54. sin(3y) = x+y, 2 - JI IT 4 In Exercises 55-62, find an equation of the tangent line at the given point. 55. xy+ x2y2 = 6, (2, 1) 56. x2/3 + 213 = 2, (1, 1) 57. x2 + siny = xy2 + 1, (1, 0) 58. sin(x - y) = x cos (y + "), (2,") 59. 2x1/2 + 4y-1/2 = xy, (1, 4) 60. x2 ey + yet = 4, (2, 0) 61. e2x-y = (2, 4) 62. y2ex-16 _ xy-1 = 2, (4, 2) 63. Find the points on the graph of y2 = x3 - 3x + 1 (Figure 9) the tangent line is horizontal, as follows: (a) First show that 2yy' = 3x2 - 3, where y' = dy/dx. (b) Do not solve for y'. Rather, set y' = 0 and solve for x. This yi values of x where the slope may be zero. that the positive value of x does not correspond to two points on the g188 CHAPTER 3 DIFFERENTIATION 70. GU Plot x3 + y' = 3xy + b for several values of bay scribe how the graph changes as b - 0. Then compute dy/dx at the por (b /,0). How does this value change as b - co? Do your plots conf. this conclusion? 71. Find the x-coordinates of the points where the tangent line is he zontal on the trident curve xy = x3 - 5x2 + 2x - 1, so named by lu Newton in his treatise on curves published in 1710 (Figure 12). Hint: 2x3 - 5x2 + 1 = (2x - 1)(x2 - 2x - 1). -2+ FIGURE 9 Graph of y2 = x3 - 3x + 1. 20 64. Show, by differentiating the equation, that if the tangent line at a point (x. y) on the curve x y - 2x + 8y = 2 is horizontal, then xy = 1. Then -2 substitute y = x- in xy - 2x + 8y = 2 to show that the tangent line is horizontal at the points (2. 2) and ( - 4, -4). -20 65. Find all points on the graph of 3x2 + 4y2 + 3xy = 24 where the tan- gent line is horizontal (Figure 10). FIGURE 12 Trident curve: xy = x3 - 5x2 + 2x - 1. 72. find an equation of the tangent line at each of the four points on the curve (x2 + y2 - 4x)? = 2(x2 + y?) where x = 1. This curve (Figure 13) is an example of a limacon of Pascal, named after the father of the French philosopher Blaise Pascal, who first described it in 1650. FIGURE 10 Graph of 3x2 + 4y2 + 3xy = 24. 66. Show that no point on the graph of x2 - 3xy + y? = 1 has a horizon- tal tangent line. -3+ 67. Figure I shows the graph of y + xy = x - x + 2. Find dy/dx at the two points on the graph with x-coordinate 0 and find an equation of the FIGURE 13 Limacon: (x2 + y2 - 4x)2 = 2(x2 + y?). tangent line at each of those points. 68. Folium of Descartes The curve x + y'= 3xy (Figure 1 1) was 73. Find the derivative, dy/dx, at the points where x = 1 on the folium first discussed in 1638 by the French philosopher-mathematician Rene (x2 + y?) = 4xy'. See Figure 14. Descartes, who called it the folium (meaning "leaf). Descartes's scien- tific colleague Gilles de Roberval called it the jasmine flower. Both men believed incorrectly that the leaf shape in the first quadrant was repeated in each quadrant, giving the appearance of petals of a flower. Find an equa- tion of the tangent line at the point (3. }). N -2+ FIGURE 14 Folium curve: (x2 + y2)2 = = Txy2. -2+ 74. CAS Plot (x2 + y?)? = 12(x2 - y?) + 2 for x and y between -4 FIGURE 11 Folium of Descartes: x3 + y3 = 3xy. and 4 using a computer algebra system. How many horizontal tangent lines does the curve appear to have? Find the points where these occur. 69. Find a point on the folium x' + y' = 3xy other than the origin at 75. Calculate dx/dy for the equation y# + 1 = y? + x2 and find the which the tangent line is horizontal. points on the graph where the tangent line is vertical