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1 CHALLENGE PROBLEMS: CHALLENGE PROBLEMS 1 CHAPTER A Click here for answers S Click here for solutions A 1 Find points P and Q on the parabola 1 so that the triangle ABC formed b the -ais and the tangent lines at P and Q is an equilateral triangle ; Find the point where the curves 4 and are tangent to each other, that is, have a common tangent line Illustrate b sketching both curves and the common tangent Suppose f is a function that satisfies the equation P Q f f f B 0 C for all real numbers and Suppose also that FIGURE FOR PROBLEM 1 lim l0 f 1 FIGURE FOR PROBLEM 4 (a) Find f 0 (b) Find f 0 (c) Find f 4 A car is traveling at night along a highwa shaped like a parabola with its verte at the origin (see the figure) The car starts at a point 100 m west and 100 m north of the origin and travels in an easterl direction There is a statue located 100 m east and 50 m north of the origin At what point on the highwa will the car s headlights illuminate the statue? d n 5 Prove that d n sin4 cos 4 4 n1 cos4 n 6 Find the nth derivative of the function f n 1 7 The figure shows a circle with radius 1 inscribed in the parabola Find the center of the circle = If f is differentiable at a, where a 0, evaluate the following limit in terms of f a: 9 The figure shows a rotating wheel with radius 40 cm and a connecting rod AP with length 1 m The pin P slides back and forth along the -ais as the wheel rotates counterclockwise at a rate of 60 revolutions per minute (a) Find the angular velocit of the connecting rod, ddt, in radians per second, when (b) Epress the distance OP in terms of (c) Find an epression for the velocit of the pin P in terms of lim l a f f a s sa A O å P (, 0)

2 CHALLENGE PROBLEMS 10 Tangent lines and are drawn at two points and on the parabola and the intersect at a point P Another tangent line T is drawn at a point between P 1 and P ; it intersects T 1 at Q 1 and T at Show that Q T 1 T P 1 P PQ1 PQ 1 PP1 PP T 0 N T N N P T 11 Let T and N be the tangent and normal lines to the ellipse at an point P on the ellipse in the first quadrant Let T and T be the - and -intercepts of T and N and N be the intercepts of N As P moves along the ellipse in the first quadrant (but not on the aes), what values can T, T, N, and N take on? First tr to guess the answers just b looking at the figure Then use calculus to solve the problem and see how good our intuition is sin sin 9 1 Evaluate lim l 0 1 (a) Use the identit for tan (see Equation 14b in Appendi A) to show that if two lines L and intersect at an angle, then L 1 FIGURE FOR PROBLEM 11 m m1 tan 1 m 1m where m 1 and m are the slopes of L 1 and L, respectivel (b) The angle between the curves C 1 and C at a point of intersection P is defined to be the angle between the tangent lines to C 1 and C at P (if these tangent lines eist) Use part (a) to find, correct to the nearest degree, the angle between each pair of curves at each point of intersection (i) and (ii) and Let P 1, 1 be a point on the parabola 4p with focus F p, 0 Let be the angle between the parabola and the line segment FP, and let be the angle between the horizontal line 1 and the parabola as in the figure Prove that (Thus, b a principle of geometrical optics, light from a source placed at F will be reflected along a line parallel to the -ais This eplains wh paraboloids, the surfaces obtained b rotating parabolas about their aes, are used as the shape of some automobile headlights and mirrors for telescopes) = P(, ) å 0 F( p, 0) =4p A Q C R O P 15 Suppose that we replace the parabolic mirror of Problem 14 b a spherical mirror Although the mirror has no focus, we can show the eistence of an approimate focus In the figure, C is a semicircle with center O A ra of light coming in toward the mirror parallel to the ais along the line PQ will be reflected to the point R on the ais so that PQO OQR (the angle of incidence is equal to the angle of reflection) What happens to the point R as P is taken closer and closer to the ais? 16 If f and t are differentiable functions with f 0 t0 0 and t0 0, show that lim l 0 f t f 0 t0 FIGURE FOR PROBLEM 15 sina sina sin a 17 Evaluate lim l 0 18 Given an ellipse a b 1, where a b, find the equation of the set of all points from which there are two tangents to the curve whose slopes are (a) reciprocals and (b) negative reciprocals

3 CHALLENGE PROBLEMS 19 Find the two points on the curve 4 that have a common tangent line 0 Suppose that three points on the parabola have the propert that their normal lines intersect at a common point Show that the sum of their -coordinates is 0 1 A lattice point in the plane is a point with integer coordinates Suppose that circles with radius r are drawn using all lattice points as centers Find the smallest value of r such that an line with slope intersects some of these circles 5 A cone of radius r centimeters and height h centimeters is lowered point first at a rate of 1 cms into a tall clinder of radius R centimeters that is partiall filled with water How fast is the water level rising at the instant the cone is completel submerged? A container in the shape of an inverted cone has height 16 cm and radius 5 cm at the top It is partiall filled with a liquid that oozes through the sides at a rate proportional to the area of the container that is in contact with the liquid (The surface area of a cone is rl, where r is the radius and l is the slant height) If we pour the liquid into the container at a rate of cm min, then the height of the liquid decreases at a rate of 0 cmmin when the height is 10 cm If our goal is to keep the liquid at a constant height of 10 cm, at what rate should we pour the liquid into the container? CAS 4 (a) The cubic function f 6 has three distinct zeros: 0,, and 6 Graph f and its tangent lines at the average of each pair of zeros What do ou notice? (b) Suppose the cubic function f a b c has three distinct zeros: a, b, and c Prove, with the help of a computer algebra sstem, that a tangent line drawn at the average of the zeros a and b intersects the graph of f at the third zero

4 4 CHALLENGE PROBLEMS ANSWERS S Solutions 1 (s, 1 4 ) (a) 0 (b) 1 (c) f 1 7 (0, 5 4 ) 9 (a) 4ss11 rads (b) 40(cos s8 cos ) cm (c) 11 T,, T,, N (0, 5 ), N ( 5, 0) 1 (b) (i) 5 (or 17) (ii) 6 (or 117) 15 R approaches the midpoint of the radius AO 17 sin a 19 1,, 1, 0 1 s sin (1 cos s8 cos ) cms cm min

5 CHALLENGE PROBLEMS 5 SOLUTIONS E Eercises 1 Let a be the -coordinate of Q Since the derivative of =1 is 0 =, the slope at Q is a But since the triangle is equilateral, AO/OC = /1, so the slope at Q is Therefore, we must have that a = a = Thus, the point Q has coordinates, 1 =, 1 andbsmmetr,p has coordinates 4, 1 4 (a) Put =0and =0in the equation: f(0 + 0) = f(0) + f(0) f(0) = f(0) Subtracting f(0) from each side of this equation gives f(0) = 0 (b) f 0 f(0 + h) f(0) f(0) + f(h)+0 h +0h f(0) f(h) (0) =lim h 0 h =lim f() 0 =1 (c) f 0 f( + h) f() f()+f(h)+ h + h f() () =lim h 0 h f(h)+ h + h f(h) + + h =1+ 5 We use mathematical induction Let S n be the statement that S 1 is true because d n d n sin 4 +cos 4 =4 n 1 cos(4 + nπ/) d d sin 4 +cos 4 =4sin cos 4cos sin =4sin cos sin cos Now assume S k is true, that is, = 4sin cos cos = sin cos = sin 4 =sin( 4) =cos π ( 4) =cos π +4 =4 n 1 cos 4 + n π when n =1 d k sin 4 +cos 4 =4 k 1 cos 4 + k π d k Then d k+1 d sin 4 +cos 4 = d d k k+1 d d sin 4 +cos 4 = d k d 4 k 1 cos 4 + k π = 4 k 1 sin 4 + k π d d 4 + k π = 4 k sin 4 + k π =4 k sin 4 k π =4 k cos π 4 k π =4 k cos 4 +(k +1) π which shows that S k+1 is true Therefore, d n sin 4 +cos 4 =4 n 1 cos 4 + n π d n for ever positive integer n, b mathematical induction Another proof: First write sin 4 +cos 4 = sin +cos sin cos =1 1 sin =1 1 (1 cos 4) = cos 4 Then we have dn sin 4 +cos 4 = dn d n d + 1 cos 4 n 4 4 = 1 4 4n cos 4 + n π =4 n 1 cos 4 + n π 7 We must find a value 0 such that the normal lines to the parabola = at = ± 0 intersect at a point one unit from the points ± 0, 0 The normals to = at = ± 0 have slopes 1 and pass through ± 0, 0 ± 0 respectivel, so the normals have the equations 0 = 1 0 ( 0 ) and 0 = 1 0 ( + 0 ) The common

6 6 CHALLENGE PROBLEMS -intercept is Wewanttofind the value of 0 for which the distance from 0, to 0, 0 equals 1 The square of the distance is ( 0 0) ,the-intercept is = 5 4, so the center of the circle is at 0, 5 4 = =1 4 0 = ± For these values of Another solution: Let the center of the circle be (0,a) Then the equation of the circle is +( a) =1 Solving with the equation of the parabola, =,weget + a =1 + 4 a + a =1 4 +(1 a) + a 1=0 The parabola and the circle will be tangent to each other when this quadratic equation in hasequalroots;thatis,whenthediscriminantis0 Thus,(1 a) 4 a 1 =0 1 4a +4a 4a +4=0 4a =5,soa = 5 4 The center of the circle is 0, We can assume without loss of generalit that θ =0at time t =0,sothatθ =1πt rad [The angular velocit of the wheel is 60 rpm =60 (π rad)/(60 s) =1π rad/s] Then the position of A as a function of time is A =(40cosθ, 40 sin θ) = (40 cos 1πt, 40 sin 1πt),sosin α = 40 sin θ = = sin θ = 1 sin 1πt 1 m 10 (a) Differentiating the epression for sin α, wegetcos α dα dt = 1 1π cos 1πt =4π cos θ When θ = π,wehavesin α = 1 sin θ = 6,socos α = 1 dα dt = 4π cos π cos α = π 11/1 = 4π rad/s 6 = 11 1 and (b) B the Law of Cosines, AP = OA + OP OA OP cos θ 10 =40 + OP 40 OP cos θ OP (80 cos θ) OP 1,800 = 0 OP = 1 80 cos θ ± 6400 cos θ +51,00 =40cosθ ± 40 cos θ +8 =40 cos θ + 8+cos θ cm [since OP > 0] Asacheck,notethat OP = 160 cm when θ =0and OP =80 cm when θ = π (c) B part (b), the -coordinate of P is given b =40 cos θ + 8+cos θ,so d dt = d dθ dθ dt =40 cosθ sin θ sin θ cos θ 1π = 480π sin θ 1+ cm/s 8+cos θ 8+cos θ In particular, d/dt =0cm/swhenθ =0and d/dt = 480π cm/swhenθ = π 11 It seems from the figure that as P approaches the point (0, ) from the right, T and T + AsP approaches the point (, 0) from the left, it appears that T + and T Soweguessthat T (, ) and T (, ) Itismoredifficult to estimate the range of values for N and N We might perhaps guess that N (0, ),and N (, 0) or (, 0) In order to actuall solve the problem, we implicitl differentiate the equation of the ellipse to find the equation of the tangent line: = =0,so 0 = 4 9 So at the point ( 0, 0 ) on the ellipse, an equation of the tangent line is 0 = 4 0 ( 0 ) or = This can be written as 0

7 CHALLENGE PROBLEMS = =1 =1, because (0,0) lies on the ellipse So an equation of the tangent line is Therefore, the -intercept T for the tangent line is given b 0T 9 =1 T = 9 0,andthe-intercept T is given b 0T 4 =1 T = 4 0 So as 0 takes on all values in (0, ), T takes on all values in (, ),andas 0 takes on all values in (0, ), T takes on all values in (, ) At the point ( 0, 0 ) on the ellipse, the slope of the normal line is 1 0 ( 0, 0 ) = 9 0, and its equation is 0 = 9 0 ( 0) Sothe-intercept N for the normal line is given b 0 0 = 9 ( N 0) N = = 5 0,andthe-intercept N is given b 9 0 N 0 = 9 (0 0) N = = 50 4 So as 0 takes on all values in (0, ), N takes on all values in 0, 5,andas0 takes on all values in (0, ), N takes on all values in 5, 0 1 (a) If the two lines L 1 and L have slopes m 1 and m and angles of inclination φ 1 and φ,thenm 1 =tanφ 1 and m =tanφ The triangle in the figure shows that φ 1 + α + (180 φ )=180 and so α = φ φ 1 Therefore, using the identit for tan( ), we have tan α =tan(φ φ 1 )= tan φ tan φ 1 1+tanφ tan φ 1 and so tan α = m m1 1+m 1m (b) (i) The parabolas intersect when =( ) =1If =,then 0 =, so the slope of the tangent to = at (1, 1) is m 1 =(1)=If =( ),then 0 =( ), so the slope of the tangent to =( ) at (1, 1) is m =(1 ) = Therefore, tan α = m m1 = 1+m 1m 1+( ) = 4 and so α =tan (or 17 ) (ii) =and 4 + +=0intersect when 4 + +=0 ( ) = 0 =0or,but0 is etraneous If =,then = ±1If =then 0 =0 0 = / and 4 + += =0 0 = At (, 1) the slopes are m 1 =and m =0,sotan α = = α 117 At(, 1) the slopes are m 1 = and m =0, so tan α = 0 ( ) 1+( )(0) = α 6 (or 117 )

8 8 CHALLENGE PROBLEMS 15 Since ROQ = OQP = θ, the triangle QOR is isosceles, so QR = RO = BtheLawofCosines, = + r r cos θ Hence, r cos θ = r,so = (since sin θ = /r), and hence r r cos θ = r cosθ Note that as 0+, θ 0 + r cos0 = r Thus,asP is taken closer and closer to the -ais, the point R approaches the midpoint of the radius AO 17 lim 0 sin(a +) sin(a + )+sina 0 sin a cos +cosa sin sina cos cosa sin +sina 0 sin a (cos cos +1)+cosa (sin sin) 0 sin a ( cos 1 cos +1)+cosa ( sin cos sin) 0 sin a ( cos )(cos 1) + cos a ( sin )(cos 1) (cos 1)[sin a cos +cosa sin ](cos +1) 0 (cos +1) sin [sin(a + )] sin sin(a + ) = lim 0 (cos +1) 0 cos +1 = sin(a +0) (1) cos = sin a 19 = 4 0 =4 4 1 The equation of the tangent line at = a is (a 4 a a) =(4a 4a 1)( a) or =(4a 4a 1) +( a 4 +a ) and similarl for = b So if at = a and = b we have the same tangent line, then 4a 4a 1=4b 4b 1 and a 4 +a = b 4 +b Thefirst equation gives a b = a b (a b)(a + ab + b )=(a b) Assuming a 6=b, wehave1=a + ab + b The second equation gives (a 4 b 4 )=(a b ) (a b )(a + b )=(a b ) which is true if a = b Substituting into 1=a + ab + b gives 1=a a + a a = ±1 so that a =1and b = 1 or vice versa Thus, the points (1, ) and ( 1, 0) have a common tangent line As long as there are onl two such points, we are done So we show that these are in fact the onl two such points Suppose that a b 6=0Then(a b )(a + b )=(a b ) gives (a + b )= or a + b = Thus, ab = a + ab + b a + b =1 = 1,sob = 1 a Hence, a + 1 9a =,so9a4 +1=6a 0=9a 4 6a +1= a 1 Soa 1=0 a = 1 b = 1 9a = 1 = a, contradicting our assumption that a 6=b 1 Because of the periodic nature of the lattice points, it suffices to consider the points in the 5 grid shown We can

9 CHALLENGE PROBLEMS 9 see that the minimum value of r occurs when there is a line with slope 5 which touches the circle centered at (, 1) and the circles centered at (0, 0) and (5, ) Tofind P, the point at which the line is tangent to the circle at (0, 0), we simultaneousl solve + = r and = = r = 4 r = r, = 5 rtofind Q, we either use smmetr or solve ( ) +( 1) = r and 1= 5 ( ) Asabove,weget = r, =1+ 5 r Now the slope of the line PQis,so 5 m PQ = 1+ 5 r 5 r r r = r 4 r = + 10r 4r = r =6 8r 58r = r = So the minimum value of r for which an line with slope 58 5 intersects circles with radius r centered at the lattice points on the plane is r = B similar triangles, r 5 = h 16 Equating the two epressions for dv dt l = 5 8 gives us k = r = 5h The volume of the cone is 16 5h V = 1 πr h = 1 π h = 5π h,so dv dt = 5π dh 56 h dt Nowtherateof change of the volume is also equal to the difference of what is being added ( cm /min) and what is oozing out (kπrl,whereπrl is the area of the cone and k is a proportionalit constant) Thus, dv dt and substituting h =10, dh dt = kπrl = 0, r = 5(10) 16 81,weget 5π 56 (10) ( 0) = kπ kπ = 5 8,and l 81 = =+ 750π Solving for k π 50π To maintain a certain height, the rate of oozing, kπrl, must equal the rate of the liquid 81 beingpouredin;thatis, dv dt π =0 kπrl = 50π 81 π π = 1104 cm /min 8 18