### Differential Equations to Difference Equations

Chapter 3

Section 3.1

1. (b) $m = 1$

2. (a) $o(h)$

(b) $O(h)$

(c) $o(h)$

(d) $o(h)$

(e) $o(h)$

(f) Neither

3. (b) $R_T$ is $O(h)$, but not $o(h)$.

(c) $R_S$ is both $o(h)$ and $O(h)$.

(d) $S$ is the best affine approximation to $f$ at $9$.

(e) $\sqrt{10} \approx S(10) = \displaystyle{\frac{19}{6}} = 3.167$ to three decimal places, whereas $\sqrt{10}$ is exactly $3.162$ to three decimal places.

5. $T(x) = 4(x - 1) + 2$, $f'(1) = 4$

7. $T(x) = x - 1$, $f'(0) = 1$

Section 3.2

1. (a) $f'(2) = 4$

(c) $g'(2) = -\displaystyle{\frac{1}{4}}$

(e) $f'(1) = -\displaystyle{\frac{1}{2}}$

2. (a) Best affine approximation: $T(x) = 4(x - 2) + 5$ Equation of tangent line: $y = 4(x - 2) + 5$

(c) Best affine approximation: $\displaystyle{T(x) = -\frac{1}{4}(x - 2) + \frac{1}{4}}$ Equation of tangent line: $\displaystyle{y = -\frac{1}{4}(x - 2) + \frac{1}{4}}$

(e) Best affine approximation: $\displaystyle{T(s) = -\frac{1}{2}(s - 1) + 1}$ Equation of tangent line: $\displaystyle{y = -\frac{1}{2}(s - 1) + 1}$

3. (a) $f'(x = 4x$

(c) $\displaystyle{f'(t) = -\frac{1}{2t^{\frac{3}{2}}}}$; $f$ is not differentiable at $t = 0$

(e) $y'(t) = 2t + 4$

4. Best affine approximation: $\displaystyle{T(t) = -\frac{1}{16}(x - 4) + \frac{1}{2}}$ $\displaystyle{\frac{1}{\sqrt{3.98}} \approx T(3.98) = \frac{401}{800} = 0.50125}$ to five decimal places

7. $T(t) = -9(t - 3) - 15$

11. $f'(0) = 0$

13. $g$ is not differentiable at $x = 1$.

14. (a) $\displaystyle{\frac{ds}{dt}} = 6t$

(c) $\displaystyle{\frac{dq}{ds} = 1 + \frac{2}{s^2}}$

15. $\displaystyle{\frac{d}{dx}}(4x^2) = 8x$, $\displaystyle{\frac{d}{du}(3\sqrt{u-1}) = \frac{3}{2\sqrt{u-1}}}$

16. (a) Average velocity over $[0, 1] = 0.2 \text{ meters/second}$

(b) $v = 10 - 9.8t \text{ meters/second}$

(c) $v|_{t=1} = 0.2 \text{ meters/second}$

(d) $v = 0$ when $t = 1.0204$ seconds. $v$ is positive before this time, and negative after.

(e) $s|_{t=1.0204} = 105.10 \text{ meters}$, which must be the maximum height reached by the object.

(f) $\displaystyle{\frac{dv}{dt}} = -9.8 \text{ meters/second}^2$

(g) The object is losing velocity at the same rate at all times.

17. $\displaystyle{\frac{dA}{dr}} = 2\pi r$

18. $\displaystyle{\frac{dV}{dr}} = 4\pi r^2$

19. $\displaystyle{\frac{dA}{dx}} = 2x$

Section 3.3

1. (a) $f'(x) = 3x^2 + 6$

(c) $g'(t) = 3 - 12t$

(e) $f'(t) = 18t - 36$

2. (a) $f'(x) = 8x + 4$

(c) $\displaystyle{g'(x) = \frac{11}{(2x + 5)^2}}$

(e) $\displaystyle{f'(t) = \frac{6t^6 + 104t^3 - 6t^2 - 48}{(2t^3 + 6)^2}}$

(g) $h'(t) = -\displaystyle{\frac{3}{t^2}}$

(i) $\displaystyle{h'(z) = 24z^2 + \frac{1}{2z^2}}$

3. (a) $\displaystyle{\frac{ds}{dt}} = (t^2 - 6t +3)(32t^3 + 12t) + (8t^4 + 6t^2 - 7)(2t - 6)$

(c) $\displaystyle{\frac{dy}{dx}} = (x^2 - 2x + 3)(2x^2 + 13x - 6)(6x - 4) + (x^2 - 2x + 3)(4x + 13)(3x^2 - 4x + 1) + (2x - 2)(2x^2 + 13x - 6)(3x^2 - 4x + 1)$

4. (a) $k'(2) = 26$

(c) $k'(2) = 102$

5. (a) $\displaystyle{v(t) = -1 + \frac{1}{2}t^2}$

(c) The object is at the origin when $t = -\sqrt{6}, 0, \sqrt{6}$. The velocities are: $v(-\sqrt{6}) = 2$, $v(0) = -1$, $v(\sqrt{6}) = 2$.

(e) $-\sqrt{2} < t < \sqrt{2}$

(g) $a(t) = t$

(i) The object is moving toward the left and slowing down at time $t = 1.$

Section 3.4

1. (a) $f'(x) =16(4x + 5)^3$

(c) $\displaystyle{h'(t) = -\frac{18}{(6t - 3)^3}}$

(e) $g'(z) = 2(3z+4)^3(2z^2 + z)(4z + 1) + 9(3z + 4)^2(2z^2 + z)^2$

2. (a) $\displaystyle{\frac{ds}{dt} = 16t^3(t^2 - 1) + 8t(t^2 - 1)^2}$

(c) $\displaystyle{\frac{dq}{dt} = \frac{9t^2 - 4}{2\sqrt{3t^3 - 4t}}}$

(e) $\displaystyle{\frac{dx}{dt} = \frac{40 - 32t}{(4t + 5)^2}}$

(g) $\displaystyle{\frac{dy}{dx} = \frac{3}{5(3x - 1)^{\frac{4}{5}}}}$

3. $\displaystyle{T(x) = -\frac{33}{125}(x - 2) + \frac{6}{25}}$

4. (a) $T(x) = hx + 1$

(c) $\displaystyle{\sqrt{1.06} \approx \frac{1}{3}(0.06) + 1 = 1.02}$ Rounded to four decimal places, $\sqrt{1.06} = 1.0196$.

5. (a) $\displaystyle{y = -\frac{1}{2}(x - 2) + 3}$

(c) $\displaystyle{y = -\frac{7}{8}(x - 2) + 1}$

(e) $y = 2 - x$

6. (a) $k'(0) = 9$

(c) $k'(0) = -6$

(e) $k'(0) = 8$

8. $900 \text{ cm}^3\text{/sec}$

9. $400\pi \text{ cm}^2\text{/sec}$

11. $\displaystyle{\frac{dK}{dt}\bigg|_{v=10} = 98m}$

13. $1.6 \text{ in}^2\text{/sec}$

15. (a) $0$ units per month

(b) $-100$ units per month

17.$\displaystyle{\frac{5}{2\pi}} \text{ ft/sec}$

Section 3.5

1. (a) $f'(x) = x^2\cos(x) + 2x\sin(x)$

(c) $g'(t) = -6t\sin(2t) + 3\cos(2t)$

(e) $f'(t) = -4\sin(3)t\sin(4t) + 3\cos(3t)\cos(4t)$

(f) $g'(z) = 12\sin^2(4z)\cos(4z)$

2. (a) $\displaystyle{\frac{dy}{dx} = \frac{2x\cos(2x) - \sin(2x)}{x^2}}$

(c) $\dfrac{dx}{dt} = 8t\cos(4t^2 + 1)$

(e) $\dfrac{dz}{dt} = 2\sec(2t)\tan(2t)$

(g) $\dfrac{dy}{dx} = -2x^2\csc(2x)\cot(2x) + 2x\csc(2x)$

3. (a) $\dfrac{d}{dx}(\sin^2(2x)\cos^2(3x) = -6\sin^2(2x)\cos(3x)\sin(3x) + 4\cos^2(3x)\sin(2x)\cos(2x)$

(c) $\dfrac{d}{dq}\sec^3(q^2) = 6q\sec^3(q^2)\tan(q^2)$

(e) $\displaystyle{\frac{d}{dz}\sqrt{1 + \sin^2(z)} = \frac{\sin(z)\cos(z)}{\sqrt{1 + \sin^2(z)}}}$

5. $T(t) = 1$

7. (a) $S(x) = \dfrac{1}{2}x + 1$

(b) $T(x) = 4x$

(c) $U(x) = 2x + 1$

(e) $h = f \circ g$ and $U = S \circ T$

8. (a) $\displaystyle{\lim_{x \to 0}\frac{\sin(2x)}{x} = 2}$

(c) $\displaystyle{\lim_{x \to 0}\frac{\tan(x)}{x} = 1}$

(e) $\displaystyle{\lim_{x \to 0}\frac{\sin^2(x)}{x} = 0}$

(g) $\displaystyle{\lim_{t \to 0}\frac{\sin^2(3t)}{t^2} = 9}$

9. (a) $f$ is $O(h)$, but not $o(h)$.

(c) $g$ is $O(h)$, but not $o(h)$.

(e) $f$ is $o(h)$, and so also $O(h)$.

11. (a) $D(0.00001) = 4.000010000027032$ to 15 decimal places $D_1(0.00001) = 4.000000000026205$ to 15 decimal places $D_2(0.00001) = 3.999999999848569$ to 15 decimal places $f'(2) = 2$

(c) $D(0.00001) = 0.999999999983333$ to 15 decimal places $D_1(0.00001) = 0.999999999983333$ to 15 decimal places $D_2(0.00001) = 1.000000000000000$ to 15 decimal places $f'(0) = 1$

Section 3.6

1. (a) Three solutions: $-2.3407$, $-0.9018$, $2.4080$

(c) Two solutions: $-0.8241$, $0.8241$

(e) Two solutions $0.7135$, $2.0730$

2. (d) If $\displaystyle{L = \lim_{n \to \infty}x_n}$, then $\displaystyle{L = \lim_{n \to \infty}x_{n+1} = \lim_{n \to \infty}\frac{x_n + \dfrac{c}{x_n}}{2} = \frac{L + \dfrac{c}{L}}{2}}$, from which it follows that $L^2 = c$.

3. $1.25992$

4. $1.25850$

6. The values oscillate between $1$ and $-1$; that is, for $n = 0, 1, 2, \ldots$, $x_{2n} = 1$ and $x_{2n + 1} = -1$.

Section 3.7

1. Let $g(x) = \cos(x) - x$. Then $g(0) = 1$ and $g(1) = \cos(1) - 1 < 0$ (since $\cos(1) < 1$). Hence, by the Intermediate Value Theorem, there is at least one point $c$ in the interval $(0, 1)$ such that $g(c) = 0$. Now if there were two points, say, $c_1$ and $c_2$, in $(0, 1)$ such that $g(c_1) = 0$ and $g(c_2) = 0$, then, by Rolle's Theorem, there would be a point $d$ between $c_1$ and $c_2$ such that $g'(d) = 0$. But this cannot happen since $g'(x) = -\sin(x) - 1 < 0$ for all $x$ in $(0, 1)$. Hence there is exactly one point $c$ in $(0, 1)$ such that $g%%(c)%% = 0$. That is, there is exactly one solution to the equation $\cos(x) = x$ in the interval $(0, 1)$.

3. By the Mean Value Theorem, for any two points $u$ and $v$ in $[a, b]$, there exists a point $c$ in $(a, b)$ such that $f'(c) = \frac{f(v) - f(u)}{v - u}.$ Hence $f(v) - f(u) = f'%%(c)%%(v - u)$, and so $|f(v) - f(u)| = |f'(c)||v - u| \le M|v - u|.$

5. (a) $f$ is increasing on $(0, \infty)$ and decreasing on $(-\infty, 0)$.

(c) $h$ is decreasing on both $(-\infty, 1)$ and $(1, \infty)$. There are no intervals on which $h$ is increasing.

(e) $f$ is increasing on $(-1, 1)$ and decreasing on both $(-\infty, -1)$ and $(1, \infty)$

(g) $y$ is decreasing on $(-\infty, -2)$, $(-2, 2)$, and $(2, \infty)$. There are no intervals on which $y$ is increasing.

7. (a) Suppose $u$ and $v$ are in $[a, b]$ with $u < v$. Then, by the Mean Value Theorem, there exists a point $c$ in $(a, b)$ such that $f'(c) = \frac{f(v) - f(u)}{v - u}.$ By assumption, $f'(c) > 0$, and so $f(v) - f(u) = f'(c)(v - u) > 0.$ Hence $f(v) > f(u)$; thus $f$ is increasing on $[a, b]$.

9. Let $f(x) = 1 + \dfrac{1}{2}x - \sqrt{1 + x}.$ Then $f'(x) = \dfrac{1}{2} - \dfrac{1}{2\sqrt{1 + x}}.$ Hence $f'(x) > 0$ for all $x > 0$, and so $f$ is, by Problem 7, increasing on $[0, \infty)$. In particular, for any $x > 0$, $f(0) < f(x)$. Since $f(0) = 0$, it follows that $1 + \dfrac{1}{2}x - \sqrt{1 + x} > 0$ for all $x > 0$.

11. (a) $F(x) = x^2 + k$ is an antiderivative of $f$ for any constant $k$.

(c) $G(x) = -\cos(x) + k$ is an antiderivative of $g$ for any constant $k$.

(e) $\displaystyle{H(x) = \frac{1}{3}x^3 - \frac{3}{2}x^2 + k}$ is an antiderivative of $h$ for any constant $k$.

13. If $G$ is an antiderivative of $g$, then $G(x) = -\frac{1}{2}\cos(2x) + k$ for some constant $k$.

15. $k = -1$

Section 3.8

1. (a) $f$ has a maximum value of $12$ at $x = 4$ and a minimum value of $-4$ at $x = 0$.

(c) $g$ has a maximum value of $\sqrt{2}$ at $t = -\dfrac{\pi}{4}$ and a minimum value of $-\sqrt{2}$ at $t = \dfrac{3\pi}{4}$.

(e) $g$ has a maximum value of $0.5498$ at $x = -1.0678$ and at $x = 1.0678$, and a minimum value of $-1.6646$ at $x = -2$ and $x = 2$.

(g) $f$ has a maximum value of $3.9453$ at $t = 2.2889$ and a minimum value of $0$ at $t = 0$ and $t = \pi$.

3. The side parallel to the river should be $250$ yards long and the other two sides should be $125$ yards long.

5. $\$30.94$6. The total area is maximized when the entire length of the wire is used for the circle. The total area is minimized when a piece of length$ \dfrac{\pi}{4 + \pi} $is used for the circle and the remaining piece for the square. 7. (a)$ f $has local minimum of$ 5 $at$ x = 0 $.$ f $does not have a local maximum. (c)$ g $has a local maximum of$ 4 $at$ t = -2 $and a local minimum of$ 0 $at$ t = 0 $. (e)$ f $has a local maximum of$ 1 $at$ x = 0 $.$ f $does not have a local minimum. (g)$ g $has a local maximum of$ \displaystyle{\frac{6}{25}\sqrt{\frac{3}{5}}}$at$ \displaystyle{x = -\sqrt{\frac{3}{5}}}$and a local minimum of$ \displaystyle{-\frac{6}{25}\sqrt{\frac{3}{5}}}$at$ \displaystyle{x = \sqrt{\frac{3}{5}}}$. (i)$ g $has a local maximum of$ 0 $at$ t = 0 $and a local minimum of$ -\dfrac{256}{81} $at$ t = \dfrac{4}{3} $. 8. (a)$ f''(x) = 6 $(c)$ \displaystyle{\frac{d^2s}{dt^2} = \frac{8}{(2t - 1)^3}}$(e)$ \displaystyle{\frac{d^2x}{dt^2} = -20\sin(2t)\cos(4t) - 16\cos(2t)\sin(4t)}$9.$ f $has a maximum value on$ (0, \infty) $of$ \dfrac{1}{4} $at$ x = 2 $.$ f $does not have a minimum value on$ (0, \infty) $. 11.$ 10 \text{ feet} \times 10 \text{ feet} \times 10 \text{ feet} $13. The base of the bin should be$ 11.45 \text{ feet} \times 11.45 \text{ feet} $and the height should be$ 7.63 \text{ feet} $. 15.The radius of the can should be$ \displaystyle{\sqrt{\frac{250}{\pi}}}$and the height should be$ \displaystyle{4\sqrt{\frac{250}{\pi}}}$. 16. The radius of the can should be$ \sqrt{62.5} $and the height should be$ \displaystyle{\frac{8}{\pi}\sqrt{62.5}}$. 20. (b)$ L $is maximized when$ \displaystyle{\theta = \frac{11}{20}}$. The probabilty of$ AA $is$ 0.3025 $, the probability of$ Aa $is$ 0.4950 $, and the probabilty of$ aa $is$ 0.2025 $. 22. (a)$ v(t) = 3\pi\cos(\pi t) $(b)$ a(t) = -3\pi^2\sin(\pi t) $(c) From time$ t = 0 $until time$ t = \dfrac{1}{2} $, the object moves to the right from$ x = 0 $to$ x = 3 $and is slowing down. From time$ t = \dfrac{1}{2} $until time$ t = 1 $, the object moves to the left from$ x = 3 $to$ x = 0 $and is speeding up. From time$ t = 1 $until time$ t = \dfrac{3}{2} $, the object moves to the left from$ x = 0 $to$ x = -3 $and is slowing down. From time$ t = \dfrac{3}{2} $until time$ t = 2 $, the object moves to the right from$ x = -3 $to$ x = 0 $and is speeding up. Section 3.9 1. (a)$ f $is decreasing on$ \displaystyle{\left(-\infty, \frac{1}{2}\right)}$and increasing on$ \displaystyle{\left(\frac{1}{2}, \infty\right)}$. The graph of$ f $is concave up on$ (-\infty, \infty) $.$ f $has a local minimum of$ -\dfrac{1}{4} $at$ x = \dfrac{1}{2} $. The graph of$ f $does not have any inflection points or asymptotes. (c)$ g $is increasing on$ (-\infty, -2) $and$ (0, \infty) $, and decreasing on$ (-2, 0) $. The graph of$ g $is concave down on$ (-\infty, -1) $and concave up on$ (-1, \infty) $.$ g $has a local maximum of$ 4 $at$ x = -2 $and a local minimum of$ 0 $at$ x = 0 $.$ (-1, 2) $is an inflection point. The graph of$ g $does not have any asymptotes. (e)$ f $is increasing on$ (-\infty, -1) $and$ (1, \infty) $, and decreasing on$ (-1, 1). $The graph of$ f $is concave down on$ (-\infty, 0) $and concave up on$ (0, \infty) $.$ f $has a local maximum of$2$at$ x = -1 $and a local minimum of$ -2 $at$ x = 1 $.$ (0, 0) $is an inflection point. The graph of$ f $does not have any asymptotes. (g)$ h $is increasing on$ \displaystyle{\left(-\infty, -\sqrt{\frac{3}{5}}\right)}$and$ \displaystyle{\left(\sqrt{\frac{3}{5}}, \infty\right)}$, and decreasing on$ \displaystyle{\left(-\sqrt{\frac{3}{5}}, 0\right)}$and$ \displaystyle{\left(0, \sqrt{\frac{3}{5}}\right)}$. The graph of$ h $is concave down on$ \displaystyle{\left(-\infty, -\sqrt{\frac{3}{10}}\right)}$and$ \displaystyle{\left(0, \sqrt{\frac{3}{10}}\right)}$, and concave up on$ \displaystyle{\left(-\sqrt{\frac{3}{10}}, 0\right)}$and$ \displaystyle{\left(\sqrt{\frac{3}{10}}, \infty\right)}$.$ h $has a local maximum of$ \displaystyle{\frac{6}{25}\sqrt{\frac{3}{5}}}$at$ \displaystyle{x = -\sqrt{\frac{3}{5}}}$and a local minimum of$ \displaystyle{-\frac{6}{25}\sqrt{\frac{3}{5}}}$at$ \displaystyle{x = \sqrt{\frac{3}{5}}}$.$ \displaystyle{\left(-\sqrt{\frac{3}{10}}, \frac{21}{100}\sqrt{\frac{3}{10}}\right)}$,$ (0, 0) $, and$ \displaystyle{\left(\sqrt{\frac{3}{10}}, -\frac{21}{100}\sqrt{\frac{3}{10}}\right)}$are inflection points. The graph of$ h $does not have any asymptotes. (i)$ g $is decreasing on$ (-\infty, 1) $and$ (1, \infty) $. The graph of$ g $is concave down on$ (-\infty, 1) $and concave up on$ (1, \infty) $. The line$ z = 1 $is a vertical asymptote and the line$ y = 0 $(that is, the$ z $-axis) is a horizontal asymptote.$ g $does not have any local extreme values and the graph of$ g $does not have any inflection points. (k)$f$is decreasing on$ (-\infty, 0) $and$ \displaystyle{\left(0, \frac{3}{2}\right)}$, and increasing on$ \displaystyle{\left(\frac{3}{2}, 0\right)}$. The graph of$ f $is concave up on$ (-\infty, 0) $and$ (1, \infty) $, and concave down on$ (0, 1) $.$ f $has a local minimum of$ -\dfrac{27}{16} $at$ x = \dfrac{3}{2} $.$ (0,0) $and$ (1, -1) $are inflection points. The graph of$ f $does not have any asymptotes. (m)$ h $is decreasing on$ (-\infty, -2) $,$ (-2, 2) $, and$ (2, \infty) $. The graph of$ h $is concave down on$ (-\infty, -2) $and$ (0, 2) $, and concave up on$ (-2, 0) $and$ (2, \infty) $.$ (0, 0) $is an inflection point. The lines$ t = -2 $and$ t = 2 $are vertical asymptotes and the line$ y = 0 $(that is, the$ t $-axis) is a horizontal asymptote.$ h $does not have any local extreme values. (o)$ f $is decreasing on$ (-\infty, 0) $and increasing on$ (0, \infty) $. The graph of$ f $is concave down on$ \displaystyle{\left(-\infty, -\frac{1}{\sqrt{3}}\right)}$and$ \displaystyle{\left(\frac{1}{\sqrt{3}}, \infty\right)}$, and concave up on$ \displaystyle{\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)}$.$ f $has a local minimum of$ 0 $at$ x = 0 $.$ \displaystyle{\left(-\frac{1}{\sqrt{3}}, \frac{1}{4}\right)}$and$ \displaystyle{\left(\frac{1}{\sqrt{3}}, \frac{1}{4}\right)}$are inflection points. The line$ y = 1 $is a horizontal asymptote. (q)$ x $is decreasing on$ (-\infty, 1) $and$ (1, \infty) $. The graph of$ x $is concave down on$ (-\infty, 1) $and concave up on$ (1, \infty) $. The line$ t = 1 $is a vertical asymptote and the line$ y = 2 $is a horizontal asymptote.$ x $does not have any local extreme values. The graph of$ x \$ does not have any inflection points.

3. (c) There is only one function which satisfies these conditions.

4. (c) There is only one function which satisfies these conditions.

5. (c) There is only one function which satisfies these conditions.

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