Question
| r (centimeters) | 0 | 1 | 2 | 2.5 | 4 |
| f (r) (milligrams per square centimeter) | 1 | 2 | 6 | 10 | 18 |
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 \(2\pi \int_{0}^{4}rf(r)dr.\) Approximate the value of \(2\pi \int_{0}^{4}rf(r)dr.\) 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 ≤ r ≤ 4, is modeled by the function g defined by \(g(r)=2-16 \left ( cos(1.57\sqrt{r}) \right )^{3}.\) 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
Ans:
(a)
\(f'(2.25)\approx \frac{f(25)-f(2)}{2.5-2}=\frac{10-6}{0.5}=\frac{4}{0.5}= 8\) milligrams per square centimeter per centimer.
The density of bacteria changes at a rate of approximately 8 milligrams per square centimeter per centimeter at distance r = 2.25 centimeters from the center of the dish.
(b)
\(2\pi \int_{0}^{4}r.f(r)dr\approx 2\pi \cdot (2\cdot 1+1+6\cdot 1\cdot 2+10\cdot 0.5\cdot 2+18\cdot 4)\)
\(=2\pi (2+12+12.5+108)=2\pi (134.5)=269\pi milligrams\)
(c)
As a rule, Right riemann sums are always an over estimate for functions with positive slope, and underestimates for functions with negative slope. The slope of r.f(r) is equal to r’.f(r) + r.f'(r). Since r, r’, f(r), and f'(r) are always positive on the interval [0, 4], r.f(r) always has a positive slope on that interval. Since it’s a positive sloped function, the right riemann sum for r.f(r) from 0 to 4 is an over estimate.
(d)
Avg of \(g(r)=\frac{1}{4-1}\int_{1}^{4}\left ( 2-16(cos (1.57\sqrt{r}) \right )^{3}dr=\frac{29.627}{3}=9.876\)
\(g(k)=2-16\left ( cos (1.57\sqrt{r}) \right )^{3}=9.876\)
k = 2.497
Question
| h (feet) | 0 | 2 | 5 | 10 |
| A(h) (square feet) | 50.3 | 14.4 | 6.5 | 2.9 |
A tank has a height of 10 feet. The area of the horizontal cross section of the tank at height h feet is given by the function A, where A (h) is measured in square feet. The function A is continuous and decreases as h increases. Selected values for A (h) are given in the table above.
(a) Use a left Riemann sum with the three subintervals indicated by the data in the table to approximate the volume of the tank. Indicate units of measure.
(b) Does the approximation in part (a) overestimate or underestimate the volume of the tank? Explain your reasoning.
(c) The area, in square feet, of the horizontal cross section at height h feet is modeled by the function f given by \(f(h)=\frac{50.3}{e^{0.2h}+h}\) Based on this model, find the volume of the tank. Indicate units of measure.
(d) Water is pumped into the tank. When the height of the water is 5 feet, the height is increasing at the rate of 0.26 foot per minute. Using the model from part (c), find the rate at which the volume of water is changing with respect to time when the height of the water is 5 feet. Indicate units of measure.
Answer/Explanation
Ans:
(a) \(Volume = \int_{0}^{10}A(h)dh\)
≈ (2 – 0) · A(0) + (5 -2) · A(2) + (10 – 5) · A(5)
= 2 · 50.3 + 3.14.4 + 5 · 6.5
= 176.3 cubic feet
(b) The approximation in part (a) is an overestimate because a left Riemann sum is used and A is decreasing.
(c) \(\int_{0}^{10}f(h)dh=101.325338\)
The volume is 101.325 cubic feet.
(d) Using the model, \(V(h)=\int_{0}^{h}f(x)dx.\)
\(\frac{dV}{dt}|_{h=5}=\left [ \frac{dV}{dh}\cdot \frac{dh}{dt} \right ]_{h=5}\)
\(=\left [f(h)\cdot \frac{dh}{dt} \right ]_{h=5}\)
When h = 5, the volume of water is changing at a rate of 1.694 cubic feet per minute.
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
| t (minutes) | 0 | 12 | 20 | 24 | 40 |
| v (t) (meters per minute) | 0 | 200 | 240 | -220 | 150 |
Johanna jogs along a straight path. For 0 ≤ t ≤ 40, Johanna’s velocity is given by a differentiable function v. Selected values of v(t), where t is measured in minutes and v(t) is measured in meters per minute, are given in the table above.
(a) Use the data in the table to estimate the value of v'(16).
(b) Using correct units, explain the meaning of the definite integral \(\int_{0}^{40}|v(t)|dt\) in the context of the problem.
Approximate the value of \(\int_{0}^{40}|v(t)|dt\) using a right Riemann sum with the four subintervals indicated in the table.
(c) Bob is riding his bicycle along the same path. For 0 ≤ t ≤ 10, Bob’s velocity is modeled by B(t) = t3 – 6t2 + 300, where t is measured in minutes and B (t) is measured in meters per minute.
Find Bob’s acceleration at time t = 5.
(d) Based on the model B from part (c), find Bob’s average velocity during the interval 0 ≤ t ≤ 10.
Answer/Explanation
Ans:
(a) \(v'(16)\approx \frac{240-200}{20-12}=5 meters/min^{2}\)
(b) \(\int_{0}^{40}|v(t)|dt\) is the total distance Johana jogs, in meters, over the time interval 0 ≤ t ≤ 40 minutes.
\(\int_{0}^{40}|v(t)|dt\approx 12\cdot |v(12)|+8\cdot |v(20)|+4\cdot |v(24)|+16\cdot |v(40)|\)
= 12 · 200+ 8 · 240 + 4 · 220 + 16 · 150
=2400 + 1920 + 880 + 2400
= 7600 meters
(c) Bob’s acceleration is B'(t) = 3t2 – 12t.
B'(5) = 3(25) – 12(5) = 15 meters/min2
(d) Avg vel \(\frac{1}{10}\int_{0}^{10}(t^{3}-6t^{2}+300)dt\)
\(\frac{1}{10}\left [ \frac{t^{4}}{4}-2t^{3}+300t \right ]_{0}^{10}\)
\(\frac{1}{10}\left [ \frac{10000}{4}-2000+3000 \right ]=350 meters/min\)
Question
| t (minutes) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| C(t) (ounces) | 0 | 5.3 | 8.8 | 11.2 | 12.8 | 13.8 | 14.5 |
Hot water is dripping through a coffeemaker, filling a large cup with coffee. The amount of coffee in the cup at time t, 0 ≤ t ≤ 6, is given by a differentiable function C, where t is measured in minutes. Selected values of C(t), measured in ounces, are given in the table above.
(a) Use the data in the table to approximate C'(3.5). Show the computations that lead to your answer, and indicate units of measure.
(b) Is there a time t, 2 ≤ t ≤ 4 at which C'(t) = 2 ? Justify your answer.
(c) Use a midpoint sum with three subintervals of equal length indicated by the data in the table to approximate the value of \(\frac{1}{6}\int_{0}^{6}C(t)dt.\) Using correct units, explain the meaning of \(\frac{1}{6}\int_{0}^{6}C(t)dt\) in the context of the problem.
(d) The amount of coffee in the cup, in ounces, is modeled by B(t) = 16 – 16e-0.4t. Using this model, find the rate at which the amount of coffee in the cup is changing when t = 5.
Answer/Explanation
Ans:
(a) \(C'(3.5)\approx \frac{C(4)-C(3)}{4-3}=\frac{12.8-11.2}{1}=1.6 ounces/min\)
(b) C is differentiable ⇒ C is continuous (on the closed interval) \(\frac{C(4)-C(2)}{4-2}=\frac{12.8-8.8}{2}=2\)
Therefore, by the Mean Value Theorem, there is at least
one time t, 2 < t < 4, for which C'(t)=2.
(c) \(\frac{1}{6}\int_{0}^{6}C(t)dt\approx \frac{1}{6}\left [ 2\cdot C(1)+2\cdot C(3)+2\cdot C(5) \right ]\)
\(= \frac{1}{6}(2\cdot 5.3+2\cdot 11.2+2\cdot 13.8)\)
\(= \frac{1}{6}(60.6)= 10.1 ounces\)
\(\frac{1}{6}\int_{0}^{6}C(t)dt\) is the average amount of coffee in the cup, in ounces, over the time interval 0 ≤ t ≤ 6 minutes.
(d) \(B'(t)=-16(-0.4)e^{-0.4t}=6.4e^{-0.4t}\)
\(B'(5)=6.4e^{-0.4(5)}=\frac{6.4}{e^{2}}ounces/min\)
Question
There is an area A bounded by the curve \(f(x)=x^{2}\) and the x-axis.
(A) Use a right-hand Riemann sum to find area A on the interval from x = 0 to x = 4, using 4 subdivisions of equal length.
(B) Find the area A on the interval from x = 0 to x = 4 using 8 subdivisions of equal length.
(C) Now find area A by integrating f (x) over the interval (0, 4).
Answer/Explanation
(A) The size of the subdivision is given by \(\Delta x_{i}=\frac{4-0}{4}=1\).Then make a table of values for f (x).Now compute \(A=\sum_{i=1}^{4}f(c_{i})\Delta x_{i}=1(1)+4(1)+9(1)+16(1)=30\).
(B)Now \(\Delta x_{i}=\frac{4-0}{8}=\frac{1}{2}.\) Then \(A=\sum_{i=1}^{8}f(c_{i})\Delta x_{i}=\frac{1}{4}\left ( \frac{1}{2} \right )+1\left ( \frac{1}{2} \right )+\frac{9}{4}\left ( \frac{1}{2} \right )+4\left ( \frac{1}{2} \right )+\frac{25}{4}\left ( \frac{1}{2} \right )+9\left ( \frac{1}{2} \right )+\frac{49}{4}\left ( \frac{1}{2} \right )+16\left ( \frac{1}{2} \right )=\frac{1}{8}+\frac{4}{8}+\frac{9}{8}+\frac{16}{8}+\frac{25}{8}+\frac{36}{8}+\frac{49}{8}+\frac{64}{8}=25\frac{1}{2}\)
(C) The area \(A=\int_{0}^{4}x^{2}dx=\frac{1}{3}x^{3}|^{4}_{0}=\frac{64}{3}-0=21\frac{1}{3}\)
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
| t (minutes) | 0 | 2 | 5 | 9 | 10 |
| H(t) (degrees Celsius) | 66 | 60 | 52 | 44 | 43 |
As a pot of tea cools, the temperature of the tea is modeled by a differentiable function H for 0 ≤ t ≤ 10, where time t is measured in minutes and temperature H(t) is measured in degrees Celsius. Values of H(t) at selected values of time t are shown in the table above.
(a) Use the data in the table to approximate the rate at which the temperature of the tea is changing at time t = 3.5. Show the computations that lead to your answer.
(b) Using correct units, explain the meaning of \(\frac{1}{10}\int_{0}^{10}H(t)dt\) in the context of this problem. Use a trapezoidal sum with the four subintervals indicated by the table to estimate \(\frac{1}{10}\int_{0}^{10}H(t)dt.\)
(c) Evaluate \(\int_{0}^{10}H'(t)dt.\) Using correct units, explain the meaning of the expression in the context of this problem.
(d) At time t = 0, biscuits with temperature 100 0C were removed from an oven. The temperature of the biscuits at time t is modeled by a differentiable function B for which it is known that
B'(t) = – 13.84e-0.173t . Using the given models, at time t = 10, how much cooler are the biscuits than the tea?
Answer/Explanation
Ans:
(a)
\(\frac{H(5)-H(2)}{5-2}=\frac{-8}{3}\frac{0_{C}}{min}\)
(b)
\(\frac{1}{10}\int_{0}^{10}H(t)dt\approx \frac{\left [2(\frac{66+180}{2}) +3(\frac{52+60}{2})+4(\frac{44+52}{2})+1(\frac{43+44}{2}) \right ]}{10}=52.95\)
This represents the average temperature in degree Celsius of the tea over the interval 0 ≤ t ≤ 10
(c)
\(\int_{0}^{10}H'(t)dt= H(10)-H(0)=43-66=-23^{0}C\)
This expression shows the total change in temperature in degree Celsius from t = 0 to t = 10.
(d)
\(B'(t)=-13.84e^{-.173t}\)
\(B(10)=\int_{0}^{10}-13.84e^{-.173t}+100 = 100-65.817 = 34.1827\)
43-341827 = 8.817
= 88170C Celsius