Chapter 7 Challenge Problems Solutions

SOLUTION

The region is bounded by three straight sides of lengths \(2, 1,\) and \(1 + e^{-2}.\) It is bounded above by the curve \(y = 1 + e^{-2x},\) whose derivative is \[\deriv{y}{x} = -2e^{-2x} \pd\] Thus, the curve's length over \(0 \leq x \leq 1\) is \[L = \int_0^1 \sqrt{1 + \par{-2e^{-2x}}^2} \di x = \int_0^1 \sqrt{1 + 4e^{-4x}} \di x \approx 1.368 \pd\] So the region's perimeter is \[2 + 1 + \par{1 + e^{-2}} + 1.368 \approx \boxed{5.503}\]

SOLUTION We model the light bulb's longevity using an exponential probability density function: \[ f(t) = \bc 0 & t \lt 0 \nl ke^{-kt} & t \geq 0 \cma \ec \] where \(t\) is time in hours. The mean of this function is given to be \(\mu = 800.\) Recall that the mean is \(\mu = 1/k,\) so we have \[ f(t) = \bc 0 & t \lt 0 \nl \dfrac{e^{-t/800}}{800} & t \geq 0 \pd \ec \] Accordingly, the probability that a light bulb lasts more than \(900\) hours is given by the improper integral \[\int_{900}^\infty \frac{e^{-t/800}}{800} \di t \pd\] Using the methods of Section 6.5, we obtain \[ \ba P(T \geq 900) &= \lim_{b \to \infty} \int_{900}^b \frac{e^{-t/800}}{800} \di t \nl &= \lim_{b \to \infty} \par{- e^{-t/800}} \intEval_{900}^b \nl &= \par{\lim_{b \to \infty} - e^{-b/800}} - \par{-e^{-900/800}} \nl &= e^{-9/8} \approx \boxed{0.325} = 32.5 \% \pd \ea \]

SOLUTION The total surplus \(s_T\) is the area between the curves \(D\) and \(S\)—namely, \[s_T = \int_0^x \parbr{D(t) - S(t)} \di t \pd\] (We use \(t\) simply as a dummy variable.) To maximize \(s_T,\) we find the critical numbers—the values of \(x\) for which \(\textderiv{s_T}{x} = 0.\) (See Section 3.6 to review the process of optimization.) By Part I of the Fundamental Theorem of Calculus, \[\deriv{s_T}{x} = D(x) - S(x) \pd\] So we see \[D(X) - S(X) = 0 \implies D(X) = S(X) \pd\] Observe that \(D(x) - S(x) \gt 0\) when \(x \lt X\) and \(D(x) - S(x) \lt 0\) when \(x \gt X.\) Hence, the First-Derivative Test asserts that \(s_T\) has a relative maximum at \(x = X.\) Note that \(s_T = 0\) when \(x = 0.\) Also, \(s_T\) is increasing on \([0, X]\) and decreasing on \([X, \infty),\) so the absolute maximum must occur at \(x = X,\) meaning total surplus is maximized at the equilibrium point.

SOLUTION The integrand of the arc length function takes the form \(\sqrt{1 + [f'(t)]^2}.\) We therefore see \[ \ba \sqrt{1 + [f'(t)]^2} &= \sqrt{t^2 - 2t + 2} \nl 1 + [f'(t)]^2 &= t^2 - 2t + 2 \nl [f'(t)]^2 &= t^2 - 2t + 1 = (t - 1)^2 \nl \implies f'(t) &= t - 1 \pd \ea \] [We choose the positive solution instead of \(f'(t) = -(t - 1)\) because then \(f\) is increasing and so \(f'\) must be positive.] Solving this differential equation, we have \[f(t) = \tfrac{1}{2} t^2 - t + C \pd\] Then substituting the initial condition \(f(2) = 5\) shows \[f(2) = \tfrac{1}{2} (2)^2 - 2 + C = 5 \implies C = 5 \pd\] Hence, the identity of \(f\) is \[f(t) = \boxed{\tfrac{1}{2} t^2 - t + 5}\]

SOLUTION Picking the downward direction to be the positive \(y\)-axis, we note that the rectangle is bounded between \(y = m\) and \(y = m + m\) \( = 2m.\) At each \(y,\) a horizontal strip of the rectangle is a distance of \(h(y) = y\) beneath the surface. With a constant width of \(w(y) = 1/m^2,\) the hydrostatic force is \[ \ba F &= \gamma \int_m^{2m} y \frac{1}{m^2} \di y = \frac{\gamma}{m^2} \par{\frac{y^2}{2}} \intEval_m^{2m} = \frac{3 \gamma}{2} \pd \ea \] So \(F\) does not depend on \(m.\) Accordingly, the hydrostatic force against the rectangle is independent of \(m.\)

SOLUTION The lamina is balanced at its mean value \(\mu\) if and only if its centroid is at \(\overline x = \mu.\) (See Section 7.4 to review finding centroids.) The lamina's area is \(\int_{-\infty}^\infty f(x) \di x = 1\) (since \(f\) is a probability density function), so the \(x\)-coordinate of the centroid is \[ \ba \overline x &= \frac{\ds \int_{-\infty}^\infty x f(x) \di x}{1} \nl &= \int_{-\infty}^\infty x f(x) \di x \nl &= \mu \cma \ea \] as requested.

SOLUTION It is most convenient to take the downward direction to be the positive \(y\)-axis. A thin, horizontal approximating rectangle at any \(y\) is located a distance \(h(y) = y\) beneath the surface. By similar triangles, \[\frac{w(y)}{y - k} = \frac{4}{4} \implies w(y) = y - k \pd\] The body is bounded between \(y = k\) and \(y = k + 4,\) so the hydrostatic force is \[ \ba F &= 62.5 \int_k^{k + 4} y (y - k) \di y = 62.5 \int_k^{k + 4} \par{y^2 - ky} \di y \nl &= 62.5 \par{\frac{y^3}{3} - \frac{ky^2}{2}} \intEval_k^{k + 4} = 62.5 \par{\frac{(k + 4)^3}{3} - \frac{k(k + 4)^2}{2} - \frac{k^3}{3} + \frac{k^3}{2}} \nl &= 62.5 \par{\frac{k^3 + 12k^2 + 48k + 64}{3} - \frac{k^3 + 8k^2 + 16k}{2} - \frac{k^3}{3} + \frac{k^3}{2}} \nl &= 62.5 \par{8k + \frac{64}{3}} = \frac{125}{2} \par{8k + \frac{64}{3}} = 500k + \frac{4000}{3} \cma \ea \] as requested.

SOLUTION The integral expression for the arc length from \(x = a\) to \(x = \infty\) is given by \[L = \int_a^\infty \sqrt{1 + [f'(x)]^2} \di x\] if \(f'(x)\) is continuous for \(x \geq a.\) Because \([f'(x)]^2 \geq 0,\) we observe that \[\sqrt{1 + [f'(x)]^2} \geq \sqrt 1 \gt 0 \pd\] We see \[\int_a^\infty \sqrt 1 \di x = \int_a^\infty \dd x = x \intEval_a^\infty = \infty \cma\] so \(\int_a^\infty \sqrt 1 \di x\) diverges. By the Comparison Test for Improper Integrals (from Section 6.5), the integral \(\int_a^\infty \sqrt{1 + [f'(x)]^2} \di x\) also diverges. Hence, the arc length \(L\) is infinite.

SOLUTION Since \(\textderiv{y}{x} = -e^{-x},\) the lateral surface area of the generated solid is \[S = \int_0^\infty 2 \pi e^{-x} \sqrt{1 + \par{-e^{-x}}^2} \di x = 2 \pi \int_0^\infty e^{-x} \sqrt{1 + e^{-2x}} \di x \pd\] Substituting \(u = e^{-x},\) we get \(\dd u = -e^{-x} \di x.\) When \(x = 0,\) \(u = 1;\) when \(x = \infty,\) \(u = 0.\) So the integral becomes \[S = - 2 \pi \int_1^0 \sqrt{1 + u^2} \di u = 2 \pi \int_0^1 \sqrt{1 + u^2} \di u \pd\] Letting \(u = \tan \theta\) results in \(\dd u = \sec^2 \theta \di \theta.\) When \(u = 0,\) \(\theta = 0;\) when \(u = 1,\) \(\theta = \pi/4.\) Hence, we attain \[ \ba S &= 2 \pi \int_0^{\pi/4} \par{\sec \theta} \sec^2 \theta \di \theta = 2 \pi \int_0^{\pi/4} \sec^3 \theta \di \theta \nl &= 2 \pi \par{\tfrac{1}{2} \sec \theta \tan \theta + \tfrac{1}{2} \ln \abs{\sec \theta + \tan \theta} } \intEval_0^{\pi/4} \nl &= 2 \pi \parbr{\tfrac{1}{2} \par{\sqrt 2} (1) + \tfrac{1}{2} \ln \par{\sqrt 2 + 1}} - 0 \nl &= \boxed{\pi \parbr{\sqrt 2 + \ln \par{\sqrt 2 + 1}}} \approx 7.212 \pd \ea \]

SOLUTION Converting the measurements into feet, we find the height to be \(5/12 \un{ft}\) and the base radius to be \(3/12\) \(= 1/4 \un{ft}.\) Let's define a coordinate system such that the origin is at the bottom-center of the cup's cross section. We model this parabola using \(y = ax^2.\) Substituting the coordinates of the top tip—that is, \(x = 1/4\) and \(y = 5/12\)—shows \[a = \frac{5/12}{(1/4)^2} = \frac{20}{3} \pd\] In addition, the top of the cup—that is, the surface level of the milk—is the line \(y = 5/12.\) A thin, horizontal approximating rectangle to the cup is therefore a depth of \(h(y) = 5/12 - y\) below the surface. Its width is \(2x,\) where \[x = \sqrt{\frac{y}{a}} = \sqrt{\frac{y}{20/3}} = \sqrt{\frac{3y}{20}} \pd\] Instead of using \(\gamma = 62.5,\) as for water, we use \(\gamma = 64.3.\) Doing so, we calculate the hydrostatic force to be \[ \ba F &= 64.3 \int_0^{5/12} \par{\frac{5}{12} - y} \cdot 2 \sqrt{\frac{3y}{20}} \di y \nl &= 64.3 \sqrt{\frac{3}{5}} \int_0^{5/12} \par{\tfrac{5}{12} \sqrt y - y^{3/2}} \di y \nl &= 64.3 \sqrt{\frac{3}{5}} \par{\tfrac{5}{18} y^{3/2} - \tfrac{2}{5} y^{5/2}} \intEval_0^{5/12} \nl &\approx \boxed{1.488 \un{lb}} \ea \]

SOLUTION The region is bounded from \(x = 1\) to \(x = \infty.\) By using the Disk Method (from Section 5.3), the horn's volume is \[ \ba V &= \int_1^\infty \pi \par{\frac{1}{x}}^2 \di x = \lim_{k \to \infty} \int_1^k \frac{\pi}{x^2} \di x \nl &= \lim_{k \to \infty} \par{-\frac{\pi}{x}} \intEval_1^k = - \par{0 - \pi} = \pi \pd \ea \] Thus, Gabriel's Horn has a finite volume. Now we move on to its lateral surface area: With \(\textderiv{y}{x}\) \(= -1/x^2,\) the horn's surface area is given by \[ \ba S &= \int_1^\infty 2 \pi \par{\frac{1}{x}} \sqrt{1 + \par{-\frac{1}{x^2}}^2} \di x \nl &= \int_1^\infty 2 \pi \par{\frac{1}{x}} \sqrt{1 + \frac{1}{x^4}} \di x \pd \ea \] Because \(\sqrt{1 + 1/x^4} \geq\) \(\sqrt 1 = 1,\) we have \[\int_1^\infty 2 \pi \par{\frac{1}{x}} \sqrt{1 + \frac{1}{x^4}} \di x \geq \int_1^\infty 2 \pi \par{\frac{1}{x}} \di x \pd \] The right-hand integral becomes the limit \[\lim_{k \to \infty} \par{2 \pi \ln x} \intEval_1^k = \infty \pd\] Since \(\int_1^\infty 2 \pi \par{\frac{1}{x}} \di x\) diverges, the Comparison Test for Improper Integrals (from Section 6.5) asserts that \(\int_1^\infty 2 \pi \par{\frac{1}{x}} \sqrt{1 + \frac{1}{x^4}} \di x\) also diverges. Accordingly, Gabriel's Horn has an infinite surface area.

SOLUTION Region \(S\) balances on the \(x\)-axis if and only if the \(y\)-coordinate of its centroid is \(0.\) This can only occur if \(L \lt 0.\) The area of region \(R\) is \(A_R = \int_0^a f(x) \di x,\) and let \(A_S\) be the area of region \(S.\) The \(y\)-coordinate of the centroid of region \(S\) is therefore given by \[ \overline y = \frac{1}{A_S} \int_0^a \par{\tfrac{1}{2} [f(x)]^2 - \tfrac{1}{2} L^2} \di x \pd \] Equating this expression to \(0,\) we get \[ \ba \frac{1}{A_S} \int_0^a \par{\tfrac{1}{2} [f(x)]^2 - \tfrac{1}{2} L^2} \di x &= 0 \nl \int_0^a \tfrac{1}{2} [f(x)]^2 \di x - \int_0^a \tfrac{1}{2} L^2 \di x &= 0 \nl \int_0^a \tfrac{1}{2} [f(x)]^2 \di x - \tfrac{1}{2} aL^2 &= 0 \nl \int_0^a \tfrac{1}{2} [f(x)]^2 \di x &= \tfrac{1}{2} aL^2 \pd \ea \] Dividing both sides by \(A_R\) gives \[ \ba \frac{1}{A_R} \int_0^a \tfrac{1}{2} [f(x)]^2 \di x &= \frac{aL^2}{2A_R} \nl y_c &= \frac{aL^2}{2A_R} \pd \ea \] Solving for \(L\) (and choosing the negative solution) yields \[L = -\sqrt{\frac{2 y_c}{a} A_R} = \boxed{-\sqrt{\frac{2 y_c}{a} \int_0^a f(x) \di x}}\]