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-5.3- Determining Intervals on Which a Function is Increasing or Decreasing
(d) Topic- 8.1- Finding the Average Value of a Function on an Interval
The density of a bacteria population in a circular petri dish at a distance \(r\) centimeters from the center of the dish is given by an increasing, differentiable function \(f\) , where \(f(r)\) is measured in milligrams per square centimeter. Values of \(f (r)\) for selected values of \(r\) are given in the table above.
(a) Use the data in the table to estimate \(f'(2.25)\). Using correct units, interpret the meaning of your answer in the context of this problem.
(b) The total mass, in milligrams, of bacteria in the petri dish is given by the integral expression \(2\pi \int_{0}^{4}r f(r)dr.\) Approximate the value of \(2\pi \int_{0}^{4}r f(r)dr\) using a right Riemann sum with the four subintervals indicated by the data in the table.
(c) Is the approximation found in part (b) an overestimate or underestimate of the total mass of bacteria in the petri dish? Explain your reasoning.
(d) The density of bacteria in the petri dish, for \(1 \leq r \leq 4\), is modeled by the function \(g\) defined by \(g(r)=2-16(cos (1.57\sqrt{r}))^{3}.\) For what value of \(k\) , \(1 < k < 4\), is \(g (k)\) equal to the average value of \(g (r)\) on the interval \(1 ≤ r ≤ 4\) ?
▶️Answer/Explanation
1(a)
\(f'(2.25)\approx \dfrac{f(2.5)-f(2)}{2.5-2}=\dfrac{10-6}{0.5}=\dfrac{4}{0.5}=8\)
At a distance of \(r=2.25\) centimeters from the center of the petri dish, the density of the bacteria population is increasing at a rate of \(8\) milligrams per square centimeter per centimeter.
1(b)
\(2\pi \int_{0}^{4} r f(r) \, dr \approx 2\pi \big( 1 \cdot f(1) \cdot (1 – 0) + 2 \cdot f(2) \cdot (2 – 1)+ 2.5 \cdot f(2.5) \cdot (2.5 – 2) + 4 \cdot f(4) \cdot (4 – 2.5) \big)\)
\(= 2\pi \big( 1 \cdot 2.1 + 2 \cdot 6.1 + 2.5 \cdot 10.5 + 4 \cdot 18 \cdot 1.5 \big)\)
\(= 269\pi = 845.088\)
1(c)
\(\frac{d}{dr} \big( r f(r) \big) = f(r) + r f'(r)\)
Because \(f\) is nonnegative and increasing, \(\frac{d}{dr} \big( r f(r) \big) > 0\) on the interval \(0 \leq r \leq 4\). Thus, the integrand \(r f(r)\) is strictly increasing.
Therefore, the right Riemann sum approximation of \(2\pi \int_{0}^{4} r f(r) \, dr\) is an overestimate.
1(d)
\(\text{Average value} = g_{\text{avg}} = \dfrac{1}{4 – 1} \int_{1}^{4} g(r) \, dr\)
\(\dfrac{1}{4 – 1} \int_{1}^{4} g(r) \, dr = 9.875795\)
\(g(k) = g_{\text{avg}} \implies k = 2.497\)
Question 2
(a) Topic-8.2- Connecting Position, Velocity, and Acceleration of Functions Using Integrals
(b) Topic-4.1- Interpreting the Meaning of the Derivative in Context
(c) Topic- 4.2- Straight-Line Motion: Connecting Position, Velocity, and Acceleration
(d) Topic-8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals
A particle, \(P\), is moving along the \(x\)-axis. The velocity of particle \(P\) at time \(t\) is given by \(v_{p}(t)= sin(t^{1.5})\) for \(0 \leq t \leq \pi\). At time \(t = 0\), particle \(P\) is at position \(x = 5\).
A second particle, \(Q\), also moves along the \(x\)-axis. The velocity of particle \(Q\) at time \(t\) is given by \(v_{Q}(t) = (t – 1.8) \cdot 1.25^t\) for \(0 \leq t \leq \pi\). At time \(t = 0\), particle \(Q\) is at position \(x = 10\).
(a) Find the positions of particles \(P\) and \(Q\) at time \(t = 1\).
(b) Are particles \(P\) and \(Q\) moving toward each other or away from each other at time \(t = 1\) ? Explain your reasoning.
(c) Find the acceleration of particle \(Q\) at time \(t = 1\). Is the speed of particle \(Q\) increasing or decreasing at time \(t = 1\) ? Explain your reasoning.
(d) Find the total distance traveled by particle \(P\) over the time interval \(0 \leq t \leq \pi\).
▶️Answer/Explanation
2(a)
\(x_P(1) = 5 + \int_{0}^{1} v_P(t) \, dt = 5.370660\)
At time \(t = 1\), the position of particle \(P\) is \(x = 5.371\) or \(or 5.370)\).
\(x_Q(1) = 10 + \int_{0}^{1} v_Q(t) \, dt = 8.564355\)
At time \(t = 1\), the position of particle \(Q\) is \(x = 8.564\).
2(b)
\(v_P(1) = \sin(1^{1.5}) = 0.841471 > 0\)
At time \(t = 1\), particle \(P\) is moving to the right.
\(v_Q(1) = (1 – 1.8) \cdot 1.25^1 = -1 < 0\)
At time \(t = 1\), particle \(Q\) is moving to the left.
At time \(t = 1\), \(x_P(1) < x_Q(1)\), so particle \(P\) is to the left of particle \(Q\).
Thus, at time \(t = 1\), particles \(P\) and \(Q\) are moving toward each other.
2(c)
\(a_Q(1) = v_Q'(1) = 1.026856\)
The acceleration of particle \(Q\) is \(1.027\) or \(1.026\) at time \(t = 1\).
\(v_Q(1) = -1 < 0 \quad\) and \(\quad a_Q(1) > 0\)
The speed of particle \(Q\) is decreasing at time \(t = 1\) because the velocity and acceleration have opposite signs.
2(d)
\(\int_{0}^{\pi} |v_P(t)| \, dt = 1.93148\)
Over the time interval \(0 \leq t \leq \pi\), the total distance traveled by particle \(P\) is \(1.931\).
Question 3
(a) Topic-6.9- Integrating Using Substitution
(b) Topic-2.8- The Product Rule
(c) Topic-8.9- Volume with Disc Method: Revolving Around the \(x\)- or \(y\)-Axis
A company designs spinning toys using the family of functions \(y = cx\sqrt{4-x^{2}},\) where \(c\) is a positive constant. The figure above shows the region in the first quadrant bounded by the \(x\)-axis and the graph of \(y = cx\sqrt{4-x^{2}},\) for some \(c\). Each spinning toy is in the shape of the solid generated when such a region is revolved about the \(x\)-axis. Both \(x\) and \(y\) are measured in inches.
(a) Find the area of the region in the first quadrant bounded by the \(x\)-axis and the graph of \(y = cx\sqrt{4-x^{2}},\) for \(c = 6\).
(b) It is known that, for \(y = cx\sqrt{4-x^{2}},\frac{dy}{dx}=\frac{c\left ( 4-2x^{2} \right )}{\sqrt{4-x^{2}}}.\) For a particular spinning toy, the radius of the largest cross-sectional circular slice is 1.2 inches. What is the value of \(c\) for this spinning toy?
(c) For another spinning toy, the volume is \(2\pi\) cubic inches. What is the value of \(c\) for this spinning toy? Ans:
▶️Answer/Explanation
3(a)
\(6x\sqrt{4 – x^2} = 0 \implies x = 0, \, x = 2\)
Area} \(= \int_{0}^{2} 6x\sqrt{4 – x^2} \, dx\)
Let \(u = 4 – x^2\).
\(du = -2x \, dx \implies -\frac{1}{2} \, du = x \, dx\)
\(x = 0 \implies u = 4 – 0^2 = 4\)
\(x = 2 \implies u = 4 – 2^2 = 0\)
\(\int_{0}^{2} 6x\sqrt{4 – x^2} \, dx = \int_{4}^{0} 6\left(-\frac{1}{2}\right)\sqrt{u} \, du= -3\int_{4}^{0} u^{1/2} \, du = 3\int_{0}^{4} u^{1/2} \, du\)\(= 2u^{3/2} \big|_{u=0}^{u=4} = 2 \cdot 8 = 16\)
The area of the region is \(16\) square inches.
3(b)
The cross-sectional circular slice with the largest radius occurs where \(y=c x \sqrt{4 – x^2}\) has its maximum on the interval \(0 < x < 2\).\(\frac{dy}{dx} = \frac{c \, (4 – 2x^2)}{\sqrt{4 – x^2}} = 0 \implies x = \sqrt{2}\)\(x = \sqrt{2} \implies y = c \sqrt{2} \sqrt{4 – (\sqrt{2})^2} = 2c \implies\)\(2c = 1.2 \implies c = 0.6\)
3(c)
\(\text{Volume} = \int_{0}^{2} \pi \big(c x \sqrt{4 – x^2}\big)^2 \, dx
= \pi c^2 \int_{0}^{2} x^2 (4 – x^2) \, dx\)
\(= \pi c^2 \int_{0}^{2} \big(4x^2 – x^4\big) \, dx\)
\(= \pi c^2 \left[ \frac{4}{3}x^3 – \frac{1}{5}x^5 \right]_{0}^{2}\)
\(= \pi c^2 \left(\frac{32}{3} – \frac{32}{5}\right) = \frac{64\pi c^2}{15}\)
\(\frac{64\pi c^2}{15} = 2\pi \implies c^2 = \frac{15}{32} \implies c = \sqrt{\frac{15}{32}}\)
Question 4
(a) Topic-5.6- Determining Concavity of Functions over Their Domains
(b) Topic-2.8- The Product Rule
(c) Topic-4.7- Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms
(d) Topic-5.1- Using the Mean Value Theorem
Let \(f\) be a continuous function defined on the closed interval \(−4 \leq x \leq 6\). The graph of \(f\), consisting of four line segments, is shown above. Let \(G\) be the function defined by \(G(x)=\int_{0}^{x}f(t)dt.\)
(a) On what open intervals is the graph of \(G\) concave up? Give a reason for your answer.
(b) Let \(P\) be the function defined by \(P (x) = G (x)\cdot f(x)\). Find \(P'(3 )\).
(c) Find \(\lim_{x\rightarrow 2}\frac{G(x)}{x^{2}-2x}.\)
(d) Find the average rate of change of G on the interval \([−4, 2]\). Does the Mean Value Theorem guarantee a value \(c\), \(−4 < c < 2\), for which \(G'(c)\) is equal to this average rate of change? Justify your answer.
▶️Answer/Explanation
4(a)
\(G'(x) = f(x)\)
The graph of \(G\) is concave up for \( -4 < x < -2\) and \(2 < x < 6\), because \(G’ = f\) is increasing on these intervals.
4(b)
\(P'(x) = G(x) \cdot f'(x) + f(x) \cdot G'(x)\)
\(P'(3) = G(3) \cdot f'(3) + f(3) \cdot G'(3)\)
Substituting \(G(3) = \int_{0}^{3} f(t)\), \(dt = -3.5\) and \(G'(3) = f(3) = -3\) into the above expression for \(P'(3)\) gives the following:
\(P'(3) = -3.5 \cdot 1 + (-3) \cdot (-3) = 5.5\)
4(c)
\(\lim_{x \to 2} \left(x^2 – 2x\right) = 0\)
Because \(G\) is continuous for \(-4 \leq x \leq 6\), \(\lim_{x \to 2} G(x) = \int_{0}^{2} f(t)\), \(dt = 0\).
Therefore, the limit \(\lim_{x \to 2} \frac{G(x)}{x^2 – 2x}\) is an indeterminate form of type \(\frac{0}{0}\).
Using L’Hopital’s Rule,
\(\lim_{x \to 2} \frac{G(x)}{x^2 – 2x} = \lim_{x \to 2} \frac{G'(x)}{2x – 2}\)
\(= \lim_{x \to 2} \frac{f(x)}{2x – 2} = \frac{f(2)}{2} = \frac{-4}{2} = -2\)
4(d)
\(G(2) = \int_{0}^{2} f(t)\), \(dt = 0 \quad\) and \(\quad G(-4) = \int_{0}^{-4} f(t)\), \(dt = -16\)
Average rate of change \(= \frac{G(2) – G(-4)}{2 – (-4)} = \frac{0 – (-16)}{6} = \frac{8}{3}\)
Yes, \(G'(x) = f(x)\) so \(G\) is differentiable on \((-4, 2)\) and continuous on \([-4, 2]\).
Therefore, the Mean Value Theorem applies and guarantees a value \(c\), \(-4 < c < 2\), such that
\(G'(c) = \frac{8}{3}\).
Question 5
(a) Topic-3.2- Implicit Differentiation
(b) Topic-2.10- Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions
(c) Topic-2.10- Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions
(d) Topic- 5.7- Using the Second Derivative Test to Determine Extrema
Consider the function \(y = f (x)\) whose curve is given by the equation \(2y2 − 6 = y sin x\) for \(y > 0\).
(a) Show that \(\frac{dy}{dx}=\frac{ycos x}{4y-sin x}.\)
(b) Write an equation for the line tangent to the curve at the point (0, \(\sqrt{3}\) ).
(c) For \(0 \leq x \leq \pi\) and \(y > 0\), find the coordinates of the point where the line tangent to the curve is horizontal.
(d) Determine whether f has a relative minimum, a relative maximum, or neither at the point found in part (c). Justify your answer.
▶️Answer/Explanation
5(a)\(\frac{d}{dx} \left(2y^2 – 6\right) = \frac{d}{dx} \left(y \sin x\right)
\implies 4y \frac{dy}{dx} = \frac{dy}{dx} \sin x + y \cos x\)
\(\implies 4y \frac{dy}{dx} – \frac{dy}{dx} \sin x = y \cos x \implies \frac{dy}{dx} \left(4y – \sin x\right) = y \cos x\)
\(\implies \frac{dy}{dx} = \frac{y \cos x}{4y – \sin x}\)
5(b)At the point \((0, \sqrt{3}), \quad \frac{dy}{dx} = \frac{\sqrt{3} \cos 0}{4\sqrt{3} – \sin 0} = \frac{1}{4}\).
An equation for the tangent line is \(y = \sqrt{3} + \frac{1}{4}x\).
5(c)\(\frac{dy}{dx} = \frac{y \cos x}{4y – \sin x} = 0 \implies y \cos x = 0 \text{ and } 4y – \sin x \neq 0\)
\(y \cos x = 0 \text{ and } y > 0 \implies x = \frac{\pi}{2}\)
\(\text{When } x = \frac{\pi}{2}, \, y \sin x = 2y^2 – 6 \implies y \sin \frac{\pi}{2} = 2y^2 – 6\)
\(\implies y = 2y^2 – 6 \implies 2y^2 – y – 6 = 0\)
\(\implies (2y + 3)(y – 2) = 0 \implies y = 2\)
\(\text{When } x = \frac{\pi}{2} \text{ and } y = 2, \, 4y – \sin x = 8 – 1 \neq 0.\)
\(\text{Therefore, the line tangent to the curve is horizontal at the point } \left(\frac{\pi}{2}, 2\right).\)
\(\frac{d^2y}{dx^2} = \frac{(4y – \sin x) \left(\frac{dy}{dx} \cos x – y \sin x\right) – (y \cos x) \left(4 \frac{dy}{dx} – \cos x\right)}{(4y – \sin x)^2} \)
\(\text{When } x = \frac{\pi}{2} \text{ and } y = 2, \)
\(\frac{d^2y}{dx^2} = \frac{(4 \cdot 2 – \sin \frac{\pi}{2}) \left(0 \cdot \cos \frac{\pi}{2} – 2 \cdot \sin \frac{\pi}{2}\right) – (2 \cos \frac{\pi}{2}) \left(4 \cdot 0 – \cos \frac{\pi}{2}\right)}{(4 \cdot 2 – \sin \frac{\pi}{2})^2} \)
\(= \frac{(7)(-2) – (0)(0)}{(7)^2} = \frac{-2}{7} < 0. \)
\(f \text{ has a relative maximum at the point } \left(\frac{\pi}{2}, 2\right) \text{ because } \frac{dy}{dx} = 0 \text{ and } \frac{d^2y}{dx^2} < 0. \)
Question 6
(a) Topic-7.3- Sketching Slope Fields
(b) Topic-7.4- Reasoning Using Slope Fields
(c) Topic-7.6- Finding General Solutions Using Separation of Variables
(d) Topic-5.7- Using the Second Derivative Test to Determine Extrema
A medication is administered to a patient. The amount, in milligrams, of the medication in the patient at time t hours is modeled by a function y = A(t ) that satisfies the differential equation \(\frac{dy}{dt}=\frac{12-y}{3}.\) At time t = 0 hours, there are 0 milligrams of the medication in the patient.
(a) A portion of the slope field for the differential equation \(\frac{dy}{dt}=\frac{12-y}{3}\) is given below. Sketch the solution curve through the point (0, 0).
(b) Using correct units, interpret the statement \(\lim_{t\rightarrow \infty }A(t)=12\) in the context of this problem.
(c) Use separation of variables to find y = A(t), the particular solution to the differential equation \(\frac{dy}{dt}=\frac{12-y}{3}\) with initial condition A(0) = 0.
(d) A different procedure is used to administer the medication to a second patient. The amount, in milligrams, of the medication in the second patient at time t hours is modeled by a function y = B(t) that satisfies the differential equation \(\frac{dy}{dt}=3 – \frac{y}{t+2}.\) At time t = 1 hour, there are 2.5 milligrams of the medication in the second patient. Is the rate of change of the amount of medication in the second patient increasing or decreasing at time t = 1 ? Give a reason for your answer.
▶️Answer/Explanation
6(a)
6(b)
Over time the amount of medication in the patient approaches \(12\) milligrams.
6(c)
\(\frac{dy}{dt} = \frac{12 – y}{3} \text{ with initial condition } A(0) = 0. \)
\(\frac{dy}{dt} = \frac{12 – y}{3} \implies \frac{dy}{12 – y} = \frac{dt}{3} \)
\(\int \frac{dy}{12 – y} = \int \frac{dt}{3} \implies -\ln|12 – y| = \frac{t}{3} + C \)
\(\ln|12 – y| = -\frac{t}{3} – C \implies |12 – y| = e^{-t/3 – C} \)
\(\implies y = 12 + K e^{-t/3} \)
\(0 = 12 + K \implies K = -12 \)
\(y = A(t) = 12 – 12e^{-t/3} \)
6(d)\(\frac{dy}{dt} = 3 – \frac{y}{t + 2} \implies \frac{d^2y}{dt^2} = (-1) \frac{\frac{dy}{dt}(t + 2) – y}{(t + 2)^2} \)
\(B'(1) = 3 – \frac{B(1)}{3} = 3 – \frac{2.5}{3} = \frac{6.5}{3} \)
\(B”(1) = -\frac{B'(1) \cdot 3 – B(1)}{3^2} = -\frac{6.5 – 2.5}{9} = -\frac{4}{9} < 0 \)
\(\text{The rate of change of the amount of medication is decreasing at time } t = 1 \)
\(\text{ because } B^{”}(1) < 0 \text{ and } \frac{d^2y}{dt^2} \text{ is continuous in an interval containing } t = 1. \)