Click here for answers. sin x cos x s2

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1 CHALLENGE PROBLEMS CHALLENGE PROBLEMS: CHAPTER A Click here for answers. S Click here for solutions. sin cos s. Show that for all.. Show that y y 6 for all numbers and y such that and. 3. Let a and b be positive numbers. Show that not both of the numbers a b and b a can be greater than. y. Find the point on the parabola y at which the tangent line cuts from the first quadrant the triangle with the smallest area. 5. Find the highest and lowest points on the curve y y. 6. Water is flowing at a constant rate into a spherical tank. Let Vt be the volume of water in the tank and Ht be the height of the water in the tank at time t. (a) What are the meanings of Vt and Ht? Are these derivatives positive, negative, or zero? (b) Is V t positive, negative, or zero? Eplain. (c) Let t, t, and t 3 be the times when the tank is one-quarter full, half full, and three-quarters full, respectively. Are the values Ht, Ht, and Ht 3 positive, negative, or zero? Why? 7. Find the absolute maimum value of the function f 8. Find a function f such that f, f 0 0, and f 0 for all, or prove that such a function cannot eist. y A y B 9. The line y m b intersects the parabola y in points A and B (see the figure). Find the point P on the arc AOB of the parabola that maimizes the area of the triangle PAB. 0. Sketch the graph of a function f such that f 0 for all, f 0 for f 0 for, and lim l f 0.,. Determine the values of the number a for which the function f has no critical number: ym+b f a a 6 cos a cos O P. Sketch the region in the plane consisting of all points, y such that y y y Thomson Brooks-Cole copyright 007 FIGURE FOR PROBLEM 9 C D A B FIGURE FOR PROBLEM 3. Let ABC be a triangle with BAC 0 and AB AC. (a) Epress the length of the angle bisector AD in terms of AB. (b) Find the largest possible value of.. (a) Let ABC be a triangle with right angle A and hypotenuse. (See the figure.) If the inscribed circle touches the hypotenuse at D, show that C (b) If, epress the radius r of the inscribed circle in terms of a and. (c) If a is fied and varies, find the maimum value of r. CD ( BC AC AB ) 5. A triangle with sides a, b, and c varies with time t, but its area never changes. Let be the angle opposite the side of length a and suppose always remains acute. (a) Epress ddt in terms of b, c,, dbdt, and dcdt. (b) Epress dadt in terms of the quantities in part (a). AD a BC 6. ABCD is a square piece of paper with sides of length m. A quarter-circle is drawn from B to D with center A. The piece of paper is folded along EF, with E on AB and F on AD, so that A falls on the quarter-circle. Determine the maimum and minimum areas that the triangle AEF could have.

2 CHALLENGE PROBLEMS island 5 km C B 3 km FIGURE FOR PROBLEM 7 D nest 7. Ornithologists have determined that some species of birds tend to avoid flights over large bodies of water during daylight hours. It is believed that more energy is required to fly over water than land because air generally rises over land and falls over water during the day. A bird with these tendencies is released from an island that is 5 km from the nearest point B on a straight shoreline, flies to a point C on the shoreline, and then flies along the shoreline to its nesting area D. Assume that the bird instinctively chooses a path that will minimize its energy ependiture. Points B and D are 3 km apart. (a) In general, if it takes. times as much energy to fly over water as land, to what point C should the bird fly in order to minimize the total energy epended in returning to its nesting area? (b) Let W and L denote the energy (in joules) per kilometer flown over water and land, respectively. What would a large value of the ratio WL mean in terms of the bird s flight? What would a small value mean? Determine the ratio WL corresponding to the minimum ependiture of energy. (c) What should the value of WL be in order for the bird to fly directly to its nesting area D? What should the value of WL be for the bird to fly to B and then along the shore to D? (d) If the ornithologists observe that birds of a certain species reach the shore at a point km from B, how many times more energy does it take a bird to fly over water than land? 8. The blood vascular system consists of blood vessels (arteries, arterioles, capillaries, and veins) that convey blood from the heart to the organs and back to the heart. This system should work so as to minimize the energy epended by the heart in pumping the blood. In particular, this energy is reduced when the resistance of the blood is lowered. One of Poiseuille s Laws gives the resistance R of the blood as R C L r where L is the length of the blood vessel, r is the radius, and C is a positive constant determined by the viscosity of the blood. (Poiseuille established this law eperimentally.) The figure shows a main blood vessel with radius r branching at an angle into a smaller vessel with radius r. C vascular branching r b A r B a (a) Use Poiseuille s Law to show that the total resistance of the blood along the path ABC is R C a b cot b csc r r where a and b are the distances shown in the figure. (b) Prove that this resistance is minimized when cos r r (c) Find the optimal branching angle (correct to the nearest degree) when the radius of the smaller blood vessel is two-thirds the radius of the larger vessel. Thomson Brooks-Cole copyright 007

3 CHALLENGE PROBLEMS 3 9. A light is to be placed atop a pole of height h feet to illuminate a busy traffic circle, which has a radius of 0 ft. The intensity of illumination I at any point P on the circle is directly proportional to the cosine of the angle (see the figure) and inversely proportional to the square of the distance d from the source. (a) How tall should the light pole be to maimize I? (b) Suppose that the light pole is h feet tall and that a woman is walking away from the base of the pole at the rate of fts. At what rate is the intensity of the light at the point on her back ft above the ground decreasing when she reaches the outer edge of the traffic circle? h d 0 P 0. If a projectile is fired with an initial velocity v at an angle of inclination from the horizontal, then its trajectory, neglecting air resistance, is the parabola y tan t v cos 0 (a) Suppose the projectile is fired from the base of a plane that is inclined at an angle,, from the horizontal, as shown in the figure. Show that the range of the projectile, measured up the slope, is given by R v cos t cos (b) Determine so that R is a maimum. (c) Suppose the plane is at an angle below the horizontal. Determine the range R in this case, and determine the angle at which the projectile should be fired to maimize R. y sin 0 0 å R c c. The speeds of sound in an upper layer and in a lower layer of rock and the thickness h of the upper layer can be determined by seismic eploration if the speed of sound in the lower layer is greater than the speed in the upper layer. A dynamite charge is detonated at a point P and the transmitted signals are recorded at a point Q, which is a distance D from P. The first signal to arrive at Q travels along the surface and takes T seconds. The net signal travels from P to a point R, from R to S in the lower layer, and then to Q, taking T seconds. The third signal is reflected off the lower layer at the midpoint O of RS and takes T 3 seconds to reach Q. (a) Epress T, T, and T 3 in terms of D, h, c, c, and. (b) Show that T is a minimum when sin c c. (c) Suppose that D km, T 0.6 s, T 0.3 s, T s. Find c, c, and h. P D Speed of soundc Q h Thomson Brooks-Cole copyright 007 R Speed of soundc Note: Geophysicists use this technique when studying the structure of Earth s crust, whether searching for oil or eamining fault lines. O S

4 CHALLENGE PROBLEMS. For what values of c is there a straight line that intersects the curve y c 3 5 in four distinct points? d B E C r F D FIGURE FOR PROBLEM 3 3. One of the problems posed by the Marquis de l Hospital in his calculus tetbook Analyse des Infiniment Petits concerns a pulley that is attached to the ceiling of a room at a point C by a rope of length r. At another point B on the ceiling, at a distance d from C (where d r), a rope of length is attached and passed through the pulley at F and connected to a weight W. The weight is released and comes to rest at its equilibrium position D. As l Hospital argued, this happens when the distance is maimized. Show that when the system reaches equilibrium, the value of is ED r d (r sr 8d ) Notice that this epression is independent of both W and.. Given a sphere with radius r, find the height of a pyramid of minimum volume whose base is a square and whose base and triangular faces are all tangent to the sphere. What if the base of the pyramid is a regular n-gon (a polygon with n equal sides and angles)? (Use the fact that the volume of a pyramid is 3 Ah, where A is the area of the base.) 5. Assume that a snowball melts so that its volume decreases at a rate proportional to its surface area. If it takes three hours for the snowball to decrease to half its original volume, how much longer will it take for the snowball to melt completely? 6. A hemispherical bubble is placed on a spherical bubble of radius. A smaller hemispherical bubble is then placed on the first one. This process is continued until n chambers, including the sphere, are formed. (The figure shows the case n.) Use mathematical induction to prove that the maimum height of any bubble tower with n chambers is sn. Thomson Brooks-Cole copyright 007

5 CHALLENGE PROBLEMS 5 ANSWERS S Solutions 5.,,, m, m. 3.5 a.5 3. (a) (b) 5. (a) tan dc (b) c dt db b dt 7. (a) About 5. km from B (b) C is close to B; C is close to D; WL s5, where (c).07; no such value (d) s.6 di 9. (a) (b), where k is the constant of proportionality dt 80kh 0s 8 ft h (a) T,, T 3 sh D Dc T h sec c D h tan c c (c) c 3.85 kms, c 7.66 kms, h 0. km 5. 3(s 3 ) h 3 b db dt c dc dt b dc dt dt c db sec sb c bc cos BC Thomson Brooks-Cole copyright 007

6 6 CHALLENGE PROBLEMS SOLUTIONS E Eercises. Let f() sin cos on [0, π] since f has period π. f 0 () cos +sin 0 cos sin tan 3π or 7π.Evaluatingf at its critical numbers and endpoints, we get f(0), f 3π, f 7π,andf(π). Sof has absolute maimum value and absolute minimum value.thus, sin cos sin cos. 3. First we show that ( ) for all.letf() ( ). Thenf 0 (). Thisis0 when and f 0 () > 0 for <, f 0 () < 0 for >, so the absolute maimum of f is f.thus, ( ) for all. Now suppose that the given assertion is false, that is, a( b) > and b( a) >. Multiply these inequalities: a( b)b( a) > 6 [a( a)] [b( b)] > 6. But we know that a( a) and b( b) [a( a)] [b( b)]. Thus, we have a contradiction, so the given assertion is proved Differentiating + y + y implicitly with respect to gives + y + dy dy +y d d 0,so dy d + y dy. At a highest or lowest point, 0 y. Substituting for y in the original +y d equation gives + ( )+( ),so3 and ±. If,theny,andif then y. Thus, the highest and lowest points are (, ) and (, ). 7. f() ( ) + + ( ) + + +( ) if <0 if 0 < if ( ) + if <0 (3 ) f 0 () ( + ) + (3 ) if 0 << ( + ) if > ( ) We see that f 0 () > 0 for <0 and f 0 () < 0 for >. For0 <<, wehave f 0 () (3 ) ( +) ( ) (3 ) ( +) (3 ) ( +),sof 0 () < 0 for 0 <<, f 0 () 0 and f 0 () > 0 for <<. Wehave shownthatf 0 () > 0 for <0; f 0 () < 0 for 0 <<; f 0 () > 0 for <<;andf 0 () < 0 for >. Therefore, by the First Derivative Test, the local maima of f are at 0and,wheref takes the value. Therefore, is the absolute maimum value of f A, and B,,where and are the solutions of the quadratic equation m + b.let P, and set A (, 0), B (, 0),andP (, 0). Letf() denote the area of triangle PAB. Then f() can be epressed in terms of the areas of three trapezoids as follows: Thomson Brooks-Cole copyright 007 f()area (A ABB ) area (A AP P ) area (B BPP ) + ( ) + ( ) + ( )

7 CHALLENGE PROBLEMS 7 After epanding and canceling terms, we get f() + + ( )+ ( )+ ( ) f 0 () + +( ). f 00 () [( )] < 0 since >. f 0 () 0 ( ) P ( + ). f( P ) ( ) + ( ) + ( + ) ( ) ( ) + ( )( + ) ( 8 ) ( )( ) 8 ( )( ) 8 ( ) 3 To put this in terms of m and b,wesolvethesystemy and y m + b,givingus m b 0 m m +b. Similarly, m + m +b. The area is then ( 8 ) 3 m +b 3 8, and is attained at the point P P, P P m, m. Note: Another way to get an epression for f() is to use the formula for an area of a triangle in terms of the coordinates of the vertices: f() f() a + a 6 cos +(a ) +cos f 0 () a + a 6 sin () + (a ). The derivative eists for all, so the only possible critical points will occur where f 0 () 0 (a )(a + 3)sin a either a or (a + 3)sin, with the latter implying that sin. Since the range of sin is [, ], this equation has no solution whenever either (a +3) (a +3) < or (a +3) >. Solving these inequalities, we get 7 <a< (a) Let y AD, AB,and/ AC,sothat AB AC. We compute the area A of ABC in two ways. First, A AB AC sin π 3 3 A (area of ABD)+(area of ACD) 3. Second, AB AD sin π 3 + AD AC sin π 3 y 3 + y(/) 3 3 y( +/) Equating the two epressions for the area, we get 3 y + 3 y +/ Another method: Use the Law of Sines on the triangles ABD and ABC. InABD, wehave A + B + D α + D 80 D 0 α. Thus, +, >0. y sin(0 α) sin 0 cos α cos 0 sin α sin α sin α 3 cos α + sin α sin α y 3 cot α +, Thomson Brooks-Cole copyright 007 andbyasimilarargumentwithabc, 3 cot α +. Eliminating cot α gives y + + y +, >0.

8 8 CHALLENGE PROBLEMS (b) We differentiate our epression for y with respect to to find the maimum: dy d + () ( +) ( 0when. This indicates a maimum by the First Derivative +) Test, since y 0 () > 0 for 0 << and y 0 () < 0 for >, so the maimum value of y is y(). 5. (a) A bh with sin θ h/c,soa bc sin θ. ButA is a constant, so differentiating this equation with respect to t,weget da dt 0 bc cos θ dθ dt + b dc db sin θ + dt dt c sin θ bc cos θ dθ dt sin θ b dc dt + c db dθ dt dt tan θ dc c dt + b db. dt (b) We use the Law of Cosines to get the length of side a in terms of those of b and c, and then we differentiate implicitly with respect to t: a b + c bc cos θ a da db dc b +c dt dt dt bc( sin θ) dθ dt + b dc db cos θ + dt dt c cos θ da dt b db a dt + c dc dθ + bc sin θ dt dt b dc db cos θ c dt dt cos θ. Now we substitute our value of a from the Law of Cosines and the value of dθ/dt from part (a), and simplify (primes signify differentiation by t): da dt bb0 + cc 0 + bc sin θ [ tan θ(c 0 /c + b 0 /b)] (bc 0 + cb 0 )(cos θ) b + c bc cos θ bb0 + cc 0 [sin θ(bc 0 + cb 0 )+cos θ(bc 0 + cb 0 )]/ cos θ b + c bc cos θ bb0 + cc 0 (bc 0 + cb 0 )sec θ b + c bc cos θ 7. (a) If k energy/km over land, then Set de d 0:.k k 5 + / energy/km over water.k. So the total energy is E.k k(3 ), 0 3, and so de d.k k. (5 + / ) Testing against the value of E at the endpoints: E(0).k(5) + 3k 0k, E(5.) 7.9k, E(3) 9.5k. Thus, to minimize energy, the bird should flytoapointabout5. km from B. (b) If W/L is large, the bird would flytoapointc that is closer to B than to D to minimize the energy used flying over water. If W/L is small, the bird would flytoapointc that is closer to D than to B to minimize the distance Thomson Brooks-Cole copyright 007 of the flight. E W L(3 ) de d W L 0when W L.By the same sort of argument as in part (a), this ratio will give the minimal ependiture of energy if the bird heads for the point km from B.

9 CHALLENGE PROBLEMS 9 (c) For flight direct to D, 3, so from part (b), W/L There is no value of W/L for which the bird should fly directly to B. But note that lim, so if the point at which E is a minimum is 0 +(W/L) close to B,thenW/L is large. (d) Assuming that the birds instinctively choose the path that minimizes the energy ependiture, we can use the equation for de/d 0from part (a) with.k c,,andk : c() 5 + / c / (a) I k cos θ d k(h/d) k h d d k h h 3 k h (600 + h ) 3/ di dh k (600 + h ) 3/ h 3 (600 + h ) / h k(600 + h ) / (600 + h 3h ) [(600 + h ) 3/ ] (600 + h ) 3/ k(600 h ) (600 + h ) 5/ [k is the constant of proportionality] Set di/dh 0: 600 h 0 h 800 h By the First Derivative Test, (b) I has a local maimum at h 0 8 ft. d dt ft/s I k cos θ d k(h ) [(h ) + ] 3/ k[(h )/d] k(h ) d d 3 k(h ) (h ) + 3/ di dt di d d dt k(h ) 3 (h ) + 5/ d dt k(h )( 3) (h ) + 5/ di 80k(h ) dt 0 [(h ) + 600] 5/ k(h ) [(h ) + ] 5/. (a) Distance rate time, sotime distance/rate. T D c, T PR c + RS c h sec θ c + D h tan θ, T 3 h + D / h + D. c c c (b) dt dθ h sec θ tan θ h sec θ 0when h sec θ tan θ sec θ c c c c sin θ c cos θ sin θ 0 c cos θ c cos θ c cos θ this gives a minimum. 0 sin θ c c. The First Derivative Test shows that (c) Using part (a) with D and T 0.6,wehaveT D c 0.6 c 3.85 km/s. T h + D 3 c h + D T 3 c h T 3 c D (0.3) (/0.6) 0. km. To find c,we Thomson Brooks-Cole copyright 007 use sin θ c h sec θ D h tan θ from part (b) and T + from part (a). From the figure, c c c

10 0 CHALLENGE PROBLEMS sin θ c c sec θ hc T c c c c and tan θ c c + D c c hc. c c c c,so c c Using the values for T [given as 0.3], h, c,andd, we can graph hc Y T and Y c c c + D c c hc c c c and find their intersection points. Doing so gives us c.0 and 7.66,butifc.0,thenθ arcsin(c /c ) 69.6, which implies that point S is to the left of point R in the diagram. So c 7.66 km/s. 3. Let a EF and b BF asshowninthefigure. Since BF + FD, FD b. Now ED EF + FD a + b r + (d ) + a r + (d ) + r r + d d + + r Let f() r + d + r d. f 0 () r / ( ) d + r d / ( d) r + d d + r d. f 0 () 0 r d d + r d r d d + r d d + r d 3 d r d 0d 3 d r + d r 0d ( d) r d 0d ( d) r ( + d)( d) 0( d) d r ( + d) But d>r>,so 6 d. Thus, we solve d r dr 0for : r ± ( r ) (d)( dr ) r ± r +8d r. Because r (d) d +8d r >r, the negative can be discarded. Thus, r + r r +8d d r r + r +8d d r + r r +8d d (r >0) Themaimumvalueof ED occurs at this value of. Thomson Brooks-Cole copyright V 3 πr3 dv dr dt πr dt.butdv dt is proportional to the surface area, so dv dt k πr for some constant k. Therefore, πr dr dt k πr dr k constant. An antiderivative of k with respect to t dt

11 CHALLENGE PROBLEMS is kt,sor kt + C. Whent 0, the radius r must equal the original radius r 0,soC r 0,andr kt + r 0.To find k we use the fact that when t 3, r 3k + r 0 and V V0 π(3k + 3 r0)3 3 πr3 0 (3k + r 0) 3 r3 0 3k + r 0 r 3 0 k 3 r0.sincer kt + r 3 0, r 3 r0 t + r 3 0. When the snowball has melted completely we have r 0 r 3 0 longer. t + r 3 0 0which gives t Hence, it takes h 33 min Thomson Brooks-Cole copyright 007