3 above fail? 13. For what value of n does the result in Example 16. Find the equation of the curve whose slope at any point is 2x + 3 and which passes through the point (2, 3). Find the equation of the tangent line at this point. Plot the curve. 17. The slope of a curve at any point is 6x2 + 4x and it passes through the origin. Find the equation of the curve and the equation of the normal line at the point of inflection. Plot the graphs of the two equations. 18. Find the cubic function whose graph has a maximum point at (1, 4) and a minimum point at (3, - 5). Find the equation of the tangent line at the point of inflection. Suggestion. Dxy = a(x − 1)(x − 3). 19. The slope of a curve at any point is 4x-3 and it passes through the point (1, 2). Find the equation of the curve and the coördinates of Since A changes as x changes and is determined when x is fixed, A is a function of x. As N moves to R, x increases by an amount Ax, A by an amount PNRQ = AA, and y by an amount SQ Ay. Since PNRQ is less than the rectangle TN RQ and greater than the rectangle PNRS, we have As Ax approaches zero, y + Ay approaches y and AA /Ax ap ΔΑ proaches DA; and as lies between the magnitudes y and Ax y + Ay we have D2A = y = f(x). Therefore we have the Theorem. The rate at which the area A changes with respect to x is equal to the right-hand ordinate of the bounding curve. Symbolically, ᎠᎪ (1) The area A can now be found as a function of x by integration, the constant of integration being determined by the fact that A = 0 when x = a. If a fixed right-hand boundary is chosen the area is determined. The method of finding the area is shown in EXAMPLE 1. Find the area bounded by the curve y = x2, the x-axis and the ordi4. nates at x 1 and x = ЦА 21 18 T&B $15 12 -9 Equation (4) gives the area under the curve starting at the ordinate MP, at x = 1, and continuing to any second ordinate. In this case the second ordinate is fixed at x = 4. Hence substituting x = 4 in (4) we have The graph of (4) is the curve LMD which crosses the x-axis at x= 1 where the area MPBN begins. The number of square units in MBPN is the same as the number of linear units in the ordinate ND, which is 21 units in length. Equation (1) may be interpreted thus: Suppose that we had y = c, whose graph is a straight line parallel to the x-axis. Then the rate of change of A with respect to x, D.A, would be uniform, since D⭑A Y = C. It would be measured by the change in A due to a unit change in x (page 48), that is, by a rectangle with base unity and altitude c. B D M A 1 N FIG. 176. Returning to the figure used in proving the theorem, let NR 1. Then the area of the rectangle PNRS = y x 1y is the amount by which A would increase when x increases by unity, provided that A increased uniformly for x>ON. The method of proving the theorem above may be used to show that an area A bounded by a curve, the y-axis, and two abscissas (one fixed, the other not), is such that (d) y = 1 to x = 1. Find the area under each of the following curves from x = (a) y = 3x2, (b) y = 2/x2, (c) y = x3, 2. Find the area bounded by the curve y x2, the x-axis and the line x = 2, and the area bounded by the curve, the y-axis and the line y 4. Check by finding the area of the rectangle formed by the axes and the lines x = 2 and y 4. parabola y 3. Find the area above the x-axis and below the x2 + 4. x2 4. If the positive intercepts of the parabola y x2 + 2x + 3 on the coördinate axes are denoted by A and B, and if a tangent parallel to the chord AB cuts the ordinates at A and B in C and D, and touches the parabola in E, show that the area of the parabolic segment ABE is twothirds that of the parallelogram ABCD. = 5. If the tangent to the parabola y 3x2 6x+8, which is parallel to the chord joining the minimum point A to the point B on the curve whose abscissa is 2, cuts the ordinates of A and B in C and D, find the relation between the area of the parabolic segment from A to B and the area of the parallelogram ABCD. 6. Find the area bounded by the following pairs of curves: 7. Find the area bounded by the curve y = x3 6x2 + 9x + 4, the tangent to the curve at the point of inflection, and the ordinates at the maximum and minimum points. 8. Show that the area under the parabola y x = h and x = k, (k>h), is A ax2 + bx + c between (k − h) (Y1 + 4Y2 + Y3), where y1, y2, Ys are the values of y at x = h, x = h+k X = k respectively. Suggestion: Sub stitute the values of y in the above expression and show that the result is the same as the area under the curve. 9. Find the area under the following parabolas and check the results by means of the formula for the area given in Exercise 8. 10. Show that the area formula in Exercise 8 is true for any cubic function y ax3 + bx2 + cx + d. 11. Find the area under the curve y = x3 + 12x + 4 from x 1 to x = 3. Check the result by means of the formula in Exercise 8. DEFINITION. The average ordinate of a curve y = f(x) from x = a to x = b is the height of the rectangle with base b - a, whose area is equal to the area under the curve from x = to x = b. a If y represents the average ordinate and A the area under the curve from x = a a to x = b, then 12. Find the average ordinate of the following curves for the ranges 13. Given 0 t3 9t2 + 15t + 30, where is the temperature at any time t, find the average temperature for the range from maximum to mini mum temperature. ་་་ 14. The pressure p and the volume v of a gas are connected by the equation pv1.2 = k. and p 15 pounds per square inch when v = 0.5 of a cubic foot. Find the average pressure of the gas in expanding from 1 to 3 cubic feet. 15. The tension T of a spring is connected with the amount of stretching s by the equation T = 1⁄2s +3. Find the average tension as s changes from 0 to 3. 16. Find the percentage of error in the area under the curve y = x2 - 2x + 3 from x = 0 to x = 2, due to an error of 1 per cent in the 0 to x = range. 107. Motion in a Straight Line. The fundamental notions involved in the motion of a particle in a straight line are: The distance, s, measured from a convenient station on the line. The time, t, which has elapsed from a fixed time. The velocity, v, given by the equation v Dts. The acceleration, a, given by the equation a = Div. The motion may be described by an equation involving two or more of the magnitudes t, s, v, a. We have already considered motions described by an equation of the form s = f(t), and obtained the velocity and acceleration by differentiation. We shall now consider two other types in which integration is involved. If the velocity is given as a function of the time by an equation of the form v = f(t), the acceleration, a, is found by differentiating while the space, s, is found by integrating this equation. The constant involved in the integration is determined as in the following example. EXAMPLE 1. A body moves from rest in a straight line so that its velocity after t seconds is given by the law v = 4t. Assuming that a ball placed on a plane inclined at arc sin would roll according to this law, determine: (a) How far it will roll in 2 seconds. (b) Its distance from the upper end of the plane after 1.5 seconds, if it is placed 4 feet from that end. (c) Its distance from the lower end after 1 second, if it is placed 12 feet from the lower end. |