Question 1
1. From 5 A.M. to 10 A.M., the rate at which vehicles arrive at a certain toll plaza is given by \(A(t)=450\sqrt{sin(0.62t)}\), where t is the number of hours after 5 A.M. and A (t) is measured in vehicles per hour. Traffic is flowing smoothly at 5 A.M. with no vehicles waiting in line.
(a) Write, but do not evaluate, an integral expression that gives the total number of vehicles that arrive at the toll plaza from 6 A.M. (t = 1) to 10 A.M. (t = 5).
(b) Find the average value of the rate, in vehicles per hour, at which vehicles arrive at the toll plaza from 6 A.M. (t = 1) to 10 A.M. (t = 5).
(c) Is the rate at which vehicles arrive at the toll plaza at 6 A.M. (t = 1) increasing or decreasing? Give a reason for your answer.
(d) A line forms whenever A ( t) ≥ 400. The number of vehicles in line at time t, for a \(a\leq t\leq 4\), is given by \(N(t)=\int_{a}^{t}(A(x)-400)dx\) , where a is the time when a line first begins to form. To the nearest whole number, find the greatest number of vehicles in line at the toll plaza in the time interval a \(a\leq t\leq 4\). Justify your answer.
▶️Answer/Explanation
1(a) Write, but do not evaluate, an integral expression that gives the total number of vehicles that arrive at the toll plaza from 6 A.M. ( t = 1) to 10 A.M. ( t = 5 ).
The total number of vehicles that arrive at the toll plaza from
6 A.M. to 10 A.M. is given by \(\int_{1}^{5}A(t)dt\).
1(b) Find the average value of the rate, in vehicles per hour, at which vehicles arrive at the toll plaza from
6 A.M. ( t = 1) to 10 A.M. ( t = 5 ).
Average =\(\frac{1}{5-1}\int_{1}^{5} A(t) dt\) = 375.536966 5
The average rate at which vehicles arrive at the toll plaza from
6 A.M. to 10 A.M. is 375.537 (or 375.536 ) vehicles per hour.
1(c) Is the rate at which vehicles arrive at the toll plaza at 6 A.M. ( t = 1) increasing or decreasing? Give a reason for your answer.
A′(1) = 148.947272
Because A′(1) > 0, the rate at which the vehicles arrive at the toll plaza is increasing.
1(d) A line forms whenever A(t) ≥ 400. The number of vehicles in line at time t, for a ≤ t≤ 4, is given by \(N(t)=\int_{a}^{t}(A(x)-400)dx\), where a is the time when a line first begins to form. To the nearest whole number, find the greatest number of vehicles in line at the toll plaza in the time interval a ≤ t≤ 4.Justify your answer.
N′(t) = A(t) − 400 = 0
⇒ A(t) = 400 ⇒ = t 1.469372, t = 3.597713
a = 1.469372
b = 3.597713
The greatest number of vehicles in line is 71.
Question 2
2. Let f and g be the functions defined by f( x) = ln(x + 3) and \(g(x)=x^{4}+2x^{3}\) . The graphs of f and g, shown in the figure above, intersect at x = −2 and x = B, where B > 0.
(a) Find the area of the region enclosed by the graphs of f and g.
(b) For \(-2\leq x\leq B\), let h (x) be the vertical distance between the graphs of f and g. Is h increasing or decreasing at x = −0.5 ? Give a reason for your answer.
(c) The region enclosed by the graphs of f and g is the base of a solid. Cross sections of the solid taken perpendicular to the x-axis are squares. Find the volume of the solid.
(d) A vertical line in the xy-plane travels from left to right along the base of the solid described in part (c). The vertical line is moving at a constant rate of 7 units per second. Find the rate of change of the area of the cross section above the vertical line with respect to time when the vertical line is at position x = −0.5
▶️Answer/Explanation
2(a) Find the area of the region enclosed by the graphs of f and g.
\(In(x+3)=x^{4}+2x^{3}\Rightarrow x=-2,x=B=0.781975\)
\(\int_{-2}^{B}(f(x)-g(x))dx =3.603548\)
The area of the region is 3.604 (or 3.603 ).
2(b) For −2 ≤ x ≤ B , let h (x ) be the vertical distance between the graphs of f and g. Is h increasing or decreasing at x = −0.5 ? Give a reason for your answer.
h (x ) = f ( x) − g (x )
h′(x ) = f ′( x) − g′( x)
h′(−0.5) = f ′(−0.5) − g′(−0.5) = −0.6 (or −0.599 )
Since h′(−0.5) < 0, h is decreasing at x = −0.5.
2(c) The region enclosed by the graphs of f and g is the base of a solid. Cross sections of the solid taken perpendicular to the x -axis are squares. Find the volume of the solid.
\(\int_{-2}^{B}(f(x)-g(x))^{2}dx =5.340102\)
The volume of the solid is 5.340.
2(d) A vertical line in the xy -plane travels from left to right along the base of the solid described in part (c).The vertical line is moving at a constant rate of 7 units per second. Find the rate of change of the area of the cross section above the vertical line with respect to time when the vertical line is at position x = −0.5.
The cross section has area \(A(x)=(f(x)-g(x))^{2}\).
\(\frac{d}{dt}\left [ A(x) \right ]=\frac{dA}{dx}.\frac{dx}{dt}\)
\(\frac{d}{dt}\left [ A(x) \right ]|_{x=-0.5}=A'(-0.5).7 = -9.271842\)
At x = −0.5, the area of the cross section above the line is changing at a rate of −9.272 (or −9.271) square units per second.
Question 3
3. Let f be a differentiable function with f(4) = 3. On the interval \(0\leq x\leq 7\), the graph of f ‘, the derivative of f , consists of a semicircle and two line segments, as shown in the figure above.
(a) Find f( 0) and f(5).
(b) Find the x-coordinates of all points of inflection of the graph of f for \(0\leq x\leq 7\). Justify your answer.
(c) Let g be the function defined by g( x) = f(x ) − x. On what intervals, if any, is g decreasing for \(0\leq x\leq 7\) ? Show the analysis that leads to your answer.
(d) For the function g defined in part (c), find the absolute minimum value on the interval \(0\leq x\leq 7\). Justify your answer.
▶️Answer/Explanation
3(a) Find f (0 ) and f (5).
\(f(o)=f(4)+\int_{4}^{0}f'(x)dx=3-\int_{0}^{4}f'(x)dx=3+2\pi\)
\(f(5)=f(4)+\int_{4}^{5}f'(x)dx= 3+\frac{1}{2}=\frac{7}{2}\)
3(b) Find the x -coordinates of all points of inflection of the graph of f for 0 < x < 7. Justify your answer.The graph of f has a point of inflection at each of x = 2 and x = 6, because f ′( x) changes from decreasing to increasing at x = 2 and from increasing to decreasing at x = 6.
3(c) Let g be the function defined by g( x) = f ( x) − x. On what intervals, if any, is g decreasing for 0 ≤ x ≤ 7 ? Show the analysis that leads to your answer.
g′(x ) = f ′( x) − 1
f ′( x) − ≤ 10 ⇒ f ′( x) ≤ 1
The graph of g is decreasing on the interval 0 ≤ x ≤ 5 because g′(x ) ≤ 0 on this interval.
3(d) For the function g defined in part (c), find the absolute minimum value on the interval 0 ≤ x ≤ 7. Justify your answer.
g is continuous, g′(x ) < 0 for 0 < x < 5, and g′(x ) > 0 for 5 < x < 7.
Therefore, the absolute minimum occurs at x = 5, and \(g(5)=f(5)-5=\frac{7}{2}-5=-\frac{3}{2}\) is the absolute minimum value of g.
Question 4
4. An ice sculpture melts in such a way that it can be modeled as a cone that maintains a conical shape as it decreases in size. The radius of the base of the cone is given by a twice-differentiable function r, where r t( t ) is measured in centimeters and t is measured in days. The table above gives selected values of r'(t), the rate of change of the radius, over the time interval \(0\leq t\leq 12\).
(a) Approximate r “(8.5) using the average rate of change of r’ over the interval \(7\leq t\leq 10\). Show the computations that lead to your answer, and indicate units of measure.
(b) Is there a time t, \(0\leq t\leq 3\), for which r ‘(t ) = −6 ? Justify your answer.
(c) Use a right Riemann sum with the four subintervals indicated in the table to approximate the value of \(\int_{0}^{12}r'(t)dt\).
(d) The height of the cone decreases at a rate of 2 centimeters per day. At time t = 3 days, the radius is 100 centimeters and the height is 50 centimeters. Find the rate of change of the volume of the cone with respect to time, in cubic centimeters per day, at time t = 3 days. (The volume V of a cone with radius r and height h is \(V=\frac{1}{3}\pi r^{2}h .\)
▶️Answer/Explanation
4(a) Approximate r′′(8.5) using the average rate of change of r′ over the interval 7 ≤ t≤ 10 . Show the computations that lead to your answer, and indicate units of measure.
\(r”(8.5)\approx \frac{r'(10)-r'(7)}{10-7}=\frac{-3.8-(-4.4)}{10-7}\)
\(=\frac{0.6}{3}=0.2\) centimeter per day per day
4(b) Is there a time t, 0 ≤ t≤ 3, for which r ′(t ) = −6? Justify your answer.
r (t ) is twice-differentiable. ⇒ r′( t) is differentiable.
⇒ r′( t) is continuous.
r′(0 ) = −6.1 < −6 < −5.0 = r′(3 )
Therefore, by the Intermediate Value Theorem, there is a time t, 0 < t< 3, such that r′( t) = −6.
4(c) Use a right Riemann sum with the four subintervals indicated in the table to approximate the value of
\(\int_{0}^{12} r′(t) dt\).
\(\int_{0}^{12} r′(t) dt ≈ 3r′(3) + 4r′(7) + 3r′(10) + 2r′(12) \)
= 3(−5.0) + 4(−4.4) + 3(−3.8) + 2(−3.5)
= −51
4(d) The height of the cone decreases at a rate of 2 centimeters per day. At time t = 3 days, the radius is 100 centimeters and the height is 50 centimeters. Find the rate of change of the volume of the cone with
respect to time, in cubic centimeters per day, at time t = 3 days. (The volume V of a cone with radius r and height h is \(V=\frac{1}{3}\pi r^{2}h\) .)
\(\frac{dV}{dt}=\frac{2}{3}\pi rh\frac{dr}{dt}+\frac{1}{3}\pi r^{2}\frac{Dh}{dt}\)
\(\frac{dV}{dt}|_{t=3}=\frac{2}{3}\pi (100)(50)(-5)+\frac{1}{3}\pi (100)^{2}(-2)=-\frac{70,000\pi }{3}\)
The rate of change of the volume of the sculpture at t = 3 is approximately \(-\frac{70,000\pi }{3}\)cubic centimeters per day.
Question 5
5. Consider the differential equation \(\frac{dy}{dx}=\frac{1}{2}sin\left ( \frac{\pi }{2}x \right )\sqrt{y+7}.\) Let y = f ( x) be the particular solution to the differential equation with the initial condition f( 1) = 2. The function f is defined for all real numbers.
(a) A portion of the slope field for the differential equation is given below. Sketch the solution curve through the point (1, 2).
(b) Write an equation for the line tangent to the solution curve in part (a) at the point (1, 2). Use the equation to approximate f(0.8).
(c) It is known that f “( x) > 0 for \(-1\leq x\leq 1\). Is the approximation found in part (b) an overestimate or an underestimate for f(0.8) ? Give a reason for your answer.
(d) Use separation of variables to find y = f(x), the particular solution to the differential equation \(\frac{dy}{dx}=\frac{1}{2}sin\left ( \frac{\pi }{2}x \right )\sqrt{y+7}\) with the initial condition f( 1) = 2.
▶️Answer/Explanation
5(a) A portion of the slope field for the differential equation is given below. Sketch the solution curve through the point (1, 2).
5(b) Write an equation for the line tangent to the solution curve in part (a) at the point (1, 2). Use the equation to approximate f (0.8).
\(\frac{dy}{dx}|_{(x,y) = (1,2)} = \frac{1}{2}.3 sin(\frac{\pi }{2}) = \frac{3}{2}\)
An equation for the tangent line is\(y=2+\frac{3}{2}(x-1)\).
\(f(0.8) \approx 2+\frac{3}{2}(0.8-1)=1.7\)
5(c) It is known that f ′′(x ) > 0 for −1 ≤ x ≤ 1. Is the approximation found in part (b) an overestimate or an underestimate for f (0.8) ? Give a reason for your answer. Because f ′′( x) > 0, f is concave up on −1 ≤ x ≤ 1, the tangent line lies below the graph of y = f (x ) at x = 0.8, and the approximation for f (0.8) is an underestimate.
5(d) Use separation of variables to find y = f ( x), the particular solution to the differential equation \(\frac{dy}{dx}=\frac{1}{2}sin(\frac{\pi }{2}x)\sqrt{y+7}\) with the initial condition f ( 1) = 2.
Question 6
6. Particle P moves along the x-axis such that, for time t > 0, its position is given by \(x_{p}(t)= 6-4e^{-t}\). Particle Q moves along the y-axis such that, for time t > 0, its velocity is given by \(v_{Q}(t)=\frac{1}{t^{2}}\) . At time t = 1,the position of particle Q is \(y_{Q}(1)=2\) .
(a) Find \(v_{p}(t)\), the velocity of particle P at time t.
(b) Find \(a_{Q}(t)\) , the acceleration of particle Q at time t. Find all times t, for t > 0, when the speed of particle Q is decreasing. Justify your answer.
(c) Find \(y_{Q}(t)\), the position of particle Q at time t.
(d) As t → ∞, which particle will eventually be farther from the origin? Give a reason for your answer.
▶️Answer/Explanation
6(a) Find \(v_{p}(t)\) , the velocity of particle P at time t.
\(v_{p}(t)=x_{p}'(t)=4e^{-t}\)
6(b) Find \(a_{Q}(t)\), the acceleration of particle Q at time t. Find all times t, for t > 0, when the speed of particle Q is decreasing. Justify your answer.
\(a_{Q}(t)=v_{Q}'(t)=\frac{-2}{t^{3}}\)
for t > 0, \(a_{Q}(t)\) < 0 and \( v_{Q}(t)\) > 0.
Because the velocity and acceleration have opposite signs, the speed of particle Q is decreasing for all t > 0.
6(c) Find \( y_{Q}(t)\) , the position of particle Q at time t.
\(y_{Q}(t)=y_{Q}(1)+\int_{1}^{t} \frac{1}{s^{2}}ds\)
\(=2-\left ( \frac{1}{s}|_{1}^{t} \right )=2-\frac{1}{t}+1=3-\frac{1}{t} \)
6(d) As t → ∞, which particle will eventually be farther from the origin? Give a reason for your answer.
For particle P,\(\displaystyle \lim_{t \to \infty } (6-4e^{-t})=6\).
For particle Q, \(\displaystyle \lim_{t \to \infty } (3-\frac{1}{t})=3\).
Because 6 > 3, particle P will eventually be farther from the origin.