This appendix contains answers to all activities in the text. Answers for preview activities are not included.
1Understanding the Derivative 1.1How do we measure velocity? 1.1.1Position and average velocity
Activity1.1.2.
Answer.
\(AV_{[0.4,0.8]} = 12.8\) ft/sec; \(AV_{[0.7,0.8]} = 8\) ft/sec; the other average velocities are, respectively, 6.56, 6.416, 0, 4.8, 6.24, 6.384, all in ft/sec.
\(m = 12.8\) is the average velocity of the ball between \(t = 0.4\) and \(t = 0.8\text{.}\)
Like a straight line with slope about 6.4.
About 6.4 feet per second.
1.1.2Instantaneous Velocity
Activity1.1.3.
Answer.
\(AV_{[1.5,2]} = -24\) ft/sec, which is negative.
The instantaneous velocity at \(t = 1.5\) is approximately \(-16\) ft/sec; at \(t = 2\text{,}\) the instantaneous velocity is about \(-32\) ft/sec, and \(-16>-32\text{.}\)
When the ball is rising, its instantaneous velocity is positive, while when the ball is falling, its instantaneous velocity is negative.
Take shorter and shorter time intervals and draw the lines whose slopes represent average velocity. If those lines’ slopes are approaching a single number, that number represents the instantaneous velocity.
The instantaneous velocity at \(t = 2\) is greater than the average velocity on \([1.5,2.5]\text{.}\)
1.3The derivative of a function at a point 1.3.1The Derivative of a Function at a Point
Activity1.3.2.
Answer.
\(f\) is linear.
The average rate of change on \([1,4]\text{,}\)\([3,7]\text{,}\) and \([5,5+h]\) is \(-2\text{.}\)
\(f'(1)=-2\text{.}\)
\(f'(2)=-2\text{,}\)\(f'(\pi)=-2\text{,}\) and \(f'(-\sqrt{2})=-2\text{,}\) since the slope of a linear function is the same at every point.
Activity1.3.3.
Answer.
The vertex is \((\frac{1}{2},36)\text{.}\)
\(\frac{s(2)-s(1)}{2-1} = -32\) feet per second.
\(s'(1) = -16\text{.}\)
\(s'(a)\) is positive whenever \(0 \le a \lt \frac{1}{2}\text{;}\)\(s'(a)\) to be negative whenever \(\frac{1}{2} \lt a \lt 2\text{;}\)\(s'(\frac{1}{2}) = 0\text{.}\)
Activity1.3.4.
Answer.
\(AV_{[2,4]} \approx 9171\) people per decade is expected to be the average rate of change of the city’s population over the two decades from 2030 to 2050.
See the graph provided in (a) above. The magenta line has slope equal to the average rate of change of \(P\) on \([2,4]\text{,}\) while the green line is the tangent line at \((2,P(2))\) with slope \(P'(2)\text{.}\)
It appears that the tangent line’s slope at the point \((a,P(a))\) will increase as \(a\) increases.
1.4The derivative function 1.4.1How the derivative is itself a function
Activity1.4.2.
Answer.
Activity1.4.3.
Answer.
\(f'(x) = 0\text{.}\)
\(g'(t) = 1\text{.}\)
\(p'(z) = 2z\text{.}\)
\(q'(s) = 3s^2\text{.}\)
\(F'(t) = \frac{-1}{t^2}\text{.}\)
\(G'(y) = \frac{1}{2\sqrt{y}}\text{.}\)
1.5Interpreting, estimating, and using the derivative 1.5.2Toward more accurate derivative estimates
Activity1.5.2.
Answer.
\(F'(30) \approx = 3.85\) degrees per minute.
\(F'(60) \approx = 1.56\) degrees per minute.
\(F'(75) \gt F'(90)\text{.}\)
The value \(F(64) = 330.28\) is the temperature of the potato in degrees Fahrenheit at time 64, while \(F'(64) = 1.341\) measures the instantaneous rate of change of the potato’s temperature with respect to time at the instant \(t = 64\text{,}\) and its units are degrees per minute. Because at time \(t = 64\) the potato’s temperature is increasing at 1.341 degrees per minute, we expect that at \(t = 65\text{,}\) the temperature will be about 1.341 degrees greater than at \(t = 64\text{,}\) or in other words \(F(65) \approx 330.28 + 1.341 = 331.621\text{.}\) Similarly, at \(t = 66\text{,}\) two minutes have elapsed from \(t = 64\text{,}\) so we expect an increase of \(2 \cdot 1.341\) degrees: \(F(66) \approx 330.28 + 2 \cdot 1.341 = 332.962\text{.}\)
Throughout the time interval \([0,90]\text{,}\) the temperature \(F\) of the potato is increasing. But as time goes on, the rate at which the temperature is rising appears to be decreasing. That is, while the values of \(F\) continue to get larger as time progresses, the values of \(F'\) are getting smaller (while still remaining positive). We thus might say that “the temperature of the potato is increasing, but at a decreasing rate.”
Activity1.5.3.
Answer.
It costs $800 to make 2000 feet of rope.
“dollars per foot.”
\(C(2100) \approx = 835\text{,.}\)
Either \(C'(2000) = C'(3000)\) or \(C'(2000) > C'(3000)\text{.}\)
Impossible. The total cost function \(C(r)\) can never decrease.
Activity1.5.4.
Answer.
\(f'(90) \approx 0.0006\) liters per kilometer per kilometer per hour.
At 80 kilometers per hour, the car is using fuel at a rate of 0.015 liters per kilometer.
When the car is traveling at 90 kilometers per hour, its rate of fuel consumption per kilometer is increasing at a rate of 0.0006 liters per kilometer per kilometer per hour.
Velocity is increasing on \(0\lt t\lt 1\text{,}\)\(3\lt t\lt 4\text{,}\)\(7\lt t\lt 8\text{,}\) and \(10\lt t\lt 11\text{;}\)\(y = v(t)\) is decreasing on \(1\lt t\lt 2\text{,}\)\(4\lt t\lt 5\text{,}\)\(8\lt t\lt 9\text{,}\) and \(11\lt t\lt 12\text{.}\) Velocity is constant on \(2\lt t\lt 3\text{,}\)\(5\lt t\lt 7\text{,}\) and \(9\lt t\lt 10\text{.}\)
\(a(t) = v'(t)\) and \(a(t) = s''(t)\text{.}\)
\(s''(t)\) is positive since \(s'(t)\) is increasing.
increasing.
decreasing.
constant.
increasing.
decreasing.
constant.
concave up.
concave down.
linear.
Activity1.6.3.
Answer.
Degrees Fahrenheit per minute.
\(F''(30) \approx -0.119\text{.}\)
At the moment \(t = 30\text{,}\) the temperature of the potato is 251 degrees; its temperature is rising at a rate of 3.85 degrees per minute; and the rate at which the temperature is rising is falling at a rate of 0.119 degrees per minute per minute.
Increasing at a decreasing rate.
Activity1.6.4.
Answer.
1.7Limits, Continuity, and Differentiability 1.7.1Having a limit at a point
Activity1.7.2.
Answer.
\(f(-2) = 1\text{;}\)\(f(-1)\) is not defined; \(f(0) = \frac{7}{3}\text{;}\)\(f(1) = 2\text{;}\)\(f(2) = 2\text{.}\)
\(\lim_{x \to -2} f(x)\) does not exist. The values of the limits as \(x \to a\) for \(a = -1, 0, 1, 2\) are \(\frac{5}{3}, \frac{7}{3}, 3, 2\text{.}\)
\(a = -2\text{,}\)\(a = -1\text{,}\) and \(a = 1\text{.}\)
1.8The Tangent Line Approximation 1.8.2The local linearization
Activity1.8.2.
Answer.
\(L(-1) = -2\text{;}\)\(L'(-1) = 3\text{.}\)
\(g(-1) = -2\text{;}\)\(g'(-1) = 3\text{.}\)
Less.
\(g(-1.03) \approx L(-1.03) = -2.09\text{.}\)
Concave up.
The illustration below shows a possible graph of \(y = g(x)\) near \(x = -1\text{,}\) along with the tangent line \(y = L(x)\) through \((-1, g(-1))\text{.}\)
See the image below, which shows, at left, a possible graph of \(y = f(x)\) near \(x = 2\text{,}\) along with the tangent line \(y = L(x)\) through \((2, f(2))\text{.}\)
Too large.
2Computing Derivatives 2.1Elementary derivative rules 2.1.2Constant, Power, and Exponential Functions
\(I'(0.5) = \frac{50}{e^{0.5}} \approx 30.327\text{,}\)\(I'(2) = \frac{-100}{e^{2}} \approx -13.534\text{,}\) and \(I'(5) = \frac{-400}{e^5} \approx -2.695\text{,}\) each in candles per millisecond.
\(s'(1) = \frac{-2\sin(1)-4\cos(1)}{e^1} \approx -1.414\) feet per second.
\(p'(3) = 30\) and \(q'(3) = \frac{13}{8}\text{.}\)
2.4Derivatives of other trigonometric functions 2.4.1Derivatives of the cotangent, secant, and cosecant functions
Activity2.4.2.
Answer.
All real numbers \(x\) such that \(x \ne \frac{\pi}{2} + k\pi\text{,}\) where \(k = \pm 1, \pm 2, \ldots\text{.}\)
\(h'(x) = \frac{\sin(x)}{\cos^2(x)}\text{.}\)
\(h'(x) = \sec(x) \tan(x)\text{.}\)
\(h\) and \(h'\) have the same domain: all real numbers \(x\) such that \(x \ne \frac{\pi}{2}+k\pi\text{,}\) where \(k = 0, \pm 1, \pm 2, \ldots\text{.}\)
Activity2.4.3.
Answer.
All real numbers \(x\) such that \(x \ne k\pi\text{,}\) where \(k = 0, \pm 1, \pm 2, \ldots\text{.}\)
\(h'(x) = -\frac{\cos(x)}{\sin^2(x)}\text{.}\)
\(h'(x) = -\csc(x) \cot(x)\text{.}\)
\(p\) and \(p'\) have the same domain: all real numbers \(x\) such that \(x \ne k\pi\text{,}\) where \(k = 0, \pm 1, \pm 2, \ldots\text{.}\)
\((1.418697,0.543912)\text{,}\)\((-1.418697,-0.543912)\text{,}\)\((-3.63143, 1.64443)\text{,}\) and \((3.63143, -1.64443)\text{.}\)
Activity2.7.3.
Answer.
Horizontal at \(x \approx 0.42265\text{,}\) thus \((0.42265, -1.05782); (0.42265, 0.229478); (0.42265, 0.770522); (0.42265, 2.05782)\text{.}\) There are four more points where \(x \approx 1.57735\text{.}\)
When \(y = \frac{1}{2}, \frac{1 \pm \sqrt{5}}{2}\text{,}\) so one point is \((2.21028, \frac{1}{2})\text{.}\)
\(y - 1 = \frac{1}{2}(x-1)\text{.}\)
Activity2.7.4.
Answer.
\(\frac{dy}{dx}(-3y^2 - 6x) = 6y-3x^2 \) and the tangent line has equation \(y - 3 = 1(x+3)\text{.}\)
\(\frac{dy}{dx} = \frac{3x^2 + 1}{\cos(y) + 1}\) and the tangent line has equation \(y = \frac{1}{2}x\text{.}\)
\(\frac{dy}{dx} = \frac{3e^{-xy} - 3xye^{-xy}}{3x^2e^{-xy}+2y}\) and the tangent line is \(y - 1 = 0.234950(x - 0.619061)\text{.}\)
2.8Using Derivatives to Evaluate Limits 2.8.1Using derivatives to evaluate indeterminate limits of the form \(\frac{0}{0}\text{.}\)
3Using Derivatives 3.1Using derivatives to identify extreme values 3.1.1Critical numbers and the first derivative test
Activity3.1.2.
Answer.
\(x = -4\) or \(x = 1\text{.}\)
\(g\) has a local maximum at \(x = -4\) and neither a max nor min at \(x = 1\text{.}\)
\(g\) does not have a global minimum; it is unclear (at this point in our work) if \(g\) increases without bound, so we can’t say for certain whether or not \(g\) has a global maximum.
\(\lim_{x \to \infty} g'(x) = \infty\text{.}\)
A possible graph of \(g\) is the following.
3.1.2The second derivative test
Activity3.1.3.
Answer.
\(x = -1\) is an inflection point of \(g\text{.}\)
\(g\) is concave up for \(x \lt -1\text{,}\) concave down for \(-1 \lt x \lt 2\text{,}\) and concave down for \(x \gt 2\text{.}\)
\(g\) has a local minimum at \(x = -1.67857351\text{.}\)
\(g\) is a degree 5 polynomial.
Activity3.1.4.
Answer.
In the graph below, \(h(x) = x^2 + \cos(3x)\) is given in dark blue, while \(h(x) = x^2 + \cos(1.6x)\) is shown in light blue.
If \(\frac{2}{k^2} \gt 1\text{,}\) then the equation \(\cos(kx) = \frac{2}{k^2}\) has no solution. Hence, whenever \(k^2 \lt 2\text{,}\) or \(k \lt \sqrt{2} \approx 1.414\text{,}\) it follows that the equation \(\cos(kx) = \frac{2}{k^2}\) has no solutions \(x\text{,}\) which means that \(h''(x)\) is never zero (indeed, for these \(k\)-values, \(h''(x)\) is always positive so that \(h\) is always concave up). On the other hand, if \(k \ge \sqrt{2}\text{,}\) then \(\frac{2}{k^2} \le 1\text{,}\) which guarantees that \(\cos(kx) = \frac{2}{k^2}\) has infinitely many solutions, due to the periodicity of the cosine function. At each such point, \(h''(x) = 2 - k^2 \cos(kx)\) changes sign, and therefore \(h\) has infinitely many inflection points whenever \(k \ge \sqrt{2}\text{.}\)
To see why \(h\) can only have a finite number of critical numbers regardless of the value of \(k\text{,}\) consider the equation
which implies that \(2x = k\sin(kx)\text{.}\) Since \(-1 \le \sin(kx) \le 1\text{,}\) we know that \(-k \le k\sin(kx) \le k\text{.}\) Once \(|x|\) is sufficiently large, we are guaranteed that \(|2x| \gt k\text{,}\) which means that for large \(x\text{,}\)\(2x\) and \(k\sin(kx)\) cannot intersect. Moreover, for relatively small values of \(x\text{,}\) the functions \(2x\) and \(k\sin(kx)\) can only intersect finitely many times since \(k\sin(kx)\) oscillates a finite number of times. This is why \(h\) can only have a finite number of critical numbers, regardless of the value of \(k\text{.}\)
3.2Using derivatives to describe families of functions 3.2.1Describing families of functions in terms of parameters
Activity3.2.2.
Answer.
\(p\) has two critical numbers (\(x = \pm \sqrt{\frac{a}{3}}\)) whenever \(a \gt 0\) and no critical numbers when \(a \lt 0\text{.}\)
When \(a \lt 0\text{,}\)\(p\) is always increasing and has no relative extreme values. When \(a\gt 0\text{,}\)\(p\) has a relative maximum at \(x = -\sqrt{\frac{a}{3}}\) and a relative minimum at \(x = +\sqrt{\frac{a}{3}}\text{.}\)
\(p\) is CCD for \(x \lt 0\) and \(p\) is CCU for \(x\gt 0\text{,}\) making \(x = 0\) an inflection point.
If \(b\) is large and \(x\) is close to zero, \(h'(x)\) is relatively large near \(x = 0\text{,}\) and the curve’s slope will quickly approach zero as \(x\) increases. If \(b\) is small, the graph is less steep near \(x = 0\) and its slope goes to zero less quickly as \(x\) increases.
Activity3.2.4.
Answer.
\(L\) is an always increasing function.
\(L\) is concave up for all \(t \lt -\frac{1}{k} \ln \left(\frac{1}{c}\right)\) and concave up for all other values of \(t\text{.}\)
\(\lim_{t \to \infty} \frac{A}{1+ce^{-kt}} = A\text{,}\) and
The absolute minimum time the hiker can achieve is \(0.99302\) hours, which is attained by hiking about 2.2 km from \(P\) to \(Q\) and then turning into the woods for the remainder of the trip.
Activity3.4.4.
Answer.
Maximum area: \(A(\frac{5}{\sqrt{3}}) = \frac{500}{9}\sqrt{3} \approx 96.225\text{.}\) Maximum perimeter: \(P(1) = 52\text{.}\) At \(x = \frac{\sqrt{82}-1}{3}\) the absolute maximum of combined perimeter and area occurs.
Activity3.4.5.
Answer.
\(A(1.19606) \approx 2.2018\) is the absolute maximum cross-sectional area, which leads to the absolute maximum volume.
On \((0,1)\text{,}\)\(s\) is increasing because velocity is positive.
\(s(t) = 32t - 16t^2\text{.}\)
\(s(1) - s(\frac{1}{2}) = 4\text{.}\)
\(A = 4\) feet is the total distance the ball traveled vertically on \([\frac{1}{2},1]\text{.}\)
\(s(1) - s(0) = 16\) is the vertical distance the ball traveled on the interval \([0,1]\text{.}\) Equivalently, the area between the velocity curve and the \(t\)-axis on \([0,1]\) is \(A = 16\) feet.
\(s(2) - s(0) = 0\text{,}\) so the ball has zero change in position on the interval \([0,2]\text{.}\)
4.1.3When velocity is negative
Activity4.1.4.
Answer.
Total distance traveled is \(2\text{;}\) change in position is \(0\text{.}\)
\(0 \lt t \lt 1\) and \(4 \lt t \lt 8 \text{.}\)
\(s(8) - s(0) = 5 \ \mbox{m} \text{,}\) while the distance traveled on \([0,8]\) is \(D = 13\text{,}\) and thus these two quantities are different.
One time at which the instantaneous rate at which calories are burned equals the average rate on \([0,40]\) is \(t = \frac{5}{3}(6 - \sqrt{6}) \approx 5.918\text{.}\)
5Evaluating Integrals 5.1Constructing Accurate Graphs of Antiderivatives 5.1.1Constructing the graph of an antiderivative
Activity5.1.2.
Answer.
\(F\) is increasing on \((0,2)\) and \((5,7)\text{;}\)\(F\) is decreasing on \((2,5)\text{.}\)
\(F\) is concave up on \((0,1)\text{,}\)\((4,6)\text{;}\) concave down on \((1,3)\text{,}\)\((6,7)\text{;}\) neither on \((3,4)\text{.}\)
A relative maximum at \(x = 2\text{;}\) a relative minimum at \(x = 5\text{.}\)
Left endpoint rule results are overestimates; right endpoint rules are underestimates; midpoint rules are overestimates; trapezoid rules are underestimates. Simpson’s rule is exact for both \(f\) and \(g\text{,}\) while a slight overestimate of \(\int_0^1 h(x) dx\text{.}\)
6Using Definite Integrals 6.1Using Definite Integrals to Find Area and Length 6.1.1The Area Between Two Curves
If we have an existing arrangement and balancing point, moving one of the locations to the left will move the balancing point to the left; similarly, moving one of the locations to the right will move the balancing point to the right. If instead we add weight to an existing location, if that location is left of the balancing point, the balancing point will move left; the behavior is similar if on the right.
Both graphs have a vertical asymptote at \(x = 0\) and for both graphs, the \(x\)-axis is a horizontal asymptote. However, the graph of \(y = \frac{1}{x^{3/2}}\) will ’’approach the \(x\)-axis faster’’ than the graph of \(y = \frac{1}{x}\text{.}\)
The area bounded by the graph of \(y = \frac{1}{x}\text{,}\) the \(x\)-axis, and the vertical line \(x = 1\) is infinite or unbounded. However, The area bounded by the graph of \(y = \frac{1}{x^{3/2}}\text{,}\) the \(x\)-axis, and the vertical line \(x = 1\) is equal to 2.
When \(y \lt 0\) and when \(y \gt 4\text{,}\)\(y\) is a decreasing function of \(t\text{.}\) When \(0 \lt y \lt 4\text{,}\)\(y\) is a increasing function of \(t\text{.}\)
If we first think about how \(y_1\) is generated for the initial value problem \(\frac{dy}{dt} = f(t) = 2t-1, \ y(0) = 0\text{,}\) we see that \(y_1 = y_0 + \Delta t \cdot f(t_0)\text{.}\) Since \(y_0 = 0\text{,}\) we have \(y_1 = \Delta t \cdot f(t_0)\text{.}\) From there, we know that \(y_2\) is given by \(y_2 = y_1 + \Delta t f(t_1)\text{.}\) Substituting our earlier result for \(y_1\text{,}\) we see that \(y_2 = \Delta t \cdot f(t_0) + \Delta t f(t_1)\text{.}\) Continuing this process up to \(y_5\text{,}\) we get
\begin{equation*}
y_5 = \Delta t \cdot f(t_0) + \Delta t f(t_1) + \Delta t f(t_2) + \Delta t f(t_3) + \Delta t f(t_4)
\end{equation*}
This is precisely the left Riemann sum with five subintervals for the definite integral \(\int_0^1 (2t-1)~dt\text{.}\)
Solutions to this differential equation all differ by only a constant.
Activity7.3.3.
Answer.
\(y = 0\) or \(y = 6\text{;}\)\(y = 0\) is unstable, \(y = 6\) is stable.
The solution will tend to \(y = 6\text{.}\)
\(t_i\)
\(y_i\)
\(dy/dt\)
\(\Delta y\)
\(0.0\)
\(1.0000\)
\(5.0000\)
\(1.0000\)
\(0.2\)
\(2.0000\)
\(8.0000\)
\(1.6000\)
\(0.4\)
\(3.6000\)
\(8.6400\)
\(1.7280\)
\(0.6\)
\(5.3280\)
\(3.5804\)
\(0.7161\)
\(0.8\)
\(6.0441\)
\(-0.2664\)
\(-0.0533\)
\(1.0\)
\(5.9908\)
\(0.0551\)
\(0.0110\)
The value of \(y_i = 6\) for every value of \(i\text{.}\)
