Chapter 2 Challenge Problems Solutions

SOLUTION

1. The maximum height occurs when \(x = 0 \col\) \[ \ba y &= 211.49 - 20.96 \cosh 0 \nl &= 211.49 - 20.96 = \boxed{190.53 \un{m}} \ea \] (Note: The Gateway Arch's maximum height is \(192 \un{m},\) but the central arch is \(190.53 \un{m}\) tall.)

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2. Using a graphing calculator, we find the solutions of \[y = 211.49 - 20.96 \cosh 0.03291765x = 80\] to be \(x = \pm 76.647.\) Hence, the central curve is \(80 \un{m}\) tall at the points located \(76.647 \un{m}\) from the center.

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3. The derivative function is \[ \ba y' &= -20.96(0.03291765) \sinh 0.03291765x \nl &= -0.689953944 \sinh 0.03291765x \pd \ea \] So at \(x = \pm 76.747,\) the slopes are \[ \ba y'(\pm 76.647) &= -0.689953944 \sinh 0.03291765 (\pm 76.647) \nl &\approx \mp 4.273 \pd \ea \] So the slope at \((-76.747, 80)\) is approximately \(4.273,\) and the slope at \((76.747, 80)\) is approximately \(-4.273.\)

SOLUTION

The word tangent means that the line intersects \(y = g(x)\) at \(x = a\) with slope \(g'(a).\) We anticipate that the tangent line to \(f\) at \(x = 2\) is also tangent to \(g\) in the third quadrant. So we are considering only one tangent line. We see \(f(2) = 4\) and \(f'(2) = 4,\) so an equation of this tangent line is \[y - 4 = 4(x - 2) \or y = 4x - 4 \pd\] This line has slope \(4\) and must be tangent to \(g(x) = -x^2 - C,\) whose derivative is \(g'(x) = -2x.\) Equating slopes shows \[-2x = 4 \implies x = -2 \pd\] Thus, the line \(y = 4x - 4\) is tangent to \(f\) at \(x = 2\) and tangent to \(g\) at \(x = -2.\) The word tangent means that this line also intersects \(g(x)\) at \(x = -2.\) When \(x = -2,\) the line's \(y\)-value is \[4(-2) - 4 = -12 \cma\] so \(g(-2) = -12.\) Thus, we see \[g(-2) = -(-2)^2 - C = -12 \implies \boxed{C = 8}\]

SOLUTION The line \(T(x)\) is the linearization of \(f(x)\) at \(x = 3\) if and only if \(T\) is tangent to \(f\) at \(3\)—that is, \(T(3) = f(3)\) and \(T'(3) = f'(3).\) Observe that \[f(3) = 9 + C \and T(3) = 13 \pd\] So we see \[9 + C = 13 \implies C = 4 \pd\] Also, \(f'(3) = T'(3) = 6.\) Thus, the graphs of \(f\) and \(T\) are tangent at \(x = 3\) for \(\boxed{C = 4}.\) You can view an interactive plot of \(f\) and \(T\) at https://www.desmos.com/calculator/4lpyxm3m1v?lang=en.

SOLUTION The derivative of the reciprocal function is \[\deriv{}{x} \par{\frac{1}{f(x)}} = -\frac{f'(x)}{[f(x)]^2} \pd\] If the slope to \(f\) equals the slope to \(1/f,\) then their derivatives are equal: \[f'(x) = -\frac{f'(x)}{[f(x)]^2} \pd\] The only solution to this equation is \(f'(x) = 0.\) If we disregard this solution and instead cancel the \(f'(x)\)'s on both sides, then we get \[ \ba 1 &= -\frac{1}{[f(x)]^2} \nl [f(x)]^2 &= -1 \cma \ea \] which has no real solutions. Hence, the only possible value of the slope is \(\boxed{0}.\)

SOLUTION Observe that the limit definition of \(f'(a)\) is \[f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \pd\] Let's multiply the numerator and denominator of the given limit by \((\sqrt x + \sqrt a);\) doing so shows \[ \ba \lim_{x \to a} \frac{f(x) - f(a)}{\sqrt x - \sqrt a} &= \lim_{x \to a} \frac{f(x) - f(a)}{\sqrt x - \sqrt a} \frac{\par{\sqrt x + \sqrt a}}{\par{\sqrt x + \sqrt a}} \nl &= \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \par{\sqrt x + \sqrt a} \nl &= f'(a) \lim_{x \to a} \par{\sqrt x + \sqrt a} \nl &= \boxed{2 f'(a) \sqrt a} \ea \]

SOLUTION

Region \(R\) is a triangle whose hypotenuse is the tangent line. The derivative of \(1/x\) is, by the Power Rule, \[-x^{-2} = -\frac{1}{x^2} \pd\] Thus, the slope to the curve \(1/x\) at \(x = a\) is \(-1/a^2.\) The point of tangency is \((a, 1/a),\) so an equation of the tangent line is \[y - \frac{1}{a} = -\frac{1}{a^2}(x - a) \or y = -\frac{x}{a^2} + \frac{2}{a} \pd\] To find the triangle's base \(b,\) we set \(y = 0 \col\) \[0 = -\frac{b}{a^2} + \frac{2}{a} \implies b = 2a \pd\] The triangle's height \(h\) is found by letting \(x = 0 \col\) \[h = -\frac{0}{a^2} + \frac{2}{a} \implies h = \frac{2}{a} \pd\] Hence, the triangle's area is \[\tfrac{1}{2} bh = \tfrac{1}{2}(2a) \par{\tfrac{2}{a}} = \boxed{2}\] Surprisingly, the area does not depend on \(a \exclam\)

SOLUTION Note that \(e^{\cos(\pi/2)}\) \(= e^0\) \(= 1.\) So the limit is \[\lim_{x \to \pi/2} \frac{e^{\cos x} - e^{\cos(\pi/2)}}{x - \dfrac{\strut \pi}{2}} \pd\] This is the limit definition of the derivative of \(e^{\cos x}\) at \(x = \pi/2.\) Hence, the limit equals \[ \ba \deriv{}{x} \par{e^{\cos x}} \intEval_{x = \pi/2} &= \par{- e^{\cos x} \sin x} \intEval_{x = \pi/2} \nl &= \boxed{-1} \ea \]

SOLUTION By the limit definition of a derivative, \[\deriv{}{x} \sin(x + c) = \lim_{h \to 0} \frac{\sin(x + c + h) - \sin(x + c)}{h} \pd\] We now use the addition identity for sine, treating \((x + c)\) as one term. Doing so gives \[ \ba \deriv{}{x} \sin(x + c) &= \lim_{h \to 0} \frac{\sin(x + c) \cos h + \cos(x + c) \sin h - \sin(x + c)}{h} \nl &= \lim_{h \to 0} \frac{\sin(x + c) (\cos h - 1)}{h} + \lim_{h \to 0} \frac{\sin h \cos(x + c)}{h} \nl &= \sin(x + c) \lim_{h \to 0} \frac{\cos h - 1}{h} + \cos(x + c) \lim_{h \to 0} \frac{\sin h}{h} \pd \ea \] Recall that as \(h \to 0,\) \((\cos h - 1)/h \to 0\) and \(\sin h/h \to 1.\) Thus, we see \[\deriv{}{x} \sin(x + c) = \sin(x + c) \cdot 0 + \cos(x + c) \cdot 1 = \cos(x + c) \cma\] as requested.

SOLUTION For the curves of \(f(x) = x^2\) and \(g(x) = C + \ln x\) to intersect once, they must be tangent at some point \(x = a.\) Recall that the word tangent means \(f(a) = g(a)\) and \(f'(a) = g'(a).\) Observe that \(f'(x) = 2x\) and \(g'(x) = 1/x.\) Equating the derivatives at \(a\) shows \[2a = \tfrac{1}{a} \implies a = \tfrac{1}{\sqrt 2} \pd\] Now we substitute this value into the equation \(x^2 = C + \ln x,\) seeing \[ \ba \par{\tfrac{1}{\sqrt 2}}^2 &= C + \ln \tfrac{1}{\sqrt 2} \nl \implies C &= \boxed{\tfrac{1}{2} - \ln \tfrac{1}{\sqrt 2}} \ea \]

SOLUTION The bounded region is a triangle of width \(e.\) To find its height, we must calculate the \(y\)-intercept of the normal line. Recall from Section 2.2 that a normal line is perpendicular to a tangent line. The derivative of \(\ln x\) is \(1/x,\) so the tangent to the curve at \(x = e\) has slope \(1/e.\) An equation of the tangent line is \[y - \ln e = \frac{1}{e} (x - e) \or y = \frac{x}{e} \pd\] The normal's slope is the negative reciprocal of \(1/e\)—that is, \(-e.\) An equation of the normal line is therefore \[y - \ln e = -e(x - e) \or y = 1 - e(x - e) \pd\] When \(x = 0\) we see \[y = 1 - e(0 - e) = 1 + e^2 \pd\] The \(y\)-intercept \((1 + e^2)\) is the height of the bounded triangle. Thus, its area is \[\tfrac{1}{2} (e)(1 + e^2) = \boxed{\tfrac{1}{2} (e + e^3)}\]

SOLUTION Recall that \(f^{(n)}(x)\) represents the \(n\)th derivative of \(f(x).\) To find the \(n\)th derivative of \(f,\) we write out some derivatives to identify a pattern. The first four derivatives of \(f\) are as follows: \[ \ba f'(x) &= k \cos kx \nl f''(x) &= -k^2 \sin kx \nl f'''(x) &= -k^3 \cos kx \nl f^{(4)}(x) &= k^4 \sin kx \pd \ea \]

Even \(n\) The second derivative of \(f\) has a negative sign, but its fourth derivative has a positive sign. Likewise, \(n = 6\) features a negative sign and \(n = 8\) has a positive sign. The pattern for this sign alternation is represented by \[(-1)^{n/2} \pd\] Thus, we conclude that \[f^{(n)}(x) = \boxed{(-1)^{n/2} k^n \sin kx \cmaa n \textrm{ even}}\]

Odd \(n\) The first derivative of \(f\) has a positive sign, but its third derivative has a negative sign. If we kept repeating the pattern, then we would have a positive sign for \(n = 5\) and a negative sign for \(n = 7.\) The pattern for this sign alternation is represented by \[(-1)^{1 + (n + 1)/2} = (-1)^{(n - 1)/2} \pd\] Thus, we have \[f^{(n)}(x) = \boxed{(-1)^{(n - 1)/2} k^n \cos kx \cmaa n \textrm{ odd}}\]

SOLUTION

The word tangent means that the graphs of \(y = x^2 + b\) and \(y = \abs x\) intersect each other with the same slope. The graph of \(\abs x\) has slope \(-1\) for \(x \lt 0\) and slope \(1\) for \(x \gt 0.\) The derivative function for the parabola is \[\deriv{}{x} \par{x^2 + b} = 2x \pd\] We equate this expression to \(1\) and \(-1\) to find \[ \baat{2} 2x &= 1 \lspace 2x &&= -1 \nl x &= \tfrac{1}{2} \scol \lspace x &&= -\tfrac{1}{2} \pd \eaat \] At each value of \(x,\) \(\abs x = 1/2.\) Thus, the parabola must also reach the \(y\)-value \(1/2\) when \(x = 1/2\) and \(x = -1/2 \col\) \[y = \par{\pm \tfrac{1}{2}}^2 + b = \tfrac{1}{2} \implies \boxed{b = \tfrac{1}{4}}\]

SOLUTION

1. We have \(f'(x) = ke^{kx}.\) So an equation of the line tangent to \(f\) at \((a, e^{ka})\) is \[y - e^{ka} = k e^{ka} (x - a) \pd\] This line has an \(x\)-intercept at \(x = c,\) so substituting \((c, 0)\) gives \[ \ba 0 - e^{ka} &= k e^{ka} (c - a) \nl \implies c &= a - \frac{1}{k} \pd \ea \] Then \[ \ba \lim_{k \to \infty} c &= \lim_{k \to \infty} \par{a - \frac{1}{k}} \nl &= \boxed a \ea \] This result is logical: as \(k\) increases, the curve \(f\) becomes steeper and so \(\ell\) becomes closer to the vertical line \(x = a.\)

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2. As \(k \to 0^+,\) \(1/k \to \infty.\) Accordingly, as requested, \[\underbrace{a - \frac{1}{k}}_{c} \to -\infty \as k \to 0^+ \pd\] Geometrically, \(k \to 0^+\) means the curve \(f(x) = e^{kx}\) becomes flatter, so line \(\ell\) becomes more horizontal. By making \(k\) arbitrarily close to \(0\) (with \(k \gt 0\)), the \(x\)-intercept of line \(\ell\) continually shifts to the left, away from the \(y\)-axis.

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3. Differentiating both sides of \(c = a - 1/k\) with respect to \(k,\) we get \[ \ba \deriv{c}{k} &= \frac{1}{k^2} \nl \implies \dd c &= \frac{1}{k^2} \di k \pd \ea \] Taking \(k = 2,\) \(\Delta c \approx \dd c,\) and \(\di k = \Delta k\) \(= 0.1,\) we have \[\Delta c \approx \frac{1}{(2)^2} (0.1) = \boxed{\frac{1}{40}}\]

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4. Differentiating both sides of \(c = a - 1/k\) with respect to time, and noting that \(a\) is constant, we get \[ \deriv{c}{t} = \frac{1}{k^2} \deriv{k}{t} \pd \] We are given \(\textderiv{k}{t} = 2\) and \(k = 4,\) which we substitute to get \[ \ba \deriv{c}{t} \intEval_{k = 4} = \frac{1}{4^2} (2) = \boxed{\frac{1}{8}} \ea \]

SOLUTION

1. The Quotient Rule gives the derivative of \(f\) to be \[ \ba f'(x) &= \frac{(0) \par{x^2 + 4x + k} - 1(2x + 4)}{\par{x^2 + 4x + k}^2} \nl &= \frac{-2x - 4}{\par{x^2 + 4x + k}^2} \cma \ea \] as requested.

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2. If \(k = 4,\) then \(f\) becomes \[f(x) = \frac{1}{x^2 + 4x + 4} \pd\] The function \(f\) has a vertical asymptote when its denominator is \(0\) but its numerator is nonzero. So we equate the denominator to \(0\) to find \[ \ba x^2 + 4x + 4 &= 0 \nl (x + 2)^2 &= 0 \nl \implies x &= -2 \pd \ea \] Thus, the only vertical asymptote to the graph of \(f\) for \(k = 4\) is \(\boxed{x = -2}.\)

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3. The graph of \(f\) has a horizontal tangent at some \(x\) if and only if \(f'(x) = 0.\) We see \[f'(x) = \frac{-2x - 4}{\par{x^2 + 4x + k}^2} = 0 \implies \boxed{x = -2}\] Thus, the family of graphs \(f(x) = 1/(x^2 + 4x + k)\) has a horizontal tangent at \(x = -2\) if \(k \ne 4,\) but has a vertical asymptote if \(k = 4.\)

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4. Observe that the denominator of \(f'\) has a higher degree than the numerator of \(f'.\) Hence, the denominator grows faster than the numerator as \(x \to \infty\) or \(x \to -\infty,\) pushing the value of \(f'\) toward \(0.\) Accordingly, \[\lim_{x \to -\infty} f'(x) = 0 \and \lim_{x \to \infty} f'(x) = 0 \pd\] These limits tell us that the slope to \(f\) approaches \(0\) as \(x \to \infty\) or \(x \to -\infty.\) The graph of \(f\) has one horizontal asymptote, \(y = 0,\) so the slope approaching \(0\) is expected as \(x\) approaches infinity or negative infinity.

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5. The graph of \(f(x) = 1/(x^2 + 4x + k)\) has no vertical asymptotes if and only if the quadratic equation \(x^2 + 4x + k = 0\) has no real solutions. The discriminant of the quadratic is \[(4)^2 - 4(1)(k) = 16 - 4k \pd\] (See Section 0.7 to review how the discriminant shows how many real zeros a quadratic has.) If the quadratic equation has no real solutions, then the discriminant must be negative: \[16 - 4k \lt 0 \implies k \gt 4 \pd\] Thus, \(f(x) = 1/(x^2 + 4x + k)\) has no vertical asymptotes for \(k \gt 4\) because its denominator is never \(0.\)

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You can view the graph of \(f\) at https://www.desmos.com/calculator/lbzuhjdpo9?lang=en.

SOLUTION The functions' derivatives are, respectively, \[ \ba f'(x) &= x^3 + 12 \cma \nl g'(x) &= 3x^2 + 4x \cma \nl h'(x) &= 4x + 27 \pd \ea \] The tangents to all three graphs are parallel if and only if they have the same slope, namely, \(f' = g' = h'.\) There are three cases we need to satisfy: \(f' = g',\) \(f' = h',\) and \(g' = h'.\)

1. Solving \(f' = g',\) we see \[ \ba x^3 + 12 &= 3x^2 + 4x \nl x^3 - 3x^2 - 4x + 12 &= 0 \nl x^2 (x - 3) - 4(x - 3) &= 0 \nl (x - 3) (x^2 - 4) &= 0 \nl \implies x &= -2 \cma x = 2 \cma x = 3 \pd \ea \]
2. Equating \(f' = h',\) we see \[ \ba x^3 + 12 &= 4x + 27 \nl x^3 - 4x - 15 &= 0 \pd \ea \] Out of all the solutions from the first case—that is, \(x = -2,\) \(x = 2,\) and \(x = 3\)—only \(x = 3\) solves this equation.
3. We solve \(g' = h',\) as follows: \[ \ba 3x^2 + 4x &= 4x + 27 \nl x^2 &= 9 \nl \implies x &= -3 \cma x = 3 \pd \ea \]

Hence, the only solution to \(f' = g' = h'\) is \(\boxed{x = 3}.\)

SOLUTION

Let's take the \(x\)-axis to be the ground and the \(y\)-axis to be the cable's axis of symmetry. Then the cable's shape is a catenary: \[f(x) = a \cosh \par{\frac{x}{a}} + c\] for any constants \(a\) and \(c.\) Our axis orientation provides the points \(f(0) = 0.7,\) \(f(-2.5) = 3,\) and \(f(2.5) = 3.\) Substituting \(f(0) = 0.7,\) we have \begin{align} f(0) = a \cosh 0 + c &= 0.7 \nonum \nl a + c &= 0.7 \pd \label{eq:chain-theta-1} \end{align} Likewise, substituting \(f(2.5) = 3\) gives \begin{equation} a \cosh \par{\frac{2.5}{a}} + c = 3 \pd \label{eq:chain-theta-2} \end{equation} From \(\eqref{eq:chain-theta-1},\) \(c = 0.7 - a;\) substituting this expression into \(\eqref{eq:chain-theta-2}\) gives \[a \cosh \par{\frac{2.5}{a}} + 0.7 - a = 3 \pd\] By graphing, we find \(a = 1.642;\) then \(c = 0.7 - 1.642\) \(= -0.942.\) So the shape of the hanging cable is \[f(x) = 1.642 \cosh \par{\frac{x}{1.642}} - 0.942 \pd\] The slope of the tangent to the cable at the right end is \[f'(2.5) \approx 2.183 \pd\] By similar triangles (see the figure), the angle is \[\theta = \atan \par{\frac{1}{2.183}} \approx 0.430 = \boxed{24.6 \degree}\]

SOLUTION The problem becomes trivial if the identity of line \(\ell\) is known; this is our objective. First let's express the coordinates of point \(C(0, y_C)\) in terms of \(r.\) Note that point \(R\) is a point of tangency, so at \(R\) the graph \(y = \abs x\) and the circle share the same slope, \(1.\) Accordingly, an equation of the normal line (see Section 2.2) at \(R\) is \[y - y_C = -\frac{1}{1} (x - 0) \or y = y_C - x \pd\] The circle and \(y = \abs x\) also have the same \(y\)-coordinate at \(R,\) so we see \[y_C - x = x \implies x = \tfrac{1}{2} y_C \pd\] Hence, the coordinates of \(R\) are \(\par{\tfrac{1}{2} y_C, \tfrac{1}{2} y_C}.\) By the Distance Formula, we have \[ \ba r &= \length{CR} \nl &= \sqrt{\par{0 - \tfrac{1}{2} y_C}^2 + \par{y_C - \tfrac{1}{2} y_C}^2} \nl &= \frac{y_C}{\sqrt 2} \pd \ea \] It follows that \(y_C = r \sqrt 2.\) Hence, the identity of line \(\ell\) is \[ \ba y &= y_C + r \nl &= r \sqrt 2 + r \nl &= r (1 + \sqrt 2) \pd \ea \] Line \(\ell\) therefore intersects \(y = \abs x\) in the first quadrant at \(Q(r(1 + \sqrt 2), r(1 + \sqrt 2))\) and in the second quadrant at \(P(-r(1 + \sqrt 2), r(1 + \sqrt 2)).\) Therefore, \(\Delta OPQ\) has a width of \(2r(1 + \sqrt 2)\) and a height of \(r(1 + \sqrt 2).\) So its area is \[ \ba A &= \tfrac{1}{2} \parbr{2r(1 + \sqrt 2)} \parbr{r(1 + \sqrt 2)} \nl &= r^2 (1 + \sqrt 2)^2 \nl &= \boxed{r^2 \par{2 + 2 \sqrt 2 + 1}} \approx 5.828 r^2 \pd \ea \]

SOLUTION Observe that \(f(x) \to (\pi/2)^+\) as \(x \to \infty.\) Consequently, \(\inv f(x) \to \infty\) as \(x \to (\pi/2)^+.\) We have \[ \baf &&f'(x) &= -\frac{1}{x^2} + \frac{1}{x^2 + 1} = -\frac{1}{x^4 + x^2} &\nl \laWord{and} &&\lim_{x \to (\pi/2)^+} \par{\inv f}'(x) &= \lim_{x \to (\pi/2)^+} \frac{1}{f' \par{\inv f(x)}} \pd \eaf \] To evaluate this limit, we substitute \(u = \inv f(x).\) Then \(u \to \infty\) as \(x \to (\pi/2)^+,\) so our limit becomes \[ \ba \lim_{u \to \infty} \frac{1}{f'(u)} &= \lim_{u \to \infty} \frac{1}{-\dfrac{1}{u^4 + u^2}} \nl &= \lim_{u \to \infty} - \par{u^4 + u^2} \nl &= -\infty \pd \ea \]