Schaum's Outline of Theory and Problems of Beginning Calculus

226 ANTIDERIVATIVES [CHAP. 2929.5 A rocket is shot straight up into the air with an initial velocity of 256 feet per second. (a) Whendoes it reach its maximum height? (6) What is its maximum height? (c) When does it hit theground again? (d) What is its speed when it hits the ground?In free-fall problems, U = I a dt and s = 5 U dt because, by definition, a = du/dt and U = ds/dt. Sincea = - 32 feet per second per second (when up is positive),U = 1-32 dt = -32t + C,t2s = ji-32t + C,) dt = (-32) + C,t + C, = - 16t2 + C,t + C,in which the values of C1 and C, are determined by the initial conditions of the problem. In the presentcase, it is given that 40) = 256 and 40) = 0. Hence, 256 = 0 + C, and 0 = 0 + 0 + C, , so that(a) For maximum height, ds/dt = o = -32t + 256 = 0. So,when the maximum height is reached.(b) Substituting t = 8 in (2),U = -32t+256 (1)s = - 16t2 + 256t (2)256t=--32- 8 seconds(c) Setting s = 0 in (2),- 16t2 + 256t = 0-16t(t - 16) = 0t=O or t=16The rocket leaves the ground at t = 0 and returns at t = 16.(d) Substituting t = 16 in (Z), 416) = -32(16) + 256 = -256 feet per second. The speed is the magnitudeof the velocity, 256 feet per second.29.6 Find an equation of the curve passing through the point (2, 3) and having slope 3x3 - 2x + 5 ateach point (x, y).The slope is given by the derivative. So,Hence,Since (2,3) is on the curve,Thus, C = - 15, anddY- dx = 3x3 - 2x + 53Y = l (32- 2x + 5) dx = - x4 - x2 + 5x + C433 = - (2)4- (2), + 5(2) + C = 12 - 4 + 10 + C = 18 + C4