Chapter 9 Challenge Problems Solutions

SOLUTION

1. Let \(t\) be time in seconds; also let \(x\) and \(y\) (measured in inches) be the horizontal and vertical distances, respectively, from the serving player. Note that the ball's initial height is the height of the player, \(6(12)\) \(= 72 \un{in}.\) The parametric equations are therefore \[ \ba x &= \boxed{880(\cos \alpha) t} \nl y &= 72 + 880(\sin \alpha) t - \tfrac{1}{2}(386) t^2 \nl &= \boxed{72 + 880(\sin \alpha) t - 193 t^2} \ea \]

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2. We eliminate the parameter \(t\) from our parametric equations: Solving for \(t\) gives \[t = \frac{x}{880 \cos \alpha} \cma\] from which we obtain \[ y = 72 + 880(\sin \alpha) \par{\frac{x}{880 \cos \alpha}} - 193 \par{\frac{x}{880 \cos \alpha}}^2 \pd \] Simplifying shows that the volleyball's trajectory takes the shape \[\boxed{y = 72 + x \tan \alpha - \frac{193 x^2}{880^2 \cos^2 \alpha}}\]

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3. Because the volleyball must be hit above the \(92\)-inch net, we need \(y \gt 92\) when \(x = 315.\) We seek the solutions of \(\alpha\) to \[72 + 315 \tan \alpha - \frac{193 (315)^2}{880^2 \cos^2 \alpha} \gt 92 \pd\] Because \[\frac{1}{\cos^2 \alpha} = \sec^2 \alpha = 1 + \tan^2 \alpha \cma\] the inequality becomes \[ \ba 72 + 315 \tan \alpha - \frac{193 (315)^2}{880^2} (1 + \tan^2 \alpha) &\gt 92 \cma \nl -20 + 315 \tan \alpha - \frac{193 (315)^2}{880^2} - \frac{193 (315)^2}{880^2} \tan^2 \alpha &\gt 0 \nl -44.729 + 315 \tan \alpha - 24.729 \tan^2 \alpha &\gt 0 \pd \ea \] The solution to this quadratic inequality is \[0.144 \lt \tan \alpha \lt 12.594 \or 0.143 \lt \alpha \lt 1.492 \pd\] Converting this interval to degrees, we conclude that the volleyball is successfully served for \[\boxed{8.194 \degree \lt \alpha \lt 85.46 \degree}\] This result is logical: when volleyball players serve the ball, they must worry primarily about hitting the ball hard enough and straight across the net—not about the angle from the ground.

SOLUTION Let \(v_0\) be the velocity at which the football is kicked, and let \(\alpha\) be the angle to the horizontal from which it is kicked. Note that its starting height is \(0\) and its final height is \(0.\) At any time \(t,\) the football's trajectory follows the parametric equations \[ \ba x &= v_0 (\cos \alpha) t \cma \nl y &= v_0 (\sin \alpha) t - \tfrac{1}{2} gt^2 \pd \ea \] We want the time at which the football touches the ground, so we equate \(y\) to \(0 \col \) \[ \ba v_0 (\sin \alpha) t - \tfrac{1}{2} gt^2 &= 0 \nl \implies t &= \frac{2 v_0 \sin \alpha}{g} \pd \ea \] This is the time at which the football hits the ground. We therefore see \[D = v_0 (\cos \alpha) \par{\frac{2 v_0 \sin \alpha}{g}} = \frac{2 v_0^2 \sin \alpha \cos \alpha}{g} \pd\] Because \(2 \sin \alpha \cos \alpha\) \(= \sin 2 \alpha,\) we attain \[D = \frac{v_0^2 \sin 2 \alpha}{g} \cma\] which is maximized for \(\alpha = 45 \degree.\) (When \(\alpha = 45 \degree,\) \(\sin 2 \alpha = 1,\) the maximum value of sine.) Thus, the football should be kicked at \(\boxed{45 \degree}\) to the horizontal to maximize the range of its trajectory.

SOLUTION Do not confuse the graphs of \(y = f(t)\) and \(y = g(t)\) with the curve \(C.\)

1. The function \(f\) is linear. In the given figure, the graph of \(y = f(t)\) passes through the points \((0, 6)\) and \((6, 0).\) Thus, the identity of \(f\) is given by the line \[f(t) = \boxed{6 - t}\]

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2. The function \(g\) is a quadratic. In the given figure, we see that \(y = g(t)\) passes through the origin and the point \((4, 4).\) Thus, \[g(t) = \boxed{\tfrac{1}{4} t^2}\]

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3. The curve \(C\) is parameterized by \[ \baat{2} x &= f(t) &&= 6 - t \cma \nl y &= g(t) &&= \tfrac{1}{4} t^2 \pd \eaat \] From the first equation \((x = 6 - t),\) we get \(t = 6 - x.\) Substituting this expression into the second equation \((y = \tfrac{1}{4} t^2)\) gives \[\boxed{y = \tfrac{1}{4} (6 - x)^2}\] The shape of the curve \(C\) follows this Cartesian equation.

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4. 

   When \(x = 6,\) we see \[y = \tfrac{1}{4} (6 - 6)^2 = 0 \pd\] Also, when \(x = 0\) \[y = \tfrac{1}{4} (0 - 6)^2 = 9 \pd\] Thus, the curve \(C\) is a quadratic with vertex \((6, 0)\) and \(y\)-intercept \((0, 9).\) As \(t\) increases from \(0,\) \(x = f(t)\) decreases but \(y = g(t)\) increases. We therefore draw arrows that point left and upward along the curve.

SOLUTION

The curve intersects the \(y\)-axis when \(t \cos t = 0,\) which first occurs when \(t = \pi/2;\) at this time, \(y = \ln(\pi/2).\) As \(t \to 0^+,\) \(\ln t \to -\infty.\) This area is therefore unbounded. In Cartesian coordinates, the improper integral (see Section 6.5) that represents this area is \[\int_{y = -\infty}^{y = \ln(\pi/2)} x \di y \pd\] As \(y \to -\infty,\) \(t \to 0^+;\) when \(y = \ln(\pi/2),\) \(t = \pi/2.\) Also, \(\dd y = (1/t) \di t.\) The improper integral therefore becomes \[ \ba \int_0^{\pi/2} (t \cos t) \par{\tfrac{1}{t}} \di t &= \int_0^{\pi/2} \cos t \di t \nl &= \sin t \intEval_0^{\pi/2} \nl &= \sin \tfrac{\pi}{2} - \sin 0 \nl &= \boxed 1 \ea \] as requested.

SOLUTION The ball's height \(y\) above the ground at any time \(t\) is \[ \ba y &= 20 + 16 (\sin 45 \degree) t - \tfrac{1}{2}(9.8)t^2 \nl &= 20 + \par{8 \sqrt 2} t - 4.9 t^2 \pd \ea \] To maximize \(y,\) we locate its critical points. Since the trajectory is a parabola, the absolute maximum of \(y\) must occur at the critical point. (See Section 3.1 to review the arguments of locating extrema.) We differentiate to attain \[\deriv{y}{t} = 8 \sqrt 2 - 9.8 t = 0 \implies t = \frac{8 \sqrt 2}{9.8} \pd\] Thus, the maximum height is \[ \ba \maxSub y &= 20 + 8 \sqrt 2 \par{\frac{8 \sqrt 2}{9.8}} - 4.9 \par{\frac{8 \sqrt 2}{9.8}}^2 \nl &= 20 + \frac{64}{9.8} \nl &\approx \boxed{26.531 \un{m}} \ea \]

SOLUTION The horse's angular speed is \(\textderiv{\theta}{t} =\) \(8 \undiv{rad}{min}.\) Our objective is to connect \(\textderiv{r}{t}\) to the quantities we can find: \(\textderiv{r}{\theta}\) and \(\textderiv{\theta}{t} = 8.\) (See Section 2.7 to review the procedure of working with related rates.) By the Chain Rule, \[ \ba \deriv{r}{t} &= \deriv{r}{\theta} \deriv{\theta}{t} \nl &= 8 \deriv{r}{\theta} \pd \ea \] We obtain \[\deriv{r}{\theta} = - 2 \sin \theta \cma\] so \(\textderiv{r}{\theta} |_{\theta = 3 \pi/2}\) \(= 2.\) Hence, \[\deriv{r}{t} \intEval_{\theta = 3\pi/2} = 8(2) = 16 \pd\] Also, \(r(3\pi/2) = 2.\) Since \(r\) and \(\textderiv{r}{t}\) have the same sign, the horse is moving away from the bell tower.

SOLUTION Because \[\deriv{y}{x} = \frac{\textderiv{y}{t}}{\textderiv{x}{t}} \cma \] we see \[ \ba \deriv{x}{t} &= 2t - 4 \nl \deriv{y}{t} &= 3t^2 - 6t + 2 \pd \ea \] Solving each differential equation reveals \[ \ba x &= t^2 - 4t + C_1 \cma \nl y &= t^3 - 3t^2 + 2t + C_2 \ea \] for any constants \(C_1\) and \(C_2.\) Substituting the initial conditions \(x(2) = 1\) and \(y(2) = 3\) shows \[ \ba x &= (2)^2 - 4(2) + C_1 = 1 \implies C_1 = 5 \nl y &= (2)^3 - 3(2)^2 + 2(2) + C_2 = 3 \implies C_2 = 3 \pd \ea \] Thus, the identities of \(f\) and \(g\) are \[ f(t) = \boxed{t^2 - 4t + 5} \lspace g(t) = \boxed{t^3 - 3t^2 + 2t + 3} \]

SOLUTION One arch of the cycloid is traced over \(0 \leq \theta \leq 2 \pi.\) We differentiate both parametric functions to see \[\deriv{x}{\theta} = r(1 - \cos \theta) \and \deriv{y}{\theta} = r \sin \theta \pd\] Observe that \[ \ba \par{\deriv{x}{\theta}}^2 + \par{\deriv{y}{\theta}}^2 &= [r(1 - \cos \theta)]^2 + (r \sin \theta)^2 \nl &= r^2 (1 - 2 \cos \theta + \cos^2 \theta + \sin^2 \theta) \nl &= r^2 (2 - 2 \cos \theta) \pd \ea \] So the arc length is \[L = \int_0^{2 \pi} \sqrt{r^2 (2 - 2 \cos \theta)} \di \theta = r \int_0^{2 \pi} \sqrt{2 - 2 \cos \theta} \di \theta \pd\] Using the identity \(\sin^2 x = \tfrac{1}{2} (1 - \cos 2x)\) with \(2x = \theta,\) we have \[\sin^2 \frac{\theta}{2} = \tfrac{1}{2} (1 - \cos \theta) \iffArrow 1 - \cos \theta = 2 \sin^2 \frac{\theta}{2} \pd\] The integral therefore becomes \[ \ba L &= r \int_0^{2 \pi} \sqrt{4 \sin^2 \frac{\theta}{2}} \di \theta \nl &= r \int_0^{2 \pi} \abs{2 \sin \frac{\theta}{2}} \di \theta \nl &= 2r \int_0^{2 \pi} \sin \frac{\theta}{2} \di \theta \nl &= 4 r \par{-\cos \frac{\theta}{2}} \intEval_0^{2 \pi} \nl &= \boxed{8r} \ea \]

SOLUTION Let \(\theta = m\) be the line that connects the origin to the particle. Since the particle travels counterclockwise with an angular speed of \(4\) radians per minute, this is the rate at which \(m\) increases with time. Mathematically, if \(t\) is time in minutes, then \(\textderiv{m}{t} = 4.\) In addition, the rate at which the particle sweeps out area with respect to \(m\) is given by \[\deriv{A}{m} = \deriv{}{m} \int_0^m \tfrac{1}{2} (1 + \sin \theta)^2 \di \theta \pd\] By Part I of the Fundamental Theorem of Calculus (see Section 4.3), this equation becomes \[\deriv{A}{m} = \tfrac{1}{2} (1 + \sin m)^2 \pd\] We want \(\textderiv{A}{t}\) when \(m = \pi/6;\) by the Chain Rule, we have \[\deriv{A}{t} = \deriv{A}{m} \deriv{m}{t} = 4 \cdot \tfrac{1}{2} (1 + \sin m)^2 = 2 (1 + \sin m)^2 \pd\] (See Section 2.7 to review the procedure of working with related rates.) Substituting \(m = \pi/6,\) we have \[ \deriv{A}{t} \intEval_{m = \pi/6} = 2 \par{1 + \sin \frac{\pi}{6}}^2 \nl = \boxed{\frac{9}{2} \un{in}^2/\un{min}} \]

SOLUTION Parametric equations for the cannonball's motion in the horizontal and vertical directions are, respectively, \[ \ba x &= (20 \cos 57 \degree) t \approx 10.893 t \cma \nl y &= 10 + (20 \sin 57 \degree) t - \tfrac{1}{2} (9.8) t^2 \approx 10 + 16.773 t - 4.9 t^2 \pd \ea \] We see \[\deriv{x}{t} = 10.893 \and \deriv{y}{t} = 16.773 - 9.8 t \pd\] The ball hits the ground when \(y = 0\)—that is, when \[t = \frac{-16.773 - \sqrt{16.773^2 - 4(-4.9)(10)}}{2(-4.9)} \approx 3.941 \un{sec} \pd\] The arc length is therefore \[ \ba L &= \int_0^{3.941} \sqrt{(10.893)^2 + (16.773 - 9.8t)^2} \di t \nl &\approx \boxed{60.451 \un{m}} \ea \]

SOLUTION

Let \(x\) be the projectile's horizontal position and \(y\) be its vertical position. If \(t\) is time, then parametric equations for the projectile are \[ \baat{2} x &= v_0 (\cos \alpha) t \cma \nl y &= v_0 (\sin \alpha) t - \tfrac{1}{2} gt^2 \pd \eaat \] The projectile's distance from the launch point to a point on its trajectory (see the figure) is, by the Pythagorean Theorem, \[ \ba r &= \length{OP} \nl &= \sqrt{x^2 + y^2} \nl &= \sqrt{[v_0 (\cos \alpha) t]^2 + \parbr{v_0 (\sin \alpha) t - \tfrac{1}{2} gt^2}^2} \nl &= \sqrt{\par{v_0}^2 \par{\cos^2 \alpha} t^2 + \par{v_0}^2 \par{\sin^2 \alpha} t^2 - v_0 (\sin \alpha) gt^3 + \tfrac{1}{4} g^2 t^4} \nl &= \sqrt{\tfrac{1}{4} g^2 t^4 - v_0 (\sin \alpha) gt^3 + \par{v_0}^2 t^2} \pd \ea \] The distance \(r\) is always increasing, so we need \(\textderiv{r}{t} \gt 0\) for all \(t \gt 0.\) Differentiating gives \[ \deriv{r}{t} = \frac{g^2 t^3 - 3 v_0 (\sin \alpha) gt^2 + 2 \par{v_0}^2 t}{2 \sqrt{\tfrac{1}{4} g^2 t^4 - v_0 (\sin \alpha) gt^3 + \par{v_0}^2 t^2}} \pd \] This function must not have any critical points; that is, \(\textderiv{r}{t} \ne 0.\) We therefore see \[ \ba \frac{g^2 t^3 - 3 v_0 (\sin \alpha) gt^2 + 2 \par{v_0}^2 t}{2 \sqrt{\tfrac{1}{4} g^2 t^4 - v_0 (\sin \alpha) gt^3 + \par{v_0}^2 t^2}} &\ne 0 \nl g^2 t^3 - 3 v_0 (\sin \alpha) gt^2 + 2 \par{v_0}^2 t &\ne 0 \nl t \parbr{g^2 t^2 - 3 v_0 (\sin \alpha) gt + 2 \par{v_0}^2} &\ne 0 \pd \ea \] By the Quadratic Formula, the expression in brackets equals \(0\) when \[ \ba t &= \frac{3 v_0 (\sin \alpha) g + \sqrt{\parbr{- 3 v_0 (\sin \alpha) g}^2 - 4 (g^2) [2 \par{v_0}^2]}}{2 g^2} \nl &= \frac{3 v_0 (\sin \alpha) g + \sqrt{9 \par{v_0}^2 g^2 \sin^2 \alpha - 8 g^2 \par{v_0}^2}}{2 g^2} \pd \ea \] No real solutions of \(t\) exist if the discriminant, the expression within the square root, is negative. So we see \[ \ba 9 \par{v_0}^2 g^2 \sin^2 \alpha - 8 g^2 \par{v_0}^2 \lt 0 &\iffArrow 9 \sin^2 \alpha - 8 \lt 0 \nl &\iffArrow \sin^2 \alpha \lt \frac{8}{9} \nl &\iffArrow \sin \alpha \lt \frac{\sqrt 8}{3} \nl &\iffArrow \alpha \lt \boxed{\asin \par{\frac{\sqrt 8}{3}}} \approx 70.429 \degree \pd \ea \] This result is independent of \(g\) and the projectile's initial speed!