Question 1
(a)-Topic-6.2- Approximating Areas with Riemann Sums
(b)-Topic-5.1- Using the Mean Value Theorem
(c)-Topic-8.2-Connecting Position, Velocity, and Acceleration of Functions Using Integrals
(d)-Topic-8.3- Using Accumulation Functions and Definite Integrals in Applied Contexts
1. A customer at a gas station is pumping gasoline into a gas tank. The rate of flow of gasoline is modeled by a differentiable function f , where f(t) is measured in gallons per second and t is measured in seconds since
pumping began. Selected values of f(t) are given in the table.
(a) Using correct units, interpret the meaning of \(\int_{60}^{135}f(t) dt\) in the context of the problem. Use a right Riemann sum with the three subintervals [60, 90], [90, 120], and [120, 135] to approximate the value of \(\int_{60}^{135}f(t) dt\).
(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, in gallons per second, can also be modeled by \(g(t)=\left ( \frac{t}{500} \right )cos\left ( (\frac{t}{120})^{2} \right )\) for \(0\leq t\leq 150\) . Using this model, find the average rate of flow of gasoline over the time interval \(0\leq t\leq 150\). Show the setup for your calculations.
(d) Using the model g defined in part (c), find the value of g’ (140). Interpret the meaning of your answer in the context of the problem.
▶️Answer/Explanation
1(a) Using correct units, interpret the meaning of \(\int_{60}^{135}f(t)dt\) in the context of the problem. Use a right
Riemann sum with the three subintervals [60, 90 ,] [90, 120 ,] and [120, 135] to approximate the value of \(\int_{60}^{135}f(t)dt\).
\(\int_{60}^{135}f(t)dt\) represents the total number of gallons of gasoline pumped into the gas tank from time t = 60 seconds to time t = 135 seconds.
\(\int_{60}^{135}f(t)dt\)
\(\approx f(90)(90-60)+f(120)(120-90)+f(135)(135-120)\)
\(=(0.15)(30)+(0.1)(30)+(0.05)(15)=8.25\)
1(b) Must there exist a value of c, for 60 < c< 120 , such that f ′(c) = 0 ? Justify your answer.f is differentiable. ⇒ f is continuous on [60, 120] .
\(\frac{f(120)-f(60)}{120-60}=\frac{0.1-0.1}{60}=0\)
By the Mean Value Theorem, there must exist a c, for 60< c < 120, such that f ′( c) = 0.
1(c) The rate of flow of gasoline, in gallons per second, can also be modeled by \(g(t)=\left ( \frac{t}{500} \right )cos\left ( (\frac{t}{120})^{2} \right )\) for 0 ≤ t ≤150. Using this model, find the average rate of flow of gasoline over the time interval 0 ≤ t ≤ 150. Show the setup for your calculations.
\(\frac{1}{150-0}\int_{0}^{150}g(t)dt\)
= 0.0959967
1(d) Using the model g defined in part (c), find the value of g′(140 .) Interpret the meaning of your answer in the context of the problem.
g′(140) ≈ −0.004908
g′(140) = − 0.005 (or −0.004 )
The rate at which gasoline is flowing into the tank is decreasing at a rate of 0.005 (or 0.004 ) gallon per second per second at time t = 140 seconds
Question 2
(a)-Topic-4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration
(b)-Topic-4.3 Rates of Change in Applied Contexts Other Than Motion
(c)-Topic-4.6 Approximating Values of a Function Using Local Linearity and Linearization
(d)-Topic-4.3 Rates of Change in Applied Contexts Other Than Motion
2. Stephen swims back and forth along a straight path in a 50-meter-long pool for 90 seconds. Stephen’s velocity is modeled by \(v(t)=2.38e^{-0.02t}sin\left ( \frac{\pi }{56}t \right )\) where t is measured in seconds and v(t) is measured in meters per second.
(a) Find all times t in the interval 0 < t < 90 at which Stephen changes direction. Give a reason for your answer.
(b) Find Stephen’s acceleration at time t = 60 seconds. Show the setup for your calculations, and indicate units of measure. Is Stephen speeding up or slowing down at time t = 60 seconds? Give a reason for your
answer.
(c) Find the distance between Stephen’s position at time t = 20 seconds and his position at time t = 80 seconds. Show the setup for your calculations.
(d) Find the total distance Stephen swims over the time interval \(0\leq t\leq 90\) seconds. Show the setup for your calculations.
▶️Answer/Explanation
2(a) Find all times t in the interval 0 < t < 90 at which Stephen changes direction. Give a reason for your answer.
For 0 < t < 90, v (t ) = 0 ⇒ t = 56.
Stephen changes direction when his velocity changes sign. This occurs at t = 56 seconds.
2(b) Find Stephen’s acceleration at time t = 60 seconds. Show the setup for your calculations, and indicate units of measure. Is Stephen speeding up or slowing down at time t = 60 seconds? Give a reason for your answer.
v’ (60) = a(60) = −0.0360162
Stephen’s acceleration at time t = 60 seconds is −0.036 meter per second per second.
2(c) Find the distance between Stephen’s position at time t = 20 seconds and his position at time t = 80 seconds. Show the setup for your calculations.
\(\int_{20}^{80}v(t)dt\)
=23.383997
The distance between Stephen’s positions at t = 20 seconds and t = 80 seconds is 23.384 (or 23.383 ) meters.
2(d) Find the total distance Stephen swims over the time interval 0 ≤ t ≤ 90 seconds. Show the setup for your calculations.
\(\int_{0}^{90}\left | v(t)\right |dt\)
=62.164216
The total distance Stephen swims over the time interval 0 ≤ t ≤ 90 seconds is 62.164 meters.
Question 3
(a)-Topic-7.3 Sketching Slope Fields
(b)-Topic-7.4 Reasoning Using Slope Fields
(c)-Topic-7.8 Exponential Models with Differential Equations
(d)-Topic-7.6 Finding General Solutions Using Separation of Variables
3. A bottle of milk is taken out of a refrigerator and placed in a pan of hot water to be warmed. The increasing function M models the temperature of the milk at time t, where M(t) is measured in degrees Celsius (°C) and t is the number of minutes since the bottle was placed in the pan. M satisfies the differential equation \(\frac{dM}{dt}=\frac{1}{4}(40-M)\) At time t = 0, the temperature of the milk is 5°C. It can be shown that M(t) < 40 for all values of t.
(a) A slope field for the differential equation \(\frac{dM}{dt}=\frac{1}{4}(40-M)\) 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 of the milk at time t = 2 minutes.
(c) Write an expression for \(\frac{d^{2}M}{dt^{2}}\) in terms of M . Use \(\frac{d^{2}M}{dt^{2}}\) to determine whether the approximation from part (b) is an underestimate or an overestimate for the actual value of M(2). Give a reason for your answer.
(d) Use separation of variables to find an expression for M(t), the particular solution to the differential equation \(\frac{dM}{dt}=\frac{1}{4}(40-M)\) with initial condition M(0) = 5.
▶️Answer/Explanation
3(a) A slope field for the differential equation \(\frac{dM}{dt}=\frac{1}{4}(40-M)\) is shown. Sketch the solution curve through the point (0, 5) .
3(b) Use the line tangent to the graph of M at t = 0 to approximate M (2 ,) the temperature of the milk at time t = 2 minutes.
\(\frac{DM}{dt} |_{t=0}=\frac{1}{4}(40-5)=\frac{35}{4}\)
The tangent line equation is\(y=5+\frac{35}{4}(t-0)\).
\(M(2)\approx 5+\frac{35}{4}.2=22.5\)
The temperature of the milk at time t = 2 minutes is approximately \(22.5^{\circ}\) Celsius.
3(c) Write an expression for \(\frac{d^{2M}}{dt^{^{2}}}\) in terms of M. Use \(\frac{d^{2M}}{dt^{^{2}}}\) to determine whether the approximation from part (b) is an underestimate or an overestimate for the actual value of M (2 .) Give a reason for your answer.
\(\frac{d^{2M}}{dt^{^{2}}}=\frac{1}{4}\frac{DM}{dt}=\frac{1}{4}\left ( \frac{1}{4}(40-M) \right )=\frac{1}{16}(40-M)\)
Because \(M(t)< 40,\frac{d^{2}M}{dt^{2}}< 0\) , so the graph of M is concave down. Therefore, the tangent line approximation of M (2) is an overestimate.
3(d) Use separation of variables to find an expression for M (t ), the particular solution to the differential equation \(\frac{DM}{dt}=\frac{1}{4}(40-M)\) with initial condition M (0 )=5.
\(\frac{dm}{40 -M} = \frac{1}{4}dt\)
\(\int \frac{dm}{40 -M} = \int \frac{1}{4}dt\)
\(-In|40 – M| = \frac{1}{4}t +C\)
\(-In |40 – 5| = 0+C \Rightarrow C = -In 35\)
\(M(t) < 40 \Rightarrow 40 – M > 0 \Rightarrow |40 -M| = 40 -M\)
\(-In(40-M)=\frac{1}{4}t-In35\)
\(In(40-M)= -\frac{1}{4}t-In35\)
\(40-M= 35e^{-t/4}\)
\(M=40- 35e^{-t/4}\)
Question 4
(a)-Topic-5.9 Connecting a Function, Its First Derivative, and Its Second Derivative
(b)-Topic-5.6 Determining Concavity of Functions over Their 2 Domains
(c)-Topic-4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms
(d)-Topic-5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points
4. The function f is defined on the closed interval [−2, 8] and satisfies f(2) = 1. The graph of f ‘ , the derivative of f , consists of two line segments and a semicircle, as shown in the figure.
(a) Does f have a relative minimum, a relative maximum, or neither at x = 6 ? Give a reason for your answer.
(b) On what open intervals, if any, is the graph of f concave down? Give a reason for your answer.
(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 the closed interval [−2, 8]. Justify your answer.
▶️Answer/Explanation
4(a) Does f have a relative minimum, a relative maximum, or neither at x = 6 ? Give a reason for your answer.
f ′(x ) > 0 on (2, 6) and f ′( x) > 0 on (6, 8 ).
f ′(x ) does not change sign at x = 6, so there is neither a relative maximum nor a relative minimum at this location.
4(b) On what open intervals, if any, is the graph of f concave down? Give a reason for your answer.The graph of f is concave down on ( 2, 0) − and (4, 6) because f ′ is decreasing on these intervals.
4(c) Find the value of \(\lim_{ x\to 2}\frac{6f(x)-3x}{x^{2}-5x+6}\),or show that it does not exist. Justify your answer
Because f is differentiable at x = 2, f is continuous at x = 2,
so \(\displaystyle \lim_{x \to 2}f(x)=f(2)=1\)
\(\displaystyle \lim_{x \to 2}(6f(x)-3x)=6.1-3.2=0\)
\(\displaystyle \lim_{x \to 2}(x^{2}-5x+6)=0\)
Because \(\lim_{ x\to 2}\frac{6f(x)-3x}{x^{2}-5x+6}\) is of indeterminate form \(\frac{0}{0}\)
L’Hospital’s Rule can be applied.
Using L’Hospital’s Rule,
\(\displaystyle \lim_{x \to 2}\frac{6f(x)-3x}{x^{2}-5x+6}=\displaystyle \lim_{x \to 2}\frac{6f'(x) – 3}{2.2 – 5} = 3.\)
4(d) Find the absolute minimum value of f on the closed interval [−2, 8 .] Justify your answer.
f′(x ) = 0 ⇒ x =− 1, x = 2, x = 6
The function f is continuous on [−2, 8 ,] so the candidates for the location of an absolute minimum for f are x = −2, x = −1, x = 2, x = 6, and x = 8.
The absolute minimum value of f is f (2)= 1 .
Question 5
(a)-Topic-5.3 Determining Intervals on Which a Function Is Increasing or Decreasing
(b)-Topic-5.6 Determining Concavity of Functions over Their Domains
(c)-Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema
(d)-Topic-5.7 Using the Second Derivative Test to Determine Extrema
5. The functions f and g are twice differentiable. The table shown gives values of the functions and their first derivatives at selected values of x.
(a) Let h be the function defined by h(x) = f(g(x)). Find h'(7). Show the work that leads to your answer.
(b) Let k be a differentiable function such that k'(x) = \(\left ( f(x) \right )^{2}\) · g(x). Is the graph of k concave up or concave down at the point where x = 4 ? Give a reason for your answer.
(c) Let m be the function defined by m(x) = \(5x^{3}+\int_{0}^{x}f'(t)dt \). Find m(2). Show the work that leads to your answer.
(d) Is the function m defined in part (c) increasing, decreasing, or neither at x = 2 ? Justify your answer.
▶️Answer/Explanation
5(a) Let h be the function defined by h(x) = f(g(x)). Find h′(7 ). Show the work that leads to your answer.
\(h'(x) = f'(g(x)). g'(x)\)
\(h'(7) = f'(g(7)).g'(7)\)
\(= f'(0).8 = \frac{3}{2}.8 = 12\)
5(b) Let k be a differentiable function such that \(k'(x) = (f(x))^{2} .g(x)\). Is the graph of k concave up or concave down at the point where x = 4 ? Give a reason for your answer.
\(k”(x) = 2f(x) .f'(x) .g(x) + (f(x))^{2} .g'(x)\)
\(k”(4) = 2f(4) .f'(4) .g(4) + (f(4))^{2} .g'(4)\)
\(= 2.4.3.(-3) + 4^{2}.2 = -72 +32 = -40\)
The graph of k is concave down at the point where x = 4 because k′′(4 0 ) < and k′′ is continuous.
5(c) Let m be the function defined by \(m(x) = 5x^{3} + \int_{0}^{x} f'(t) dt\). Show the work that leads to your answer.
\(m(2) = 5.8 +\int_{0}^{2} f'(t) dt = 4= + (f(2) – f(0))\)
\(= 40 + (7-10) = 37\)
5(d) Is the function m defined in part (c) increasing, decreasing, or neither at x = 2 ? Justify your answer.
\(m'(x) = 15x^{2} + f'(x)\)
\(m'(2) = 15.4 + f'(2)= 60 +(-8) = 52\)
Question 6
(a)-Topic-7.2 Verifying Solutions for Differential Equations
(b)-Topic-7.3 Sketching Slope Fields
(c)-Topic-7.4 Reasoning Using Slope Fields
(d)-Topic-7.4 Reasoning Using Slope Fields
6. Consider the curve given by the equation \(6xy=2+y^{3}\) .
(a) Show that \(\frac{dy}{dx}=\frac{2y}{y^{2}-2x}\).
(b) Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists.
(c) Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists.
(d) A particle is moving along the curve. At the instant when the particle is at the point\(\left ( \frac{1}{2′}-2 \right )\),its horizontal position is increasing at a rate of \(\frac{dx}{dt}=\frac{2}{3}\) unit per second. What is the value of \(\frac{dy}{dt}\) , the rate of change of the particle’s vertical position, at that instant?
▶️Answer/Explanation
6(a) \(\frac{dy}{dx} = \frac{2y}{y^{2} – 2x}.\)
\(\frac{d}{dx}(6xy) = \frac{d}{dx}(2+y^{3}) \Rightarrow 6y +6x\frac{dy}{dx} = 3y^{2}\frac{dy}{dx}\)
\(\Rightarrow 2y = \frac{dy}{dx}(y^{2} -2x) \Rightarrow \frac{dy}{dx} = \frac{2y}{y^{2}- 2x}\)
6(b) Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists.
For the line tangent to the curve to be horizontal, it is necessary that 2 0 y = (so y = 0 ) and that \(y^{2} – 2x \neq 0\)
Substituting y = 0 into \(6xy= 2-y^{3}\) yields the equation \(6x .0, x ⋅ = 2\) which has no solution.
Therefore, there is no point on the curve at which the line tangent to the curve is horizontal.
6(c) Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists.
For a line tangent to this curve to be vertical, it is necessary that \(2y\neq 0\) and that \(y^{2} – 2x = 0 (so x = \frac{y^{2}}{2})\).
Substituting \(x = \frac{y^{2}}{2}\) into \(6xy= 2-y^{3}\) yields the equation \(3y^{2} .y = 2 + y^{3} \Rightarrow 2y^{3} = 2\Rightarrow y = 1.\)
Substituting y = 1 in \(6xy= 2-y^{3}\) yields 6x = 2+1, or x \(\frac{1}{2}\). The tangent line to the curve is vertical at the point \((\frac{1}{2}, 1)\).
6(d) A particle is moving along the curve. At the instant when the particle is at the point \((\frac{1}{2}, -2)\), its horizontal position is increasing at a rate of \(\frac{dx}{dt}\) = \(\frac{2}{3}\) unit per second. What is the value of \(\frac{dy}{dt}\), the rate of change of the particle’s vertical position, at that instant?
\(6y\frac{dx}{dt} + 6x\frac{dy}{dt} = 0 + 3y^{2}\frac{dy}{dt}\)
At the point (x, y) = \((\frac{1}{2}, -2)\),
\(6(-2)(\frac{2}{3}) + 6(\frac{1}{2})\frac{dy}{dt} = 3(-2)^{2} \frac{dy}{dt}\)
\(\Rightarrow -8 + 3\frac{dy}{dt} = 12\frac{dy}{dt}\)
\(\Rightarrow \frac{dy}{dt} = -\frac{8}{9}\) unit per second