Question
Researchers on a boat are investigating plankton cells in a sea. At a depth of h meters, the density of plankton cells, in millions of cells per cubic meter, is modeled by \(p(h)=0.2h^{2}e^{-0.0025h^{2}}\) for 0 ≤ h ≤ 30 and is modeled by f(h) for h ≥ 30. The continuous function f is not explicitly given.
(a) Find p'(25 ). Using correct units, interpret the meaning of p ‘(25) in the context of the problem.
(b) Consider a vertical column of water in this sea with horizontal cross sections of constant area 3 square meters. To the nearest million, how many plankton cells are in this column of water between h = 0 and h = 30 meters?
(c) There is a function u such that 0 ≤ f(h) ≤ u(h) for all h ≥ 30 and \(\int_{30}^{\infty }u(h)dh=105.\) The column of water in part (b) is K meters deep, where K > 30. Write an expression involving one or more integrals that gives the number of plankton cells, in millions, in the entire column. Explain why the number of plankton cells in the column is less than or equal to 2000 million.
(d) The boat is moving on the surface of the sea. At time t ≥ 0, the position of the boat is (x(t), y(t)), where x'(t)= 662 sin (5t) and y ‘ (t) = 880 cos (6t). Time t is measured in hours, and x(t) and y (t) are measured in meters. Find the total distance traveled by the boat over the time interval 0 ≤ t ≤ 1.
Answer/Explanation
Ans:
(a) p′(25) = – 1.179
At a depth of 25 meters, the density of plankton cells is changing at a rate of −1.179 million cells per cubic meter per meter.
(b) \(\int_{0}^{30}3p(h)dh=1675.414936\)
There are 1675 million plankton cells in the column of water between h = 0 and h = 30 meters.
(c) \(\int_{30}^{k}3f(h)dh\) represents the number of plankton cells, in millions, in the column of water from a depth of 30 meters to a depth of K meters. The number of plankton cells, in millions, in the entire column of water is given by \(\int_{0}^{30}3p(h)dh+\int_{30}^{k}3f(h)dh.\)
Because 0 ≤ f(h) ≤ u(h) for all h ≥30,
\(3\int_{30}^{k}f(h)dh\leq 3\int_{30}^{k}u(h)dh\leq 3\int_{30}^{\infty }=3\cdot 105=315.\)
The total number of plankton cells in the column of water is bounded by 1675.415 +315 =1990.415 ≤ 2000 million.
(d) \(\int_{0}^{1}\sqrt{(x'(t))^{2}+(y'(t))}dt=757.455862\)
The total distance traveled by the boat over the time interval 0 ≤ t ≤ 1 is 757.456 (or 757.455) meters.
Question
People enter a line for an escalator at a rate modeled by the function r given by
\(r(t)=\left\{\begin{matrix}
44 \left ( \frac{t}{100} \right )^{3}\left ( 1-\frac{t}{300} \right )^{7}&~for~ 0\leq t\leq 300 & \\ 0
&~for~ t> 300 &
\end{matrix}\right.\)
where r (t) is measured in people per second and t is measured in seconds. As people get on the escalator, they exit the line at a constant rate of 0.7 person per second. There are 20 people in line at time t = 0.
(a) How many people enter the line for the escalator during the time interval 0 ≤ t ≤ 300 ?
(b) During the time interval 0 ≤ t ≤ 300, there are always people in line for the escalator. How many people are in line at time t = 300 ?
(c) For t > 300, what is the first time t that there are no people in line for the escalator?
(d) For 0 ≤ t ≤ 300, at what time t is the number of people in line a minimum? To the nearest whole number, find the number of people in line at this time. Justify your answer.
Answer/Explanation
Ans:
(a) \(\int_{0}^{300}r(t)dt=270\)
According to the model, 270 people enter the line for the escalator during the time interval 0 ≤ t ≤ 300.
(b) \(20+\int_{0}^{300}(r(t)-0.7)dt=20+\int_{0}^{300}r(t)dt-0.7\cdot 300=80\)
According to the model, 80 people are in line at time t = 300.
(c) Based on part (b), the number of people in line at time t = 300 is 80. The first time t that there are no people in line is
\(300+\frac{80}{0.7}=414.286 (or 414.285)seconds.\)
(d) The total number of people in line at time t, 0 ≤ t ≤ 300 is modeled by
\(20+\int_{0}^{t}r(x)dx-0.7t.\)
r(t) – 0.7 = 0 ⇒ t1 = 33.013298, t2 = 166.574719
| t | People in line for escalator |
0 t1 t2 300 | 20 3.803 158.070 80 |
The number of people in line is a minimum at time t = 33.013 seconds, when there are 4 people in line.
Question
| 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
Ans:
(a) \(R'(2)\approx \frac{R(3)-R(1)}{3-1}=\frac{950-1190}{3-1}=-120 liters/hr^{2}\)
(b) The total amount of water removed is given by \(\int_{0}^{8}R(t)dt.\)
\(\int_{0}^{8}R(t)dt\approx 1\cdot R(0)+2\cdot R(1)+3\cdot R(3)+2\cdot R(6)\)
= 1(1340) + 2(1190) + 3(950) +2(740)
=8050 liters
This is an overestimate since R is a decreasing function.
(c)
\(Total \approx 50000+\int_{0}^{8}W(t)dt-8050\)
=50000 + 7836.195325 – 8050 ≈ 49786 liters
(d) W(0) – R(0) > 0, W(8) – R(8) < 0, and W(t) – R(t) is continuous.
Therefore, the Intermediate value Theorem guarantees at least one time t, 0 < t < 8, for which W(t) – R(t) = 0, or W(t) = R(t).
For this value of t, 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
Grass clippings are placed in a bin, where they decompose. For 0 ≤ t ≤ 30 , the amount of grass clippings remaining in the bin is modeled by A(t) = 6.687 (0.931)t , where A(t) is measured in pounds and t is measured in days.
(a) Find the average rate of change of A(t) over the interval 0 ≤ t ≤ 30. Indicate units of measure.
(b) Find the value of A'(15) . Using correct units, interpret the meaning of the value in the context of the problem.
(c) Find the time t for which the amount of grass clippings in the bin is equal to the average amount of grass clippings in the bin over the interval 0 ≤ t ≤ 30 .
(d) For t >30, L (t), the linear approximation to A at t =30, is a better model for the amount of grass clippings remaining in the bin. Use L (t) to predict the time at which there will be 0.5 pound of grass clippings remaining in the bin. Show the work that leads to your answer.
Answer/Explanation
Ans:
(a) \(\frac{A(30)-A(0)}{30-0}=-0.197 (or-0.196) lbs/day)\)
(b) A'(15) = -0.164 (or – 0.163)
The amount of grass clippings in the bin is decreasing at a rate of 0.164 (or 0.163) lbs/day at time t = 15 days.
(c) \(A(t)=\frac{1}{30}\int_{0}^{30}A(t)dt\Rightarrow t=12.415 (or 12.414)\)
(d) L(t) = A(30) + A'(30) • (t-30)
A'(30) = – 0.055976
A(30) = 0.782928
L(t) = 0.5 ⇒ t = 35.054
Question
Let \(g(t)=\int_{0}^{t}f(x)dx\) , and consider the graph of f shown below.
(A) Evaluate g(0), g(2), and g(6).
(B) On what interval(s) is g increasing (if any)? Justify your answer.
(C) At what value(s) of t does g have a minimum value? Justify your answer.
(D) On what interval(s) is g concave down? Justify your answer.
Answer/Explanation
(A)\(g(0)=\int_{0}^{0}f(x)dx=0\)
\(g(2)=\int_{0}^{2}f(x)dx=\int_{0}^{2}(4-4x)dx=\left [ 4x-2x^{2} \right ]^{2}_{0}=0\)
\(g(6)=\int_{0}^{6}f(x)dx=\int_{0}^{2}(4-4x)dx+\int_{3}^{2}(2x-8)dx+\int_{3}^{5}(4x-14)dx+\int_{5}^{6}6dx\)
\(=0+(-3)+4+6=7\)
(B) g is increasing on \([0,1)\cup (\frac{7}{2},6]\),since g'(t)=f(t)>0 on these intervals. Note that \(\frac{7}{2}\) is the solution to 4x -14 = 0.
(C) At t=7/2,g has a minimum value. g(0)=0, g(7/2)=-7/2, g(6)=7
(D) Since g′(t) = f(t) is decreasing only on (0, 2), you see that g′′(x)<0 on this interval. Therefore, g is concave down only on (0, 2).
Question
Let f(x) = e2x . Let R be the region in the first quadrant bounded by the graph of f, the coordinate axes, and the vertical line x = k, where k > 0. The region R is shown in the figure above.
(a) Write, but do not evaluate, an expression involving an integral that gives the perimeter of R in terms of k.
(b) The region R is rotated about the x-axis to form a solid. Find the volume, V, of the solid in terms of k.
(c) The volume V, found in part (b), changes as k changes. If \(\frac{dk}{dt}=\frac{1}{3}, determine \frac{dV}{t}when k=\frac{1}{2}.\)
Answer/Explanation
Ans:
(a)
\(\frac{dy}{dx}=2e^{2x}\)
\(p = 1 + k + e^{2x}+ \int_{0}^{k}\sqrt{1+(2e^{2x})^{2}}dx\)
(b)
u = 4x
\(\frac{du}{dx}=4\)
\(v = \int_{0}^{k}\pi r^{2}dx=\int_{0}^{k}\pi (e^{2x})^{2}dx=\pi \int_{0}^{k}e^{4x}dx\)
\(= \pi \int_{0}^{k}\frac{1}{4}e^{4}\cdot du=\frac{\pi }{4}\left [ e^{4x} \right ]_{0}^{k}=\frac{\pi }{4}\left ( e^{4k} -e^{4(0)}\right )\)
\(=\frac{\pi }{4}e^{4k}-1\left ( \frac{\pi }{4} \right )\)
\(v=\frac{\pi }{4}e^{4k}- \frac{\pi }{4}\)
(c)
\(\frac{dV}{dt}=\frac{dV}{dk}\cdot \frac{dk}{dt}=\frac{d}{dk}\left ( \frac{\pi }{4}e^{4k}-\frac{\pi }{4} \right )\cdot \left ( \frac{1}{3} \right )=\frac{\pi }{4}(4)e^{4k}\cdot \left ( \frac{1}{3} \right )\)
\(\frac{dV}{dt}=\frac{\pi }{3}e^{4k}|_{k=\frac{1}{2}}=\frac{\pi }{3}\cdot e^{2}\)
Question
| t (minutes) | 0 | 4 | 9 | 15 | 20 |
| W(t) (degrees Fahrenheit) | 55.0 | 57.1 | 61.8 | 67.9 | 71.0 |
The temperature of water in a tub at time t is modeled by a strictly increasing, twice-differentiable function W, where W(t) is measured in degrees Fahrenheit and t is measured in minutes. At time t =0, the temperature of the water is 55 0F. The water is heated for 30 minutes, beginning at time t = 0. Values of W (t) at selected times t for the first 20 minutes are given in the table above.
(a) Use the data in the table to estimate W'(12) . Show the computations that lead to your answer. Using correct units, interpret the meaning of your answer in the context of this problem.
(b) Use the data in the table to evaluate \(\int_{0}^{20}W'(t)dt.\) Using correct units, interpret the meaning of \(\int_{0}^{20}W'(t)dt\) in the context of this problem.
(c) For 0 ≤ t ≤ 20, the average temperature of the water in the tub is \(\frac{1}{20}\int_{0}^{20}W'(t)dt.\) Use a left Riemann sum with the four subintervals indicated by the data in the table to approximate \(\frac{1}{20}\int_{0}^{20}W'(t)dt.\) Does this approximation overestimate or underestimate the average temperature of the water over these 20 minutes? Explain your reasoning.
(d) For 20 ≤ t ≤ 25, the function W that models the water temperature has first derivative given by \(W'(t)=0.4\sqrt{t}cos (0.06t).\) Based on the model, what is the temperature of the water at time t=25 ?
Answer/Explanation
Ans:
(a) \(W'(12)\approx \frac{W(15)-W(9)}{15-9}=\frac{67.9-61.8}{6}\)
= 1.017 (or 1.016)
The water temperature is increasing at a rate of approximately 1.017 F° per minute at time t = 12 minutes.
(b) \(\int_{0}^{20}W'(t)dt=W(20)-W(0)=71.0-55.0=16\)
The water has warmed by 16 F° over the interval from t = 0 to t = 20 minutes.
(c) \(\frac{1}{20}\int_{0}^{20}W(t)dt\approx \frac{1}{20}(4\cdot W(0)+5\cdot W(4)+6\cdot W(9)+5\cdot W(15))\)
\(= \frac{1}{20}(4\cdot 55.0+5\cdot 57.1+6\cdot 61.8+5\cdot 67.9)\)
\(= \frac{1}{20}\cdot 1215.8=60.79\)
This approximation is an underestimate, because a left Riemann sum is used and the function W is strictly increasing.
(d) \(W(25)=71.0+\int_{20}^{25}W'(t)dt\)
=71.0+2.043155 = 73.043
Question
There is no snow on Janet’s driveway when snow begins to fall at midnight. From midnight to 9 A. M., snow accumulates on the driveway at a rate modeled by $f(t ) = 7te^{cost}$ cubic feet per hour, where t is measured in hours since midnight. Janet starts removing snow at 6 A. M. ( t =6) . The rate g( t), in cubic feet per hour, at which Janet removes snow from the driveway at time t hours after midnight is modeled by
(a) How many cubic feet of snow have accumulated on the driveway by 6 A. M.?
(b) Find the rate of change of the volume of snow on the driveway at 8 A. M.
(c) Let h(t ) represent the total amount of snow, in cubic feet, that Janet has removed from the driveway at time t hours after midnight. Express h as a piecewise-defined function with domain \(0\leq t\leq 9\)
(d) How many cubic feet of snow are on the driveway at 9 A. M.?
Answer/Explanation
Ans:
Ans:
(a) $\int_0^6 f(t) d t=142.274$ or $142.275$ cubic feet
(b) Rate of change is $f(8)-g(8)=-59.582$ or $-59.583$ cubic feet per hour.
(c) $h(0)=0$
For $0<t \leq 6, h(t)=h(0)+\int_0^t g(s) d s=0+\int_0^t 0 d s=0$
For $6<t \leq 7, h(t)=h(6)+\int_6^t g(s) d s=0+\int_6^t 125 d s=125(t-6)$.
For $7<t \leq 9, \quad h(t)=h(7)+\int_7^t g(s) d s=125+\int_7^t 108 d s=125+108(t-7)$
Thus, $h(t)= \begin{cases}0 & \text { for } 0 \leq t \leq 6 \\ 125(t-6) & \text { for } 6<t \leq 7 \\ 125+108(t-7) & \text { for } 7<t \leq 9\end{cases}$
(d) Amount of snow is $\int_0^9 f(t) d t-h(9)=26.334$ or $26.335$ cubic feet.