. Differential and integral calculus, an introductory course for colleges and engineering schools. 0, ±, ± as theabscissas of the points where the curve cuts OX, and the slopes at thesepoints are respectively , , — After calculating the coordi-nates of a few intermediate points the curve may be readily drawn. 2. y3=xi(2x-5)\ Solution. (a) The curve is symmetrical with respect neither to the axes nor tothe origin. (6) Writing the equation in the form 7/ = £3(2 x — 5) and differentiating it, we get , 10 3-1 , „ 10 2Z + 1 y=—-^r-> and y =——— ■ 3 33 9 33 y changes sign when
. Differential and integral calculus, an introductory course for colleges and engineering schools. 0, ±, ± as theabscissas of the points where the curve cuts OX, and the slopes at thesepoints are respectively , , — After calculating the coordi-nates of a few intermediate points the curve may be readily drawn. 2. y3=xi(2x-5)\ Solution. (a) The curve is symmetrical with respect neither to the axes nor tothe origin. (6) Writing the equation in the form 7/ = £3(2 x — 5) and differentiating it, we get , 10 3-1 , „ 10 2Z + 1 y=—-^r-> and y =——— ■ 3 33 9 33 y changes sign when x= —\ but not when3=0. In other words, by solving y = 0 weget the abscissa of a flex, but not by solvingy= oo. The flex is (-§,-3^2) and theslope of the flex-tangent is 5 \^2. It is readilyfound that the arc at the left of the flex is con-cave, and that at the right convex. (c) The solution of y = 0 and y = oo are respectively 3 = 1 and 3=0,and plainly y does change sign at these values of x. Moreover, when3=1, y >0, and the point (1,-3) is a minimum point. When x= 0,. §§65-66 MAXIMA AND MINIMA 83 y = oo, and here the second derivative gives us no information. How-ever, from the form of y we readily see that as x passes through 0, ychanges from + to —, and therefore the origin is a maximum point. Sincethe tangent at this point is perpendicular to OX (is in fact OY), the originmust be a cusp. By plotting a number of additional points the curvemay be accurately drawn. 65. Exercises. Determine convex and concave arcs, flexes,flex-tangent, and maximum and minimum points, and draw thecurves given by the following equations: 1. 2. 3. 4. 5. 6. 7. 8. 66. 14. 3 15 Sy=x(9-x2).8 y = x4 — 4 :y=8x2-x\ y={xz+l)(x-2).20y = x* yZ = X1— 1. y2= xz — = xA- = x(25-x2) = x2 (x + 5)$y=x(7-x2){x2-4)z. Problems in Maxima and Minima. l-3z 2x 1 + x2 16. b2x2 =b a2y2 = a2b2. 17. y= sin x; y = cosz; y= secx; y
Size: 1264px × 1977px
Photo credit: © Reading Room 2020 / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1910, booksubjectcalculu, bookyear1912
