Question 1
(a) Topic-2.3-Estimating Derivatives of a Function at a Point
(b) Topic-6.3-Riemann Sums Summation Notation and Definite Integral Notation
(c) Topic-6.7-The Fundamental Theorem of Calculus and Definite Integrals
(d) Topic-1.16-Working with the Intermediate Value Theorem IVT
t (hours) | 0 | 1 | 3 | 6 | 8 |
R (t) (liters / hour) | 1340 | 1190 | 950 | 740 | 700 |
Water is pumped into a tank at a rate modeled by \(W(t)= 2000e^{-t^{2}/20}\) liters per hour for 0 ≤ t ≤ 8, where t is measured in hours. Water is removed from the tank at a rate modeled by R (t) liters per hour, where R is differentiable and decreasing on 0 ≤ t ≤ 8. Selected values of R (t) are shown in the table above. At time t=0, there are 50,000 liters of water in the tank.
(a) Estimate R'(2). Show the work that leads to your answer. Indicate units of measure.
(b) Use a left Riemann sum with the four subintervals indicated by the table to estimate the total amount of water removed from the tank during the 8 hours. Is this an overestimate or an underestimate of the total amount of water removed? Give a reason for your answer.
(c) Use your answer from part (b) to find an estimate of the total amount of water in the tank, to the nearest liter, at the end of 8 hours.
(d) For 0 ≤ t ≤ 8, is there a time t when the rate at which water is pumped into the tank is the same as the rate at which water is removed from the tank? Explain why or why not.
▶️Answer/Explanation
\(\textbf{1(a) } R'(2) \approx \frac{R(3) – R(1)}{3 – 1} = \frac{950 – 1190}{3 – 1} = -120 \text{ liters/hr}^2\)
\(\textbf{1(b)} \text{The total amount of water removed is given by } \int_0^8 R(t) \, dt.\)
\(\int_0^8 R(t) \, dt \approx 1 \cdot R(0) + 2 \cdot R(1) + 3 \cdot R(3) + 2 \cdot R(6)\)
\(\quad = 1(1340) + 2(1190) + 3(950) + 2(740)\)
\(\quad = 8050 \text{ liters}\)
\(\text{This is an overestimate since } R \text{ is a decreasing function.}\)
\(\textbf{1(c)} \text{Total } \approx 50000 + \int_0^8 W(t) \, dt – 8050\)
\(\quad = 50000 + 7836.195325 – 8050 \approx 49786 \text{ liters}\)
\(\textbf{1(d)} W(0) – R(0) > 0, \, W(8) – R(8) < 0, \text{ and } W(t) – R(t) \text{ is continuous.}\)
\(\text{Therefore, the Intermediate Value Theorem guarantees at least one time } t, \, 0 < t < 8, \text{ for which } W(t) – R(t) = 0, \text{ or } W(t) = R(t).\)
\(\text{For this value of } t, \text{ the rate at which water is pumped into the tank is the same as the rate at which water is removed from the tank.}\)
Question 2
(a) Topic-8.2-Connecting Position Velocity and Acceleration of Functions Using Integrals
(b) Topic-9.1-Defining and Differentiating Parametric Equations
(c) Topic-9.6-Solving Motion Problems Using Parametric and Vector-Valued Functions
(d) Topic-9.3-Finding Arc Lengths of Curves Given by Parametric Equations
The graph of \( y \), consisting of three line segments, is shown in the figure above. \\
At \( t = 0 \), the particle is at position \( (5, 1) \).
(a) Find the position of the particle at \( t = 3 \).
(b) Find the slope of the line tangent to the path of the particle at \( t = 3 \).
(c) Find the speed of the particle at \( t = 3 \).
(d) Find the total distance traveled by the particle from \( t = 0 \) to \( t = 2 \).
▶️Answer/Explanation
\(\textbf{2(a)}\)
The position of the particle at \( t = 3 \) is \( (x(3), y(3)) \).
\(
x(3) = x(0) + \int_0^3 x'(t) \, dt \approx 5 + 9.377035 \approx 14.37703509
\)
Since the slope of the segment connecting \( (2, -1) \) and \( (4, 0) \) is \( \frac{1}{2} \), then \( y(3) = -\frac{1}{2} \).
So the position of the particle at \( t = 3 \) is
\(
\boxed{\left( 14.377, -\frac{1}{2} \right)}.
\)
\(\textbf{(b)}\)
\(
y'(3) = \frac{1}{2} \quad \text{from part (a)}.
\)
\(
x'(3) \approx 9.956375928 \quad \text{using the given expression for } \frac{dx}{dt}.
\)
\(
m_{t=3} = \frac{dy}{dx} \bigg|_{t=3} = \frac{y'(3)}{x'(3)} \approx \frac{\frac{1}{2}}{9.956375928} = \boxed{0.0502190761}.
\)
\(\textbf{(c)}\)
\(
\text{Speed}_{t=3} = \sqrt{(x'(3))^2 + (y'(3))^2} \approx \sqrt{(9.956375928)^2 + \left(\frac{1}{2}\right)^2} \approx \boxed{9.968922791}.
\)
\(\textbf{(d)}\)
\(
\text{Total Distance} = \int_0^2 \sqrt{(x'(t))^2 + (y'(t))^2} \, dt = \int_0^1 \sqrt{(x'(t))^2 + (-2)^2} \, dt + \int_1^2 \sqrt{(x'(t))^2 + 0^2} \, dt \approx \boxed{4.34987}.
\)
Question 3
(a) Topic-5.4-Using the First Derivative Test to Determine Relative Local Extrema
(b) Topic-5.4-Using the First Derivative Test to Determine Relative Local Extrema
(c) Topic-5.5-Using the Candidates Test to Determine Absolute Global Extrema
(d) Topic-1.1-Introducing Calculus Can Change Occur at an Instant
The figure above shows the graph of the piecewise-linear function f. For -4 ≤ x ≤ 12, the function g is defined by \(g(x)=\int_{2}^{x}f(t)dt.\)
(a) Does g have a relative minimum, a relative maximum, or neither at x = 10 ? Justify your answer.
(b) Does the graph of g have a point of inflection at x = 4 ? Justify your answer.
(c) Find the absolute minimum value and the absolute maximum value of g on the interval -4 ≤ x ≤ 12. Justify your answers.
(d) For -4 ≤ x ≤ 12 , find all intervals for which g (x) ≤ 0.
▶️Answer/Explanation
\(\textbf{3(a)} \text{The function } g \text{ has neither a relative minimum nor a relative maximum at } x = 10 \text{ since } g'(x) = f(x) \text{ and } f(x) \leq 0 \text{ for } 8 \leq x \leq 12.\)
\(\textbf{3(b)} \text{The graph of } g \text{ has a point of inflection at } x = 4 \text{ since } g'(x) = f(x) \text{ is increasing for } 2 \leq x \leq 4 \text{ and decreasing for } 4 \leq x \leq 8.\)
\(\textbf{3(c) } g'(x) = f(x) \text{ changes sign only at } x = -2 \text{ and } x = 6.\)
\(
\begin{array}{|c|c|}
\hline
x & g(x) \\
\hline
-4 & -4 \\
-2 & -8 \\
6 & 8 \\
12 & -4 \\
\hline
\end{array}
\)
\(\text{On the interval } -4 \leq x \leq 12, \text{ the absolute minimum value is } g(-2) = -8 \text{ and the absolute maximum value is } g(6) = 8.\)
\(\textbf{3(d) } g(x) \leq 0 \text{ for } -4 \leq x \leq 2 \text{ and } 10 \leq x \leq 12.\)
Question 4
(a) Topic-3.6-Calculating Higher-Order Derivatives
(b) Topic-5.4-Using the First Derivative Test to Determine Relative Local Extrema
(c) Topic-4.7-Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms
(d) Topic-7.5-Approximating Solutions Using Euler’s Method
Consider the differential equation
\(
\frac{dy}{dx} = x^2 – \frac{1}{2}y.
\)
(a) Find \( \frac{d^2y}{dx^2} \) in terms of \( x \) and \( y \).
(b) Let \( y = f(x) \) be the particular solution to the given differential equation whose graph passes through the point \((-2, 8)\). Does the graph of \( f \) have a relative minimum, a relative maximum, or neither at the point \((-2, 8)\)? Justify your answer.
(c) Let \( y = g(x) \) be the particular solution to the given differential equation with \( g(-1) = 2 \). Find
\(
\lim_{x \to -1} \frac{g(x) – 2}{3(x + 1)^2}.
\)
Show the work that leads to your answer.
(d) Let \( y = h(x) \) be the particular solution to the given differential equation with \( h(0) = 2 \). Use Euler’s method, starting at \( x = 0 \) with two steps of equal size, to approximate \( h(1) \).
▶️Answer/Explanation
\(\textbf{4(a)}\)
\(
\frac{d^2y}{dx^2} = 2x – \frac{1}{2} \frac{dy}{dx} = 2x – \frac{1}{2} \left( x^2 – \frac{1}{2}y \right)
\)
\(
= 2x – \frac{1}{2}x^2 + \frac{1}{4}y.
\)
\(\textbf{(b)}\)
\(
\frac{dy}{dx} \bigg|_{(-2, 8)} = (-2)^2 – \frac{1}{2}(8) = 0.
\)
\(
\frac{d^2y}{dx^2} \bigg|_{(-2, 8)} = 2(-2) – \frac{1}{2}(-2)^2 + \frac{1}{4}(8) = -4 < 0.
\)
Hence, at \((-2, 8)\), \( f \) has a \textbf{relative maximum}.
\(\textbf{(c)}\)
\(
\lim_{x \to -1} (g(x) – 2) = g(-1) – 2 = 2 – 2 = 0, \quad \lim_{x \to -1} \left( 3(x+1)^2 \right) = 0.
\)
So by L’Hospital’s Rule:
\(
\lim_{x \to -1} \frac{g(x) – 2}{3(x+1)^2} = \lim_{x \to -1} \frac{g'(x)}{6(x+1)}.
\)
\(
\lim_{x \to -1} g'(x) = (-1)^2 – \frac{1}{2}(2) = 0, \quad \lim_{x \to -1} (6(x+1)) = 0.
\)
So by L’Hospital’s Rule:
\(
\lim_{x \to -1} \frac{g'(x)}{6(x+1)} = \lim_{x \to -1} \frac{g^{”}(x)}{6} = \frac{g^{”}(-1)}{6}.
\)
\(
g^{”}(-1) = 2(-1) – \frac{1}{2}(-1)^2 + \frac{1}{4}(2) = -\frac{1}{3}.
\)
\(
\text{So } \lim_{x \to -1} \frac{g(x) – 2}{3(x+1)^2} = \boxed{-\frac{1}{3}}.
\)
\(\textbf{(d)}\)
Using Euler’s method with 2 steps from 0 to 1, the step size is \( \frac{1}{2} \).
\(
h(0) = 2.
\)
\(
h \left( \frac{1}{2} \right) = h(0) + h'(0) \left( \frac{1}{2} \right) = 2 + \left( 0 – \frac{1}{2}(2) \right) \left( \frac{1}{2} \right) = \frac{3}{2}.
\)
\(
h(1) \approx h \left( \frac{1}{2} \right) + h’ \left( \frac{1}{2} \right) \left( \frac{1}{2} \right).
\)
\(
h’ \left( \frac{1}{2} \right) = \left( \frac{1}{2} \right)^2 – \frac{1}{2} \left( \frac{3}{2} \right) = \frac{1}{4} – \frac{3}{4} = -\frac{1}{2}.
\)
\(
h(1) \approx \frac{3}{2} + \left( -\frac{1}{2} \right) \left( \frac{1}{2} \right) = \frac{5}{4}.
\)
\(
\text{So } h(1) \approx \boxed{\frac{5}{4}}.
\)
Question 5
(a) Topic-8.3-Using Accumulation Functions and Definite Integrals in Applied Contexts
(b) Topic-8.10-Volume with Disc Method Revolving Around Other Axes
(c) Topic-4.5-Solving Related Rates Problems
The inside of a funnel of height 10 inches has circular cross sections, as shown in the figure above. At height h, the radius of the funnel is given by \(r = \frac{1}{20}\left ( 3+h^{2} \right ),\) where 0 ≤ h ≤ 10. The units of r and h are inches.
(a) Find the average value of the radius of the funnel.
(b) Find the volume of the funnel.
(c) The funnel contains liquid that is draining from the bottom. At the instant when the height of the liquid is h = 3 inches, the radius of the surface of the liquid is decreasing at a rate of \(\frac{1}{5}\) inch per second. At this instant, what is the rate of change of the height of the liquid with respect to time?
▶️Answer/Explanation
\(\textbf{5(a)} \text{Average radius } = \frac{1}{10} \int_{0}^{10} \frac{1}{20} \left( 3 + h^2 \right) \, dh = \frac{1}{200} \left[ 3h + \frac{h^3}{3} \right]_0^{10}\)
\(\quad = \frac{1}{200} \left( \left( 30 + \frac{1000}{3} \right) – 0 \right) = \frac{109}{60} \, \text{in}\)
\(\textbf{5(b)} \text{Volume } = \pi \int_{0}^{10} \left( \frac{1}{20} \left( 3 + h^2 \right) \right)^2 \, dh = \frac{\pi}{400} \int_{0}^{10} \left( 9 + 6h^2 + h^4 \right) \, dh\)
\(\quad = \frac{\pi}{400} \left[ 9h + 2h^3 + \frac{h^5}{5} \right]_0^{10}\)
\(\quad = \frac{\pi}{400} \left( \left( 90 + 2000 + \frac{100000}{5} \right) – 0 \right) = \frac{2209 \pi}{40} \, \text{in}^3\)
\(\textbf{5(c) } \frac{dr}{dt} = \frac{1}{20} (2h) \frac{dh}{dt}\)
\(-\frac{1}{5} = \frac{3}{10} \frac{dh}{dt}\)
\(\frac{dh}{dt} = -\frac{1}{5} \cdot \frac{10}{3} = -\frac{2}{3} \, \text{in/sec}\)
Question 6
(a) Topic-10.14-Finding Taylor or Maclaurin Series for a Function
(b) Topic-10.4-Integral Test for Convergence
(c) Topic-10.7-Alternating Series Test for Convergence
(d) Topic-10.10-Alternating Series Error Bound
The function \( f \) has a Taylor series about \( x = 1 \) that converges to \( f(x) \) for all \( x \) in the interval of convergence.
It is known that \( f(1) = 1 \), \( f'(1) = -\frac{1}{2} \), and the \( n \)-th derivative of \( f \) at \( x = 1 \) is given by
\(
f^{(n)}(1) = (-1)^n \frac{(n-1)!}{2^n} \quad \text{for } n \geq 2.
\)
(a) Write the first four nonzero terms and the general term of the Taylor series for \( f \) about \( x = 1 \).
(b) The Taylor series for \( f \) about \( x = 1 \) has a radius of convergence of 2. Find the interval of convergence. Show the work that leads to your answer.
(c) The Taylor series for \( f \) about \( x = 1 \) can be used to represent \( f(1.2) \) as an alternating series. Use the first three nonzero terms of the alternating series to approximate \( f(1.2) \).
(d) Show that the approximation found in part (c) is within 0.001 of the exact value of \( f(1.2) \).
▶️Answer/Explanation
\(\textbf{6(a)}\)
\(
f^{(2)}(1) = (-1)^2 \frac{(2-1)!}{2^2} = \frac{1}{4}, \quad
f^{(3)}(1) = (-1)^3 \frac{(3-1)!}{2^3} = -\frac{1}{4}.
\)
\(
f(x) \approx 1 – \frac{1}{2}(x-1) + \frac{1}{4} \frac{(x-1)^2}{2!} – \frac{1}{4} \frac{(x-1)^3}{3!}
\)
\(
= 1 – \frac{x-1}{2} + \frac{(x-1)^2}{8} – \frac{(x-1)^3}{24}.
\)
or
\(
1 – \frac{x-1}{2} + \frac{(x-1)^2}{2^2 \cdot 2!} – \frac{(x-1)^3}{2^3 \cdot 3!}.
\)
So the general term, for \(n \geq 1 \text{ is } \frac{(-1)^n (x-1)^n}{2^n n}.\)
You can also find the general term by using the formula for the Taylor series centered at 1:
\(
1 – \frac{x-1}{2} + \sum_{n=2}^\infty \frac{(-1)^n (n-1)!}{2^n} \frac{(x-1)^n}{n!} = 1 – \frac{x-1}{2} + \sum_{n=2}^\infty \frac{(-1)^n (x-1)^n}{2^n n}.
\)
\(\textbf{(b)}\)
Since the series is centered at 1 and the radius of convergence is 2, then \( |x-1| < 2 \) or \( -1 < x < 3 \).
Testing the endpoints for convergence:
When \(x = 3: \quad \sum_{n=2}^\infty \frac{(-1)^n 2^n}{2^n n} = \sum_{n=2}^\infty \frac{(-1)^n}{n} \quad \text{converges since it is an alternating series whose terms decrease to 0.}\)
When \(x = -1: \quad \sum_{n=2}^\infty \frac{(-1)^n (-2)^n}{2^n n} = \sum_{n=2}^\infty \frac{1}{n} \quad \text{diverges since it is a p-series where } p \leq 1 \text{ or it is the harmonic series.}\)
So the interval of convergence is \(-1 < x \leq 3\).
\(\textbf{(c)}\)
\(
f(1.2) \approx 1 – \frac{1.2 – 1}{2} + \frac{(1.2 – 1)^2}{2^2 \cdot 2} = \frac{181}{200}.
\)
\(\textbf{(d)}\)
Since \( 1.2 \) lies within the interval of convergence, the series converges there. So by the Alternating Series Estimation Theorem, this approximation differs from the exact value by less than the absolute value of the 4th term (this is the Alternating Series Error Bound).
\(
\left| f(1.2) – 1 + \frac{1.2 – 1}{2} – \frac{(1.2 – 1)^2}{24} \right| \leq \frac{(1.2 – 1)^3}{24} = \frac{.2^3}{24} = \frac{.008}{24} = \frac{8}{1000 \cdot 24} = \frac{1}{3000} \leq \frac{1}{1000}.
\)