\,dt =1+\int_{0}^{1}e^{\cos t}\,dt \approx \boxed{3.342}. \]
No-Calc Question 3
(a) A slope field for \(\displaystyle \frac{dM}{dt}=\frac14(40-M)\) is shown. Sketch the solution curve through the point \((0,5)\).
(b) Use the line tangent to the graph of \(M\) at \(t=0\) to approximate \(M(2)\), the temperature at \(t=2\) minutes.
(c) Write an expression for \(\displaystyle \frac{d^{2}M}{dt^{2}}\) in terms of \(M\). Use \(\displaystyle \frac{d^{2}M}{dt^{2}}\) to decide whether the approximation from part (b) is an underestimate or an overestimate of \(M(2)\). Give a reason.
(d) Use separation of variables to find the particular solution \(M(t)\) to \(\displaystyle \frac{dM}{dt}=\frac14(40-M)\) with \(M(0)=5\).
Most-appropriate topic codes:
• TOPIC 4.6: Approximating Values Using Local Linearity & Linearization — part (b)
• TOPIC 5.6: Determining Concavity of Functions over Their Domains — part (c)
• TOPIC 7.7: Finding Particular Solutions Using Initial Conditions & Separation of Variables — part (d)
▶️ Answer/Explanation
Since \(\dfrac{dM}{dt}=\frac14(40-M)>0\) for \(M<40\), the solution is increasing. Also \(\dfrac{d^{2}M}{dt^{2}}<0\) (shown in part (c)), so the curve is concave down. The solution passes through \((0,5)\) and rises toward the horizontal asymptote \(M=40\) as \(t\) increases.
(b) Tangent-line approximation at \(t=0\)
Slope at \(t=0\): \[ M'(0)=\frac14\big(40-M(0)\big)=\frac14(40-5)=\frac{35}{4}=8.75. \] Tangent line at \(t=0\): \[ L(t)=M(0)+M'(0)\,(t-0)=5+\frac{35}{4}t. \] Approximate \(M(2)\): \[ M(2)\approx L(2)=5+\frac{35}{4}\cdot 2=5+\frac{35}{2}=5+17.5=\boxed{22.5^\circ\text{C}}. \]
Differentiate the DE: \[ \frac{d^{2}M}{dt^{2}}=\frac{d}{dt}\!\left[\tfrac14(40-M)\right] =-\tfrac14\,\frac{dM}{dt}. \] Using the original DE again, \[ \frac{d^{2}M}{dt^{2}}=-\tfrac14\Big(\tfrac14(40-M)\Big) =-\frac{1}{16}(40-M). \] Because \(M(t)<40\) for all \(t\), we have \(\dfrac{d^{2}M}{dt^{2}}<0\) (concave down). For an increasing, concave-down function, the tangent line at \(t=0\) lies above the curve for \(t>0\). Therefore the estimate in (b) is an overestimate of the actual \(M(2)\).
Start with \[ \frac{dM}{dt}=\frac14(40-M). \] Separate: \[ \frac{dM}{40-M}=\frac14\,dt. \] Integrate: \[ \int \frac{dM}{40-M}=\int \frac14\,dt \quad\Rightarrow\quad -\ln|40-M|=\frac{t}{4}+C. \] Write as \(\ln|40-M|=-\tfrac{t}{4}+C\). Apply \(M(0)=5\): \(\ln(40-5)=\ln 35=C\). Hence \(\ln(40-M)=-\tfrac{t}{4}+\ln 35\). Exponentiate: \[ 40-M=35\,e^{-t/4} \quad\Rightarrow\quad \boxed{\,M(t)=40-35\,e^{-t/4}\,}. \]
Calc-Ok Question 4
The function \(f\) is defined on the closed interval \([-2,8]\) and satisfies \(f(2)=1\). The graph of \(f’\) (two line segments and a semicircle) is shown.
(a) Does \(f\) have a relative minimum, a relative maximum, or neither at \(x=6\)? Give a reason.
(b) On what open interval(s), if any, is the graph of \(f\) concave down? Give a reason.
(c) Find the value of \(\displaystyle \lim_{x\to 2}\frac{6f(x)-3x}{x^{2}-5x+6}\), or show that it does not exist. Justify your answer.
(d) Find the absolute minimum value of \(f\) on \([-2,8]\). Justify your answer.
Most-appropriate topic codes (CED):
• TOPIC 5.6: Determining Concavity of Functions over Their Domains — part (b)
• TOPIC 4.7: Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms — part (c)
• TOPIC 5.5: Using the Candidates Test to Determine Absolute (Global) Extrema — part (d)
▶️ Answer/Explanation
f′(x) > 0 on (2, 6) and f′(x) > 0 on (6, 8).
The sign of f′ does not change at x = 6 → the function is increasing through x = 6.
Conclusion: Neither a relative minimum nor a relative maximum at x = 6.
f is concave down where f″(x) < 0 ⇔ where f′ is decreasing.
From the graph, f′ decreases on (-2, 0) and (4, 6).
Intervals: (-2, 0) and (4, 6).
Continuity at x = 2: \(\lim_{x\to2} (6f(x)-3x) = 6f(2)-3\cdot2 = 0\), \(\lim_{x\to2}(x^2-5x+6)=0\).
Indeterminate form 0/0 → use L’Hôpital’s Rule:
\[ \lim_{x\to2}\frac{6f'(x)-3}{2x-5} = \frac{6f'(2)-3}{-1}. \] From the graph, \(f'(2)=0\). Hence the limit is \(\displaystyle \boxed{3}\).
Candidates: endpoints x = -2, 8 and critical points where f′ = 0 → x = -1, 2, 6.
Use \(f(2)=1\) and \(f(x) = f(2) + \int_{2}^{x} f'(t)\,dt\). Compute signed areas from the graph:
• \(\displaystyle \int_{-2}^{-1} f’ = +\tfrac12(1)(2) = +1\) (triangle above axis).
• \(\displaystyle \int_{-1}^{0} f’ = -\tfrac12(1)(2) = -1\) (triangle below axis).
• \(\displaystyle \int_{0}^{2} f’ = -\tfrac12(2)(2) = -2\).
⇒ \(\displaystyle \int_{-2}^{2} f’ = -2\). Hence \(f(-2)=f(2)-(-2)=\boxed{3}\) and \(f(-1)=f(-2)+1=\boxed{4}\).
• \(\displaystyle \int_{2}^{4} f’ = +\tfrac12(2)(2) = +2\).
• \(\displaystyle \int_{4}^{6} f’\) (upper semicircle segment): area = rectangle \(2\times2\) minus quarter–circle of radius 2 → \(4-\pi\).
⇒ \(f(6)=f(2)+2+(4-\pi)=\boxed{7-\pi}\approx 3.859.\)
• By symmetry, \(\displaystyle \int_{6}^{8} f’ = 4-\pi\).
⇒ \(f(8)=f(6)+(4-\pi)=\boxed{11-2\pi}\approx 4.716.\)
Compare candidate values: \(f(-2)=3,\; f(-1)=4,\; f(2)=1,\; f(6)=7-\pi\approx 3.859,\; f(8)=11-2\pi\approx 4.716.\)
Absolute minimum: \(\boxed{f(2)=1}\) at \(x=2\).
Calc-Ok Question 5
The graphs of functions \(f\) and \(g\) are shown for \(0 \le x \le 3\). It is known that \(g(x)=\dfrac{12}{3+x}\) for \(x\ge 0\). The twice-differentiable function \(f\) (not given explicitly) satisfies \(f(3)=2\) and \(\displaystyle \int_{0}^{3} f(x)\,dx = 10\).
(a) Find the area of the shaded region enclosed by the graphs of \(f\) and \(g\).
(b) Evaluate the improper integral \(\displaystyle \int_{0}^{\infty} (g(x))^{2}\,dx\), or show that the integral diverges.
(c) Let \(h\) be defined by \(h(x)=x\,f'(x)\). Find the value of \(\displaystyle \int_{0}^{3} h(x)\,dx\).
Most-appropriate topic codes:
• TOPIC 6.13: Improper Integrals (BC) — part (b)
• TOPIC 6.11: Integration by Parts (definite form, BC) — part (c)
▶️ Answer/Explanation
On \([0,3]\), shaded area \[ A = \int_{0}^{3}\!\big(f(x)-g(x)\big)\,dx = \int_{0}^{3}\!f(x)\,dx – \int_{0}^{3}\!g(x)\,dx. \] Given \(\int_{0}^{3} f(x)\,dx = 10\). Compute \(\displaystyle \int_{0}^{3} g(x)\,dx = \int_{0}^{3} \frac{12}{3+x}\,dx = 12[\ln(3+x)]_{0}^{3} = 12(\ln 6-\ln 3) = 12\ln 2.\) Hence \(\boxed{A = 10 – 12\ln 2}.\)
\[ \int_{0}^{\infty} (g(x))^{2}dx = \lim_{b\to\infty} \int_{0}^{b} \frac{144}{(3+x)^{2}}\,dx = \lim_{b\to\infty}\Big[-\frac{144}{3+x}\Big]_{0}^{b} = \lim_{b\to\infty}\Big(48 – \frac{144}{3+b}\Big) = \boxed{48}. \]
Integration by parts: \(u=x\Rightarrow du=dx\), \(dv=f'(x)dx\Rightarrow v=f(x)\). \[ \int x f'(x)\,dx = x f(x) – \int f(x)\,dx. \] Evaluate from \(0\) to \(3\): \[ \int_{0}^{3} h(x)\,dx = [x f(x)]_{0}^{3} – \int_{0}^{3} f(x)\,dx = 3f(3) – 10 = 3\cdot 2 – 10 = \boxed{-4}. \]
Calc-Ok Question
(a) Find \(f^{(4)}(x)\), the fourth derivative of \(f\) with respect to \(x\). Write the fourth–degree Taylor polynomial for \(f\) about \(x=0\). Show the work that leads to your answer.
(b) The fourth–degree Taylor polynomial for \(f\) about \(x=0\) is used to approximate \(f(0.1)\). Given that \(\bigl|f^{(5)}(x)\bigr|\le 15\) for \(0\le x\le 0.5\), use the Lagrange error bound to show that this approximation is within \(\displaystyle \frac{1}{10^{5}}\) of the exact value of \(f(0.1)\).
(c) Let \(g\) be the function such that \(g(0)=4\) and \(g'(x)=e^{x}f(x)\). Write the second–degree Taylor polynomial for \(g\) about \(x=0\).
Most-appropriate topic codes (from the AP Calculus AB/BC CED):
• TOPIC 10.12: Lagrange Error Bound — part (b)
▶️ Answer/Explanation
Product + chain rule on \(f^{(3)}(x)=-2x\,f'(x^{2})\):
\[ f^{(4)}(x)=\frac{d}{dx}\!\big[-2x\,f'(x^{2})\big] =-2\,f'(x^{2})+(-2x)\cdot\big(f”(x^{2})\cdot 2x\big) =-2\,f'(x^{2})-4x^{2}f”(x^{2}). \] Values at \(x=0\): \(f(0)=2\), \(f'(0)=3\), \(f”(0)=-f(0)=-2\), \(f^{(3)}(0)=0\), \(f^{(4)}(0)=-2\,f'(0)=-6\).
Taylor polynomial: \[ T_{4}(x)=f(0)+f'(0)x+\frac{f”(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}+\frac{f^{(4)}(0)}{4!}x^{4} =2+3x-\!x^{2}-\frac{1}{4}x^{4}. \]
With \(\displaystyle M=\max_{0\le x\le 0.5}\!\bigl|f^{(5)}(x)\bigr|\le 15\), the remainder after degree 4 satisfies \[ \bigl|f(0.1)-T_{4}(0.1)\bigr| \le \frac{M}{5!}\,(0.1)^{5} \le \frac{15}{120}\cdot 10^{-5} =\frac{1}{8}\cdot 10^{-5} =1.25\times 10^{-6} <\frac{1}{10^{5}}. \] Hence the approximation by \(T_{4}\) is within \(1/10^{5}\) of the exact value.
\(g(0)=4\). \(g'(x)=e^{x}f(x)\Rightarrow g'(0)=e^{0}f(0)=2\). \(g”(x)=e^{x}f(x)+e^{x}f'(x)=e^{x}\!\big(f(x)+f'(x)\big)\Rightarrow g”(0)=2+3=5\).
Therefore \[ T_{2,g}(x)=g(0)+g'(0)x+\frac{g”(0)}{2}x^{2} =4+2x+\frac{5}{2}x^{2}. \] (Construction of Taylor polynomials)