Chapter 3 Challenge Problems Solutions

SOLUTION The function must satisfy \(x(0) = 0.1.\) In the given function \(x(t),\) substituting \(t = 0\) and equating the result to \(0.1\) show \begin{align} x(0) = Ce^{0} \cos(0 - \phi) &= 0.1 \nonum \nl C \cos(-\phi) &= 0.1 \nonum \nl \implies C &= \frac{0.1}{\cos(-\phi)} \pd \label{eq:phi-1} \end{align} Because the object starts at rest, its initial velocity is \(0\) and so \(v(0) = 0.\) Velocity is the time derivative of position, so differentiating \(x(t)\) yields the following velocity function: \[v(t) = -4C e^{-4t} \cos(4t - \phi) - 4C e^{-4t} \sin(4t - \phi) \pd\] Substituting \(t = 0\) and equating the result to \(0,\) we attain \begin{align} v(0) = -4C e^{0} \cos(0 - \phi) - 4C e^{0} \sin(0 - \phi) &= 0 \nonum \nl -4C \cos(-\phi) - 4C \sin(-\phi) &= 0 \pd \label{eq:phi-2} \end{align} Substituting \(\eqref{eq:phi-1}\) into \(\eqref{eq:phi-2}\) shows \[ \ba -4 \par{\frac{0.1}{\cos(-\phi)}} \cos(-\phi) - 4 \par{\frac{0.1}{\cos(-\phi)}} \sin(-\phi) &= 0 \nl -0.4 - 0.4 \tan(-\phi) &= 0 \nl \tan(-\phi) &= -1 \nl \implies \phi &= \boxed{\frac{\pi}{4}} \ea \] [Another solution is \(\phi = 5 \pi/4,\) but we choose \(\phi = \pi/4\) to ensure \(C \gt 0\) in Equation (1).] Substituting this result back into \(\eqref{eq:phi-1}\) produces \[C = \frac{0.1}{\cos(-\pi/4)} = \boxed{\frac{1}{5 \sqrt 2}} \approx 0.141 \pd\]

SOLUTION The general form of a quadratic function is \[C(x) = ax^2 + bx + c \pd\] Our goal is to fit the parameters \(a,\) \(b,\) and \(c\) to satisfy \(C(0) = 500,\) \(C(100) = 700,\) and \(C(200) = 950.\)

Quadratic Fitting Substituting \(C(0) = 500\) shows \[ \ba C(0) = a(0)^2 + b(0) + c &= 500 \nl \implies c &= 500 \pd \ea \] Then substituting \(C(100) = 700\) yields \begin{align} C(100) = a(100)^2 + b(100) + 500 &= 700 \nonum \nl 10000a + 100b &= 200 \pd \label{eq:C-1} \end{align} Likewise, substituting \(C(200) = 950\) gives \begin{align} C(200) = a(200)^2 + b(200) + 500 &= 950 \nonum \nl 40000a + 200b &= 450 \pd \label{eq:C-2} \end{align} Multiplying \(\eqref{eq:C-1}\) by \(2\) and subtracting it from \(\eqref{eq:C-2},\) we get \[ \ba 40000a + 200b - 2(10000a + 100b) &= 450 - 2(200) \nl 20000a &= 50 \nl \implies a &= 0.0025 \pd \ea \] (See Section 0.5 to review solving systems of linear equations.) To solve for \(b,\) we substitute \(a = 0.0025\) back into either equation of the system—for example, into \(\eqref{eq:C-1} \col\) \[ \ba 10000(0.0025) + 100b &= 200 \nl \implies b &= 1.75 \pd \ea \] Hence, the identity of the cost function is \[C(x) = \boxed{0.0025x^2 + 1.75x + 500}\]

Predicted Cost The predicted cost of producing \(300\) units is \[C(300) = 0.0025(300)^2 + 1.75(300) + 500 = \boxed{\$1250}\]

SOLUTION

To connect \(f,\) \(f',\) and \(f'',\) we consider important features—extrema, zeros, intervals of increasing or decreasing—of any one graph and see how well they align with the other two graphs. Let's begin with graph I, whose minimum is at point \(A.\) If II or III is the derivative of I, then either graph must cross the \(x\)-axis at the \(x\)-coordinate of \(A.\) But neither II nor III crosses the \(x\)-axis when graph I attains its minimum. In addition, I is decreasing for all \(x\) to the left of \(A,\) but neither II nor III is entirely negative over the same interval. Thus, II and III cannot be derivatives of I, so graph I must be \(f''.\) Now we pay attention to another graph—for example, II, which has extrema at points \(B,\) \(C,\) and \(D.\) The question is whether II is \(f\) or \(f'.\) Graph III intersects the \(x\)-axis when II attains its extrema. Graph III is positive between \(B\) and \(C,\) the region over which II is increasing, and is negative between \(C\) and \(D,\) when II is decreasing. Therefore, III must be the derivative of II. So II is \(f,\) III is \(f',\) and I is \(f''.\)

SOLUTION Differentiating gives \(f'(x) = 3x^2.\) So Newton's Method gives \[ \ba x_2 &= x_1 - \frac{f(x_1)}{f'(x_1)} \nl &= x_1 - \frac{\par{x_1}^3 - 5}{3 \par{x_1}^2} \nl &= x_1 - \frac{x_1}{3} + \frac{5}{3 \par{x_1}^2} \nl &= \frac{2 x_1}{3} + \frac{5}{3 \par{x_1}^2} \pd \ea \] Equating \(x_2 = 1\) gives \[ \ba 1 &= \frac{2 x_1}{3} + \frac{5}{3 \par{x_1}^2} \nl 3 \par{x_1}^2 &= 2 \par{x_1}^3 + 5 \pd \ea \] By inspection, a solution to the preceding equation is \(x_1 = -1.\) Using polynomial division, we find \[2 \par{x_1}^3 - 3 \par{x_1}^2 + 5 = \par{x_1 + 1} \parbr{2 \par{x_1}^2 - 5 x_1 + 5} \pd\] The discriminant of the expression in brackets is \((-5)^2 - 4(2)(5)\) \(= -15.\) Since this value is negative, the quadratic expression has no zeros. Hence, the only solution is \(\boxed{x_1 = -1}.\)

SOLUTION To apply L'Hôpital's Rule, \(\lim_{x \to 0} q(x)\) must either be in the indeterminate form \(\indZero\) or \(\indInfty.\) Since \(p\) is differentiable, \(p\) is continuous and so \(\lim_{x \to 0} p(x) = p(0).\) The denominator approaches \(0\) as \(x\) approaches \(0,\) so \(\lim_{x \to 0} q(x)\) is in the indeterminate form \(\indZero.\) Thus, \[ \ba \lim_{x \to 0} (2[p(x)]^3 + 4) &= \lim_{x \to 0} (x^2 - 6x) = 0 \nl 2p(0)^3 + 4 &= 0 \nl \implies p(0) &= \boxed{-\sqrt[3]{2}} \nl \ea \] Applying L'Hôpital's Rule gives \[ \ba \lim_{x \to 0} q(x) &= \lim_{x \to 0} \frac{6[p(x)]^2 \cdot p'(x)}{2x - 6} \nl &= \frac{6[p(0)]^2 \cdot p'(0)}{-6} \nl &= -(2)^{2/3} \cdot p'(0) \nl \ea \] We are given \(\lim_{x \to 0} q(x) = 12,\) so \[ -(2)^{2/3} \cdot p'(0) = 12 \implies p'(0) = \boxed{\frac{-12}{\sqrt[3]{4}}} \pd \]

SOLUTION By inspection, we see \[(0)^3 + e^0 \equalsCheck 1 \cma\] so \(x = 0\) satisfies the equation. But to prove that \(x = 0\) is the only solution, we let \(f(x) = x^3 + e^x - 1\) and show that \(f\) has only one zero, \(x = 0.\) Observe that \[f'(x) = 3x^2 + e^x \gt 0 \cma\] so \(f(x)\) is increasing for all \(x.\) Hence, \(f(x) \lt f(0)\) \(= 0\) for all \(x \lt 0;\) likewise, \(f(x) \gt f(0) = 0\) for all \(x \gt 0.\) Thus, \(x = 0\) is the only root of \(f,\) meaning the equation \(x^3 + e^x = 1\) has only one solution, \(x = 0.\)

SOLUTION

1. If \(p > 0,\) then as \(x \to \infty,\) \[\ln x \to \infty \and x^p \to \infty \pd\] Therefore, \(\lim_{x \to \infty} k(x)\) is in the indeterminate form \(\indInfty.\) We then apply L'Hôpital's Rule to find \[ \ba \lim_{x \to \infty} \frac{\ln x}{x^p} &= \lim_{x \to \infty} \frac{1/x}{px^{p - 1}} \nl &= \lim_{x \to \infty} \frac{1}{px} \nl &= \boxed{0} \ea \]

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2. Let's introduce a new variable \(q\) such that \(q = -p.\) We then have \[\lim_{x \to \infty} k(x) = \lim_{x \to \infty} x^q \ln x \cmaa q \gt 0 \pd\] As \(x \to \infty,\) \(x^q\) and \(\ln x\) both approach \(\infty,\) so their product is also \(\infty.\) Therefore, \[\lim_{x \to \infty} k(x) = \infty\] (or alternatively, \(\lim_{x \to \infty} k(x)\) does not exist).

SOLUTION

Let \(x\) be the distance from the intersection to car B, and let \(y\) be the distance between the intersection and car A. Since car A travels at \(45\) miles per hour, its distance function with respect to time is \(y = 45t\) (where \(t\) is in hours after \(3\) pm). To write car B's distance function, we note that the car approaches the intersection one hour after 3 pm. We therefore write \(x = 55 - 55t.\) The Pythagorean Theorem gives the distance function to be \[d = \sqrt{x^2 + y^2} = \sqrt{(45t)^2 + (55 - 55t)^2} \pd\] We need to minimize \(d,\) but for convenience we choose to minimize \(d^2\) because both functions have the same critical numbers. Let \(q(t) = d^2;\) then \[q\,'(t) = 90(45t) - 110(55 - 55t) \pd\] Solving \(q\,'(t) = 0\) gives \[t = \frac{110(55)}{90(45) + 110(55)} = 0.599 \pd\] The domain of \(q\) is \(t \geq 0.\) Using a calculator, we find \(q(0) = 3025\) and \(q(0.599) = 1212.995.\) Also, \(\lim_{t \to \infty} q = \infty.\) This analysis asserts that \(t = 0.599\) must correspond to the absolute minimum of \(q\) (and, therefore, of \(d\)). Since \(0.599\) hour is \(0.599 \cdot 60 = 35.94\) minutes, we conclude that the cars are closest together at \(\boxed{\text{3:36}}\) pm.

SOLUTION Let's define a new function \(h(x) = f(x) - g(x).\) Since \(f'(x)\) and \(g'(x)\) exist on \((a,b),\) \(h\) is differentiable on \((a,b)\) and (because \(f\) and \(g\) are continuous) \(h\) is continuous on \([a,b].\) Also, for all \(x \in (a, b),\) \[ h'(x) = f'(x) - g'(x) \gt 0 \pd \] By the Mean Value Theorem applied to \(h\) on \([a,b],\) there exists a number \(c\in(a,b)\) such that \[ h'(c) = \frac{h(b) - h(a)}{b - a}\pd \] Because \(b - a\gt 0\) and \(h'(c)\gt 0,\) we have \(h(b) - h(a)\gt 0.\) But \(h(a) = f(a) - g(a) = 0,\) so \(h(b)\gt 0,\) meaning \[ f(b) - g(b)\gt 0 \implies f(b)\gt g(b) \cma \] as desired.

Verifying the Inequality Let \(f(x) = x\) and \(g(x) = \sin x.\) Then \(f(0) = g(0) = 0.\) Also, \(f'(x) = 1\) and \(g'(x)= \cos x.\) For \(0 \lt x \lt 2\pi,\) we have \(\cos x \lt 1\) and thus \(f'(x)\gt g'(x)\) on \((0,2\pi).\) Applying the preceding result on \([0,x]\) with \(x\in(0,2\pi)\) gives \[ f(x) \gt g(x) \implies x \gt \sin x \pd \] for all \(0 \lt x \lt 2 \pi.\)

SOLUTION As \(x \to 0,\) both the numerator and denominator approach \(0.\) This limit is therefore in the indeterminate form \(\indZero.\) Applying L'Hôpital's Rule gives the equivalent limit \[\lim_{x \to 0} \frac{\ds -\frac{1}{1 - x} - \cos x}{-2 \cos x (-\sin x)} = \lim_{x \to 0} \frac{\ds -\frac{1}{1 - x} - \cos x}{\sin 2x} \pd\] Now as \(x \to 0,\) the numerator approaches \(-2\) and the denominator approaches \(0.\) The function therefore has a vertical asymptote at \(x = 0,\) so let's consider the behavior on either side of the asymptote: As \(x \to 0^-,\) the denominator \(\sin 2x\) is negative, so the overall sign of the fraction is \((-)/(-) = (+).\) Thus, \[ \lim_{x \to 0^-} \frac{\ds -\frac{1}{1 - x} - \cos x}{\sin 2x} = \lim_{x \to 0^-} \frac{\ln(1 - x) - \sin x}{1 - \cos^2 x} = \infty \pd\] Conversely, as \(x \to 0^+\) the denominator \(\sin 2x\) is positive, so the fraction's sign is \((-)/(+) = (-).\) We therefore say \[ \lim_{x \to 0^+} \frac{\ds -\frac{1}{1 - x} - \cos x}{\sin 2x} = \lim_{x \to 0^+} \frac{\ln(1 - x) - \sin x}{1 - \cos^2 x} = -\infty \pd\] Because the one-sided limits do not equal each other, the limit \[\lim_{x \to 0} \frac{\ln(1 - x) - \sin x}{1 - \cos^2 x}\] indeed does not exist. So Cady is correct.

SOLUTION Recall that \(\text{time} = \text{distance}/\text{speed}.\) When the light ray travels from \(P\) to \(O,\) the time is \[t_1 = \frac{\sqrt{x^2 + a^2}}{v_1} \pd\] Likewise, traveling from \(O\) to \(Q\) takes time \[t_2 = \frac{\sqrt{(\ell - x)^2 + b^2}}{v_2} \pd\] Then the total time is \[ \ba T &= t_1 + t_2 \nl &= \frac{\sqrt{x^2 + a^2}}{v_1} + \frac{\sqrt{(\ell - x)^2 + b^2}}{v_2} \pd \ea \] Differentiating, we obtain \begin{equation} T'(x) = \frac{x}{v_1 \sqrt{x^2 + a^2}} - \frac{\ell - x}{v_2 \sqrt{(\ell - x)^2 + b^2}} \pd \label{eq:snell} \end{equation} We see \[\sin \theta_1 = \frac{x}{\sqrt{x^2 + a^2}} \qquad \sin \theta_2 = \frac{\ell - x}{\sqrt{(\ell - x)^2 + b^2}} \pd \] Then \(\eqRefer{eq:snell}\) becomes \[T'(x) = \frac{\sin \theta_1}{v_1} - \frac{\sin \theta_2}{v_2} \pd\] Solving \(T'(x) = 0\) then yields \[\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2} \cma\] as requested. We assume that this situation confers the absolute minimum of \(T.\)

SOLUTION Note that \(f'(x) = 3x^2.\) The graph of \(f\) therefore has a horizontal tangent when \(f'(x) = 0,\) that is, at \(x = 0.\) So \(x_1 = 0\) causes Newton's Method to fail because the tangent line never strikes the \(x\)-axis. For other values of \(x_1,\) the iterative formula is \[ \ba x_2 &= x_1 - \frac{f(x_1)}{f'(x_1)} \nl &= x_1 - \frac{\par{x_1}^3 + 2}{3 \par{x_1}^2} \nl &= x_1 - \frac{x_1}{3} - \frac{2}{3 \par{x_1}^2} \nl &= \frac{2 x_1}{3} - \frac{2}{3 \par{x_1}^2} \pd \ea \] Newton's Method fails to converge if it yields \(x_2 = 0,\) the location of the horizontal tangent. Solving shows \[ \ba 0 &= \frac{2 x_1}{3} - \frac{2}{3 \par{x_1}^2} \nl 0 &= \par{x_1}^3 - 1 \nl \implies x_1 &= 1 \pd \ea \] Newton's Method also fails if it yields \(x_2 = 1,\) since the following approximation would be \(x_3 = 0.\) The function \(2x_1/3 - 2/[3(x_1)^2]\) is increasing for positive \(x_1,\) so the solution to \[ 1 = \frac{2 x_1}{3} - \frac{2}{3 \par{x_1}^2} \] satisfies \(x_1 \gt 1\) (the solution is \(x_1 \approx 1.806\)). Newton's Method also fails if it yields \(x_2 \approx 1.806\) (because then \(x_3 = 1\) and \(x_4 = 0\)), yet the solution to \[ 1.806 = \frac{2 x_1}{3} - \frac{2}{3 \par{x_1}^2} \] satisfies \(x_1 \gt 1.806\) (the solution is \(x_1 \approx 2.834\)). This pattern is repeating—subsequent defective values of \(x_1\) are all outside \([0, 1].\) But inside \([0, 1],\) there are no other cases in which Newton's Method fails. Note that \(f'\) is nonnegative, so \(f\) is always increasing. Hence, \(f\) has only one zero, and Newton's Method cannot converge to another zero. Additionally, in Example 3.8-3 Newton's Method fails because the approximations bounce back and forth. But \(f\) is increasing, so the tangent lines always have nonnegative slope, thus avoiding the bouncing issue. Hence, the defective starting approximations \(x_1 \in [0, 1]\) are \(\boxed{x_1 = 0}\) and \(\boxed{x_1 = 1}.\)