Question 1
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 \(g(t)=(\frac{1}{500})cos\left ( (\frac{t}{120}^{2}) \right )\) 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 \(g(t)=(\frac{1}{500})cos\left ( (\frac{t}{120}^{2}) \right )\) for\\\ \(0\leq t\leq 150\) Using this model, find the average rate of flow of gasoline over the time interval 0 ≤ 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.
\(g'(140)\approx -0.004908\)
\(g'(140)=-0.005(or-0.004) \)
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
2. For \(0\leq t\leq \pi \), a particle is moving along the curve shown so that its position at time t is ( x(t) , ( t)y), where x( t) is not explicitly given and y(t ) = 2 sin t. It is known that \(\frac{dx}{dt}=e^{cos\\\ t}\). At time t = 0, the particle is at position (1, 0)
(a) Find the acceleration vector of the particle at time t = 1. Show the setup for your calculations.
(b) For \(0\leq t\leq \pi \), 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 of the particle 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 the time interval \(0\leq t\leq \pi \). Show the setup for your calculations.
▶️Answer/Explanation
2(a) Find the acceleration vector of the particle at time t = 1. Show the setup for your calculations.
\(x”(1)=\frac{d}{dt}(e^{cos t})|_{t=1}=-1.444407\)
\(y(t) =2sint\Rightarrow y'(t)=2cos t\)
\(y” (1)=\frac{d}{dt}(2cos t)|_{t=1}=-1.682942\)
The acceleration vector at time t = 1 is
\(a(1)=\left< -1.444,-1.683(or-1.682)\right>\)
2(b) For 0 ≤ t≤ π, find the first time t at which the speed of the particle is 1.5. Show the work that leads to your answer.
Speed =\(\sqrt{\left ( \frac{dx}{dt}\right )^{2}+\left ( \frac{dy}{dt} \right )^{2}}=\sqrt{(e^{cos t})+(2cos t)^{2}}\)
\(0\leq t\leq \pi and \sqrt{(e^{cos t})+(2cos t)^{2}}=1.5\)
\(\Rightarrow t=1.254472,t=2.358077\)
The first time at which the speed of the particle is 1.5 is t = 1.254.
2(c) Find the slope of the line tangent to the path of the particle 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.
\(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2cos t}{e^{cos t}}\)
\(\frac{dy}{dx}|_{t=1}=\frac{2 cos 1}{e^{cos 1}} =0.629530\)
The slope of the line tangent to the curve at t = 1 is 0.630 (or 0.629 ).
\(x(1)=x(0)+\int_{0}^{1}\frac{dx}{dt}dt=1+\int_{0}^{1}e^{cos t}dt=3.341575\)
The x -coordinate of the position at t = 1 is 3.342 (or 3.341).
2(d) Find the total distance traveled by the particle over the time interval 0 ≤ t≤ π. Show the setup for your calculations.
\(\int_{0}^{\pi }\sqrt{(e^{cos t})^{2}+(2 cos t)^{2}} dt\)
= 6.034611
The total distance traveled by the particle over 0 ≤ t≤ π is 6.035 (or 6.034 ).
Question 3
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\(\frac{d^{2}M}{dt^{2}}\) 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.
\(\frac{d^{2}M}{dt^{2}}=\frac{1}{4}\frac{dM}{dt}=-\frac{1}{4}\left ( \frac{1}{4}(40-M) \right )=-\frac{1}{16}(40-M)\)
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).
Question 4
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 \(\displaystyle \lim_{x \to 2}\frac{6f(x)-3x}{x^{2}-5x+6}\),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 \(\displaystyle \lim_{x \to 2}\frac{6f(x)-3x}{x^{2}-5x+6}\) is of indeterminate form \(\frac{0}{0}\),
L’Hospital’s Rule can be applied.
Using L’Hospital’s Rule,
\(\displaystyle \lim_{x \to 2}\frac{6f(x)-3x}{x^{2}-5x+6} =\displaystyle \lim_{x \to 2}\frac{6f'(x)-3}{2x-5}=\frac{6.0-3}{2.2-5}=3\)
4(d) Find the absolute minimum value of f on the closed interval [−2, 8 .] Justify your answer.
f ‘(x ) = 0 ⇒ =− 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
5. The graphs of the functions f and g are shown in the figure for \(0\leq x\leq 3\). It is known that \(g(x)=\frac{12}{3+x}\) for x ≥ 0. The twice-differentiable function f , which is not explicitly given, satisfies f(3) = 2 and \(\int_{0}^{3}f(x)dx=10\) .
(a) Find the area of the shaded region enclosed by the graphs of f and g.
(b) Evaluate the improper integral \(\int_{0}^{\infty }(g(x))^{2}dx\), or show that the integral diverges.
(c) Let h be the function defined by h(x ) = x . f ‘(x). Find the value of \(\int_{0}^{3}h(x)dx\).
▶️Answer/Explanation
5(a) Find the area of the shaded region enclosed by the graphs of f and g.
\(= \int_{0}^{3}(f(x) – g(x))dx = \int_{0}^{3} f(x)dx -\int_{0}^{3}g(x)dx\)
\(= 10 -\int_{0}^{3} \frac{12}{3+x}dx =10-12[In|3+x|]^{3}_{0}\)
\(= 10-12(In\\\ 6 – In\\\ 3) =10-12(In\\\ 2)\)
5(b) Evaluate the improper integral \(\int_{0}^{\infty } (g(x))^{2} dx\), or show that the integral diverges.
\(\int_{0}^{\infty }(g(x))^{2} dx =\displaystyle \lim_{b \to \infty }\int_{0}^{b}\frac{144}{(3+x)^{2}}dx\)
\(= \displaystyle \lim_{b \to \infty } (-\frac{144}{3+x}|^{b}_{0})\)
\(=\displaystyle \lim_{b \to \infty } (-\frac{144}{3+b} + \frac{144}{3}) = 48\)
5(c) Let h be the function defined by h(x) = x . f′(x ). Find the value of \(\int_{0}^{3}h(x)dx.\)
Question 6
6. The function f has derivatives of all orders for all real numbers. It is known that f(0) = 2, f ¢(0) = 3,\(f”(x)=-f(x^{2}),and\\\ (x)=-2x.f'(x^{2})\)
(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 \(\left | f^{(5)}(x)\right |\leq 15 for\\\ 0\leq x\leq 0.5\), use the Lagrange error bound to show that this approximation is within \(\frac{1}{10^{5}}\) 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.
▶️Answer/Explanation
6(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.
6(b) The fourth-degree Taylor polynomial for f about x = 0 is used to approximate f(0.1). Given that \(|f^{5}(x)|\leq 15 for 0 \leq x\leq 0.5,\) use the Lagrange error bound to show that this approximation is within \(\frac{1}{10^{5}}\) of the exact value of f(0.1).
By the Lagrange error bound,
6(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.
\(g”(x)=e^{x}.f(x)+e^{x}.f'(x)\)
\(g'(0) =e^{0}.f(0)=2\)
\(g”(0) =e^{0}.f(0)+e^{0}.f'(0)=2+3=5\)
The second-degree Taylor polynomial for g about x = 0 is \(T_{2}(x) =4+2x+\frac{5}{2}x^{2}\)