Chapter 5 Challenge Problems Solutions

SOLUTION Note that \(x^2 + 3 \gt \sqrt x\) for \(x \in [1, 4].\) At any \(x,\) the distance between the graphs is their vertical difference: \[d(x) = (x^2 + 3) - (\sqrt x) = x^2 + 3 - \sqrt x \pd\] The average value of \(d(x)\) over \(1 \leq x \leq 4\) is \[ \ba \avg d &= \frac{1}{4 - 1} \int_1^4 \par{x^2 + 3 - \sqrt x} \di x \nl &= \tfrac{1}{3} \par{\tfrac{1}{3} x^3 + 3x - \tfrac{2}{3} x^{3/2}} \intEval_1^4 \nl &= \boxed{\tfrac{76}{9}} \ea \]

SOLUTION

1. The line is equivalent to the function \(y = 2x - 1.\) With a top function of \(y = 2x - 1\) and a bottom function of \(y = e^{3 \sin x} \cos x,\) the area between \(x = 2\) and \(x = k\) is \[ \ba A &= \int_2^k \parbr{(2x - 1) - e^{3 \sin x} \cos x} \di x \nl &= \par{x^2 - x - \tfrac{1}{3} e^{3 \sin x}} \intEval_2^k \nl &= \boxed{k^2 - k - \tfrac{1}{3} e^{3 \sin k} - 2 + \tfrac{1}{3} e^{3 \sin 2}} \ea \]

---

2. We have \[ \ba \deriv{A}{k} &= \deriv{}{k} \par{k^2 - k - \tfrac{1}{3} e^{3 \sin k} - 2 + \tfrac{1}{3} e^{3 \sin 2}} \nl &= \boxed{2k - 1 - e^{3 \sin k} \cos k} \ea \] This quantity is the vertical distance between the two graphs at \(x = k.\)

---

3. If \(t\) is time in seconds, then we are given \(\textderiv{k}{t} = 2 \undiv{units}{sec},\) and we need \(\textderiv{A}{t}.\) By the process of Related Rates (Section 2.7), \[ \ba \deriv{A}{t} &= \deriv{A}{k} \deriv{k}{t} \nl &= 2 \deriv{A}{k} \pd \ea \] At \(k = 3,\) \[\deriv{A}{k} \intEval_{k = 3} = 2(3) - 1 - e^{3 \sin 3} \cos 3 = 5 - e^{3 \sin 3} \cos 3 \pd\] Thus, \[ \ba \deriv{A}{t} \intEval_{k = 3} &= 2 \par{5 - e^{3 \sin 3} \cos 3} \nl &= \boxed{10 - 2 e^{3 \sin 3} \cos 3} \approx 13.024 \un{units}^2/\un{sec} \pd \ea \]

SOLUTION The enclosed region \(R\) has area given by \[A = \int_0^k x^2 \di x \pd\] Conversely, the volume of the solid \(S\) has volume \[V = \int_0^k \par{x^2}^2 \di x = \int_0^k x^4 \di x \pd\] We equate \(A\) and \(V,\) finding \[ \ba \int_0^k x^2 \di x &= \int_0^k x^4 \di x \nl \tfrac{1}{3} k^3 &= \tfrac{1}{5} k^5 \nl \implies k &= \boxed{\sqrt{\frac{5}{3}}} \approx 1.291 \pd \ea \]

SOLUTION This problem requires creativity. Because the boundary functions change throughout the region, we must create subregions. One method is to split the region into two subregions, \(R_1\) and \(R_2,\) as in the figure. Note that \(y = \sqrt[3]{6 - x}\) \(\iffArrow x = 6 - y^3\) and \(x = \tfrac{1}{4} y^2\) \(\iffArrow y = \pm 2 \sqrt x.\) The curves \(y = \sqrt[3]{6 - x}\) and \(x = \tfrac{1}{4}y^2\) intersect at \((0.755, 1.737).\) By integrating with \(y,\) the area of \(R_1\)–bounded between \(x = \tfrac{1}{4} y^2\) and \(x = 0.755\)—is \[ \ba A_1 &= \int_{-1.737}^{1.737} \par{0.755 - \tfrac{1}{4} y^2} \di y \nl &\approx 1.749 \pd \ea \] Equivalently, \(A_1\) may also be calculated by integrating with \(x \col\) \[ \ba A_1 &= \int_0^{0.755} \parbr{2 \sqrt x - (-2 \sqrt x)} \di x \nl &\approx 1.749 \pd \ea \] Then by integrating with \(x,\) the area of \(R_2\)—bounded between \(y = \sqrt[3]{6 - x},\) \(y = - 2 \sqrt x,\) \(x = 0.755,\) and \(x = 5\)—is \[ \ba A_2 &= \int_{0.755}^5 \parbr{\sqrt[3]{6 - x} - (-2 \sqrt x)} \di x \nl &\approx 20.117 \pd \ea \] The total area is therefore \[A_1 + A_2 \approx \boxed{21.866}\]

areas-between-curves-calculator-12

SOLUTION

Imagine placing the bowl such that the \(y\)-axis is the central axis. Then we model its edge using a parabola \(y = ax^2;\) rotating this parabola about the \(y\)-axis generates the shape of the bowl. Since the bowl's top has a diameter of \(8\) inches, its radius is \(4\) inches. And since the bowl is \(6\) inches tall, the parabola passes through the point \((4, 6).\) Substituting this point and solving for \(a\) show \[6 = a(4)^2 \implies a = \tfrac{3}{8} \pd\] Since we rotate about the \(y\)-axis, we want to express all quantities in terms of \(y,\) so we rewrite \(y = \tfrac{3}{8} x^2\) as \(x = \sqrt{\tfrac{8}{3} y}.\) The soup level is \(2\) inches, so we calculate the volume of the solid of revolution from \(y = 0\) to \(y = 2 \col\) \[ \ba V &= \pi \int_0^2 \par{\sqrt{\tfrac{8}{3} y}}^2 \di y \nl &= \frac{4 \pi}{3} \par{y^2} \intEval_0^2 \nl &= \boxed{\frac{16 \pi}{3}} \approx 16.755 \un{in}^3 \pd \ea \]

SOLUTION Using Boyle's Law with \(V = 3 \un{ft}^3\) and \(P = 12 \un{lb/ft}^2,\) we attain \[12 = \frac{C}{3} \implies C = 36 \pd\] The work done by the gas is therefore \[W = \int_3^5 \frac{36}{V} \di V = 36 \ln \abs V \intEval_3^5 = 36 \ln \par{\frac{5}{3}} \pd\] This quantity is the amount of energy the gas uses to push out the piston; it therefore equals the work done in stretching the spring by a distance \(d.\) We express the work done in stretching the spring as \[\int_0^d 150 x \di x = 75 x^2 \intEval_0^d = 75 d^2 \pd\] Accordingly, we see \[75 d^2 = 36 \ln \par{\frac{5}{3}} \implies d = \sqrt{\frac{36}{75} \ln \par{\frac{5}{3}}} \approx \boxed{0.495 \un{ft}}\] (Of course, the negative solution for \(d\) does not make sense in this context.)

SOLUTION Let the line passing through the origin be \(y = mx.\) Then the line intersects the parabola \(y = x(2 - x)\) when \(mx = x(2 - x) \col\) \[ \ba mx &= 2x - x^2 \nl x^2 &= x(2 - m) \nl \implies x &= 0 \cma 2 - m \pd \ea \] The area of the top subregion is then \[ \ba A_{\text{top}} &= \int_0^{2 - m} \parbr{x(2 - x) - mx} \di x \nl &= \int_0^{2 - m} \parbr{-x^2 + x(2 - m)} \di x \nl &= \parbr{-\tfrac{1}{3} x^3 + \tfrac{1}{2} x^2 (2 - m)} \intEval_0^{2 - m} \nl &= -\tfrac{1}{3} (2 - m)^3 + \tfrac{1}{2} (2 - m)^2 (2 - m) - 0 \nl &= \tfrac{1}{6} (2 - m)^3 \pd \ea \] In addition, the total area of the region bounded by the parabola is \[ \ba A &= \int_0^2 x (2 - x) \di x \nl &= \int_0^2 \par{2x - x^2} \di x \nl &= \par{x^2 - \tfrac{1}{3} x^3} \intEval_0^2 \nl &= \tfrac{4}{3} \pd \ea \] The top subregion has half the area bounded by the parabola; thus, \[ \ba A_{\text{top}} &= \tfrac{1}{2} A \nl \tfrac{1}{6} (2 - m)^3 &= \tfrac{2}{3} \nl (2 - m)^3 &= 4 \nl \implies m &= \boxed{2 - \sqrt[3] 4} \approx 0.413 \pd \ea \]

SOLUTION

A sphere can be obtained by rotating a semicircle about a central axis. Since the equation of a circle of radius \(R\) is \(x^2 + y^2 = R^2,\) an equation of the right semicircle centered at the origin is \(x = \sqrt{R^2 - y^2}.\) We shade the region between this semicircle and the \(y\)-axis from \(y = -R\) to \(y = -R + h.\) Then we use the Disk Method to revolve this region around the \(y\)-axis; in doing so, we attain the shape of the bottom cap. Each disk at \(y\) has a radius of \(r(y) = \sqrt{R^2 - y^2}.\) The volume of the bottom cap is therefore \[ \ba V &= \pi \int_{-R}^{h - R} [r(y)]^2 \di y = \pi \int_{-R}^{h - R} \par{R^2 - y^2} \di y \nl &= \pi \par{R^2 y - \tfrac{1}{3} y^3} \intEval_{-R}^{h - R} \nl &= \pi \parbr{R^2 (h - R) - \tfrac{1}{3} \par{h - R}^3} + \pi \par{R^3 - \tfrac{1}{3} R^3} \nl &= \pi \par{R^2 h - R^3 - \tfrac{1}{3} h^3 + h^2 R - R^2h + \tfrac{1}{3} R^3 + R^3 - \tfrac{1}{3} R^3} \nl &= \boxed{\pi \par{h^2 R - \tfrac{1}{3} h^3}} \ea \]

SOLUTION At each \(x,\) the cross section perpendicular to the \(x\)-axis is a rectangle of base \(f(x)\) and height \(k.\) Its cross-sectional area is therefore \[A(x) = k f(x) = \frac{k}{(kx)^2 + 1} \pd\] The solid's volume is given by integrating this function from \(x = 1\) to \(x = 2,\) as follows: \[ \ba V &= \int_1^2 \frac{k}{(kx)^2 + 1} \di x \nl &= \atan kx \intEval_1^2 = \atan 2k - \atan k \pd \ea \] (We mentally performed the substitution \(u = kx\) to attain this antiderivative.) To calculate the value of \(k\) for which \(V\) is maximized, we differentiate \(V\) with respect to \(k\) and locate its critical points. (See Section 3.6 to review the process of optimization.) Doing so, we set \(\textderiv{V}{k} = 0\) and solve for \(k.\) We find the derivative to be \[ \ba \deriv{V}{k} &= \frac{2}{(2k)^2 + 1} - \frac{1}{k^2 + 1} \nl &= \frac{2k^2 + 2 - 4k^2 - 1}{\par{4k^2 + 1} \par{k^2 + 1}} \nl &= \frac{-2k^2 + 1}{\par{4k^2 + 1} \par{k^2 + 1}} \pd \ea \] So we see \[\frac{-2k^2 + 1}{\par{4k^2 + 1} \par{k^2 + 1}} = 0 \implies -2k^2 + 1 = 0 \implies k = \frac{1}{\sqrt 2} \pd\] This value corresponds to a relative maximum of \(V\) because \(\textderiv{V}{k}\) changes sign from positive to negative at \(k = 1/\sqrt 2.\) Since \(V = 0\) when \(k = 0\) and \[\lim_{k \to \infty} V = \lim_{k \to \infty} \par{\atan 2k - \atan k} = \frac{\pi}{2} - \frac{\pi}{2} = 0 \cma\] the relative maximum at \(k = 1/\sqrt 2\) must be the location of the absolute maximum of \(V.\) So the volume is maximized for \(\boxed{k = 1/\sqrt 2}.\)

SOLUTION

1. The two particles attract each other with an electrostatic force of \[ F = \frac{\par{9.0 \times 10^9} \abs{\par{20 \times 10^{-6}} \par{-15 \times 10^{-6}}}}{r^2} = \frac{2.7}{r^2} \pd \] The \(20 \muUnit C\) needs to be moved from \(x = 0\) to \(x = 3.\) Since this positive particle is attracted to the negative particle, the positive particle experiences a force acting to the right. (We therefore give \(F\) a positive sign.) The positive particle is initially located a distance of \(4 \un m\) left of the negative particle; at \(x = 3,\) this distance becomes only \(1 \un m\) left. So the work done by the electrostatic force is \[ \ba W &= \int_{-4}^{-1} \frac{2.7}{r^2} \di r = -\frac{2.7}{r} \intEval_{-4}^{-1} \nl &= 2.7 \par{\frac{1}{-4} - (-1)} = \boxed{2.025 \un J} \ea \]

---

2. The negative particle is attracted to the positive particle and so experiences a force of \(F = 2.7/r^2\) to the left. (The sign of \(F\) is therefore negative.) Initially, the negative particle is \(4 \un m\) right of the positive particle; after it is moved to \(x = 3,\) the negative particle is \(3 \un m\) right. The work done is therefore \[ \ba W &= \int_4^3 -\frac{2.7}{r^2} \di r = \frac{2.7}{r} \intEval_4^3 \nl &= 2.7 \par{\frac{1}{3} - \frac{1}{4}} = \boxed{0.225 \un J} \ea \]

---

3. At \(x = 1,\) the negative particle is only \(1 \un m\) to the right. The work done is therefore \[ \ba W &= \int_4^1 -\frac{2.7}{r^2} \di r = \frac{2.7}{r} \intEval_4^1 \nl &= 2.7 \par{1 - \frac{1}{4}} = \boxed{2.025 \un J} \ea \]

SOLUTION The electron is attracted to the center proton, so the electron experiences a force toward the center. To move the electron farther away (to a larger orbit), we must apply an equal force radially outward. By Coulomb's Law, at some distance \(r\) we must apply an outward force equal to \[F = \frac{\par{9.0 \times 10^9} \par{1.6 \times 10^{-19}}^2}{r^2} = \frac{2.304 \times 10^{-28}}{r^2} \pd\] We integrate along the electron's path—namely, from \(r = R\) to \(r = 2R.\) The energy needed equals the work done by the electrostatic force, which we calculate in joules to be \[ \ba W &= \int_R^{2R} \frac{2.304 \times 10^{-28}}{r^2} \di r = -\frac{2.304 \times 10^{-28}}{r} \intEval_R^{2R} \nl &= -2.304 \times 10^{-28} \par{\frac{1}{2R} - \frac{1}{R}} = \boxed{\frac{1.152 \times 10^{-28}}{R}} \ea \]

SOLUTION Initially, the mass of water is \(1 \un{kg};\) at the end, the mass of water is \(0.6 \un{kg}.\) Thus, the bucket's starting mass is \(3 \un{kg},\) and its final mass is \(2.6 \un{kg}.\) We define the mass as a function of height, \(y.\) Let \(y = 0\) be the bucket's initial height, and let \(y = 2 \un{m}\) be its final height. The rate of change of the bucket's mass is then \[\frac{\Delta m}{\Delta y} = \frac{2.6 - 3}{2 - 0} = -0.2 \undiv{kg}{m} \pd\] So the bucket's mass \(m\) as a function of height \(y\) is \[m(y) = 3 - 0.2y \pd\] The force of gravity, in newtons, acting on the bucket at any \(y \in [0, 2]\) is \[F(y) = m(y) \cdot g = 9.8 m(y) = 9.8(3 - 0.2 y) = 29.4 - 1.96 y \pd\] To overcome the influence of gravity, an equivalent force \(F(y)\) must be applied upward. The work done is therefore \[ \ba W &= \int_0^2 F(y) \di y \nl &= \int_0^2 (29.4 - 1.96 y) \di y \nl &= \par{29.4 y - 0.98 y^2} \intEval_0^2 \nl &= \boxed{54.88 \un J} \ea \]

SOLUTION The average value of \(f\) on \([-1, 1]\) is \[ \ba \avg f &= \frac{1}{1 - (-1)} \int_{-1}^1 \par{3x^2 + x^n \, e^{-x^6}} \di x \nl &= \tfrac{1}{2} \int_{-1}^1 \par{3x^2 + x^n \, e^{-x^6}} \di x \nl &= \tfrac{1}{2} \int_{-1}^1 3x^2 \di x + \tfrac{1}{2} \int_{-1}^1 x^n \, e^{-x^6} \di x \nl &= \tfrac{1}{2} \par{x^3} \intEval_{-1}^1 + \tfrac{1}{2} \int_{-1}^1 x^n \, e^{-x^6} \di x \nl &= 1 + \tfrac{1}{2} \int_{-1}^1 x^n \, e^{-x^6} \di x \pd \ea \] Equating \(\avg f = 1\) shows \[ \ba 1 + \tfrac{1}{2} \int_{-1}^1 x^n \, e^{-x^6} \di x &= 1 \nl \tfrac{1}{2} \int_{-1}^1 x^n \, e^{-x^6} \di x &= 0 \nl \implies \int_{-1}^1 x^n \, e^{-x^6} \di x &= 0 \pd \ea \] From the properties of symmetry for definite integrals (Section 4.4), \(\int_{-a}^a g(x) \di x = 0\) if \(g\) is odd—that is, if \(g(-x) = -g(x).\) Thus, \(g(x) = x^n \, e^{-x^6}\) must be odd. Note that \[g(-x) = (-x)^n \, e^{-(-x)^6} = (-x)^n \, e^{-x^6} \and -g(x) = -x^n \, e^{-x^6} \pd\] We only have \((-x)^n = -x^n\) only when \(n\) is an odd integer. Thus, \(\avg f = 1\) for \[n = \boxed{1 \cma 3 \cma 5 \cma \dots}\]