Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.166.ICN Textbook: Calculus Early Vectors Preliminary Edition By Stewart. Mon, Jan 13 5.7 Anti- p353: 1,4,7,12,17,22,(28),33,34,41,43, derivatives 44,46,47,48,49, 50,55,58,59,(66), 69,(74),(79) We spent the first semester learning how to differentiate a function. f(x) = x^3 + 3x + 5 f'(x) = 3x^2 + 3 f(x) = Sin[x] f'[x] = Cos[x] f(x) = Cos[x] f'[x] = -Sin[x] f(x) = Tan[x] f'[x] = Sec[x]^x f(x) = Sec[x] f'[x] = Sec[x] Tan[x] f(x) = e^x f'[x] = e^x f(x) = ln(x) f'(x) = 1/x Now we work backwards. Given a function y = f(x) we find a function y = F(x) such that F'(x) = f(x). This is called anti-differentiation or integration. Integrate: f(x) = Sin[x] F[x] = - Cos[x] n+1 x f(x) = x^n n >= 0. F[x] = ----- n+1 -2 x f(x) = x^(-3) F[x] = ----- -2 f(x) = 1/x F[x] = ln(x). My rule is that I learn the derivatives. Then to get the integral, I guess what it should be and then differentiate it to check. In the trigonometric functions, the c-functions cosine, cosecant, cotangent c-hange sign when differentiated. So to get the integral of Sin[x], I know it is +/- Cosine. So I write down Cosine and see that its derivative is -Sine. So I change my answer to -Cosine. n+1 x When integrating x^n, always write it as --------. n+1 Otherwise, you will make errors. Thus integrating 1/2x^3 + 2/3x^2 + 3x x^4 x^3 x^2 write down 1/2 ----- + 2/3 ----- + 3 ---- 4 3 2 Then simplify it to 1/8 x^4 + 2/9 x^3 + 3/2 x^2. Page 353: Integrate the following: x^3 x^2 12 x^2 + 6x -5 12 ----- + 6 ----- - 5x 3 2 x^10 x^8 x^3 6x^9 - 4x^7 + 3x^2 + 1 6------ - 4 ---- + 3 ---- + x 10 8 3 t^(3/2) Sin[x] - 2 Sqrt[t]. -Cos[x] -2 -------- 3/2 x^(5/2) x^(2/3) x^(2/3) + 2x^(-1/3) ----------- + 2 -------- 5/2 2/3 (t, -t^2, e^t) ((t^2)/2 -(t^3)/3, e^t) (Sin[t]+t, Cos[t]-t^2) (-Cos[t]+(t^2)/2, Sin[t] -(t^3)/3 ) Now we introduce the "+C". When you are integrating always finish the sentence with the phrase "+C". Theory asks us to find all of the functions y = F(x) whose derivative is y = f(x). Once you have found one of them, you have found all of them. If you assume that you know the derivative of a curve at any point, then you know the general shape of the curve, but you do not know exactly where the curve is. Notice that the set of all lines with slope 1/2 are parallel. The same is true for "parallel curves". At every point they have the same slope. So the process is to find all curves with a particular derivative. Then from among the curves, select the one that you want. Example 3 page 347. Find f(x) given than f'(x) = x^(3/2) + 5/(1+x^2) and f(0) = 2. x^(5/2) f(x) = --------- + 5 ArcTan[x] + C 5/2 f(0) = 0 + 0 + C = 2. So C = 2. The solution is x^(5/2) f(x) = --------- + 5 ArcTan[x] + 2 5/2 Example 7: A particle moves in a straight line and has acceleration given by a(t) = 6t+4. Its initial velocity v(0) = -6 cm/sec and its initial displacement s(0) = 9 cm. Find its position function s(t). s"(t) = 6t+4 s'(t) = 3t^2 + 4 t + C s'(0) = C = -6 s'(t) = 3t^2 + 4 t -6 s(t) = t^3 + 2t^2 -6t + C s(0) = C = 9 s(t) = t^3 + 2t^2 -6t + 9 Check: s(t) = t^3 + 2t^2 -6t + 9 s(0) = 9 v(t) = 3t^2 + 4t -6 v(0) = -6. a(t) = 6t + 4 page 355 problem 70. A car is traveling at 50 mi/hr when the brakes are fully applied, producing a constant deceleration of 40 ft/sec^2. What is the distance covered before the car comes to a stop. a(t) = -40 v(t) = -40 t + vo v(0) = 50/60 88 = vo v(t) = -40 t + 220/3 The time till the velocity is zero is 11/6 seconds. 2 s(t) = -40 t /2 + 220/3 t + so Starting with so = 0 at time t = 0 the formula is s(t) = -20 t^2 + 220/3 t The distance gone is s(11/6) = -20(11/6)^2 + 220/3(11/6) = s(11/6) = 605/9 = 67.222 ft Check: 2 s(t) = -20 t + 220/3 t v(t) = -40 t + 220/3 v(0) = 220/3 ft/sec or 220/3 60/88 = 50 mph a(t) = -40 ft/sec^2