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.166.05 Textbook: Calculus by Varberg, Purcell, Rigdon, eight edition. February 16 9.1 Type 0/0 p407:1-3,(4), 9-11,(12),19,20,23,(26),(30) Main Idea: Take derivatives of the numerator and denominator separately. Key Words: L'Hopital's Rule Goal: Learn easy way to take limits. -------------------------------------------------------- Previous assignment: February 9 8.5 Int by Rat fcn p397:1,(4),9,13,(16),17,18,(28),29,30,33,(42) Page 397 Problem 4 5x INT --------- dx 3 2 2x + 6x x 5/2 INT --------- dx 3 2 x + 3x x 5/2 INT --------- dx 2 x (x + 3) _ _ 1 | A B | 5/2 INT --------- dx = 5/2 INT | -------- + --------- | |_ x x+3 _| x (x + 3) 1 = A(x+3) + Bx 0 = A+B 1 = 3A A = 1/3 B = -1/3 1 | 1/3 -1/3 | 5/2 INT --------- dx = 5/2 INT | ----- + --------- | | x x+3 | x (x + 3) / \/ 1/3 x + 1 - 1/3 x = 1 1 5/2 INT --------- dx = 5/2 ( 1/3 ln(x) -1/3 ln(x+3) ) + C x (x + 3) x = 5/6 ln( ------ ) + C x+3 --------------------- D[5/6 Log[x/(x+3)],x] ----------------------- Page 397 Problem 16 x^3 - 6 x^2 + 11x -6 INT --------------------------- dx 4x^3 - 28x^2 + 56x - 32 1/4 _________________________ 4x^3 - 28x^2 + 56x - 32 | x^3 - 6 x^2 + 11x - 6 x^3 - 7 x^2 + 14x - 8 ---------------------- x^2 - 3x + 2 x^3 - 6 x^2 + 11x -6 x^2 - 3 x + 2 INT --------------------------- dx = INT 1/4 + ------------------------ 4x^3 - 28x^2 + 56x - 32 4x^3 - 28x^2 + 56x - 32 x^2 - 3 x + 2 = 1/4 x + 1/4 INT ---------------------- (x-1)(x-2)(x-4) g'(x) = (x-2)(x-4) + (x-1)(x-4) + (x-1)(x-2) g'(1) = 3 g'(2) = -2 g'(4) = 6 0 0 6 ------ ------- ------ 3 -2 6 = 1/4 x + 1/4 INT ------- + ---------- + --------- (x-1) (x-2) (x-4) = 1/4 x + 1/4 ln(x-4) + C <========= Answer / \/ x-3 D[1/4 x + 1/4 Log[x-4],x ] = ------------ 4(x-4) Which points out to an easier way. (x-1)(x-2) 1 x-3 = 1/4 x + 1/4 INT ---------------------- = 1/4 x + 1/4 INT ------- = ----------- (x-1)(x-2)(x-4) x-4 4(x-4) 1 = 1/4 x + 1/4 INT ------ x-4 = 1/4 x + 1/4 ln(x-4) ---------------------------------------------------- Page 397 Problem 28 3x+2 INT --------------- dx 2 x(x+2) + 16 x 3x+2 INT --------------- dx | 2 | x|(x+2) + 16 | | | 3x+2 A Bx+C INT --------------- dx = INT ---- + ---------------- | 2 | x 2 x| x +4x + 20 | x +4x + 20 | | 3x+2 = Ax^2 + 4 A x + 20 A Bx^2 + C x 20 A = 2 4 A + C = 3 A + B = 0 A = 1/10 B = -1/10 C = 3-4/10 = 26/10 3x+2 1/10 -1/10 x + 26/10 INT --------------- dx = INT ---- + ---------------- | 2 | x 2 x| x +4x + 20 | x +4x + 20 | | 3x+2 1/10 -1/10(x ) 26/10 INT --------------- dx = INT ---- + ---------------- + -------------- | 2 | x x^2+4x+20 x^2 + 4 x + 20 x| x +4x + 20 | | | 3x+2 1/10 -1/10(x+2 ) 26/10 +2/10 INT --------------- dx = INT ---- + ---------------- + -------------- | 2 | x x^2+4x+20 x^2 + 4 x + 20 x| x +4x + 20 | | | 3x+2 1/10 -1/20(2x+4 ) 28/10 INT --------------- dx = INT ---- + ---------------- + -------------- | 2 | x x^2+4x+20 (x+2)^2 + 16 x| x +4x + 20 | | | 3x+2 1/10 -1/20(2x+4 ) 28/10 INT --------------- dx = INT ---- + ---------------- + -------------- (1/16) | 2 | x 2 |x+2|^2 + 1 x| x +4x + 20 | x +4x+20 |---| | | | 4 | 3x+2 2 |x+2| INT --------------- dx = 1/10 ln(x) -1/20 ln(x + 4x + 20) +7/40 ArcTan |---| | 2 | | 4 | x| x +4x + 20 | ----------- | | 1/4 3x+2 2 |x+2| INT --------------- dx = 1/10 ln(x) -1/20 ln(x + 4x + 20) +7/10 ArcTan |---| + C | 2 | | 4 | x | x + 4x + 20| | | ------------------------------------------------------------- 1/10 Log[x] -1/20 Log[x^2 + 4x + 20 ] + 7/10 ArcTan[(x+2)/4] ------------------------------------------------------------- Page 397 Problem 42 In many population growth problems, there is an upper limit beyond which the population cannot grow. Let us suppose that the earth will not support a population of more than 16 billion and that there were 2 billion people in 1925 and 4 billion people in 1975. Then, if y is the population t years after 1925, an appropriate model is the differential equation dy/dt = ky(16-y) (a) Solve this differential equation (b) Find the population in 2015 (c) When will the population be 9 billion. dy ----------- = k dt y(16-y) dy ----------- = -k dt y(y-16) -dy dy ----- ----- 16 16 ------ + -------- = -k dt y y-16 -1/16 ln(y) + 1/16 ln(y-16) = -k dt + C y-16 ln(--------) = -16 k t + C y y - 16 -16 k t --------- = C e y 2 billion people in 1925 4 billion people in 1975. when t = 0 y = 2 2-16 ------- = C So C = -7 2 y-16 -16 k t ---- = -7 e y when t = 50 y = 4 4-16 -16 k 50 ------ = -7 e 4 -16 k 50 -3 = - 7 e -800 k 3/7 = e ln(3/7) = -800 k -ln(3/7) k = -------- 800 16 ln(3/7) ------------ y - 16 800 --------- = -7 e y ln(3/7) -------- t y - 16 50 --------- = -7 e y t/50 y - 16 | 3 | --------- = -7 | - | y | 7 | t = 0 y - 16 = -7 y y = 2 t = 50 y-16 = -3 y y = 4 (b) Find the population in 2015 2015 -1925 ======== 90 years (y-16)y +7 (3/7)^(90/50) == 0 {{y -> 0.0957695}, {y -> 15.9042}} (c) When will the population be 9 billion. Solve[(9-16)/9 +7 (3/7)^(t/50)==0] {{t -> 129.661}} ln(3/7) -------- t y - 16 50 --------- = -7 e y y - 16 = -7 e^(ln(3/7)/50 t) y = 0 16 y = ---------------------------- 1+7 e^(ln(3/7)/50 t) -------------------------------------------------------- f[t_] := 16/(1+7 E^(Log[3/7]/50 t)); a = Plot[f[t],{t,0,400}]; b = Plot[16,{t,0,400},PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotLabel->"Page 397 Problem 42"]; Display["42.ps",c]; ------------------------------------------------------- ============================================================ New Material: L'Hopital's rule: f(x) f'(x) Lim ------ = Lim ----- x->a g(x) x->a g'(x) But be careful. There are certain restrictions. (1) f(a)/g(a) has to be 0/0 or infinity/infinity (2) The LHS may exist but not the RHS. Example 1 Sin[x] Lim ------- x->0 x 1-Cos[x] Lim ----------- x->0 x ------------------------------------ 2 x - 9 Example 2 Lim ---------- x->3 2 x - x - 6 2 x + 3x - 10 Lim ------------ + 2 x->2 x - 4 x + 4 ---------------------------------------- Tan[2x] Example 3 Lim -------- x->0 ln[1+x] --------------------------------------- Sin[x]-x Example 4 Lim ---------- x->0 3 x ------------------------------------- 1 - Cos[x] Example 5 Lim --------------- x->0 2 x + 3 x --------------------------------------- -x e Example 6 Lim --------- x->Infinity -1 x ---------------------------------------------- Page 407 work problems 1-10. --------------------------------------------- f[x]-f[a] ----------- f'(a) Theory: f[x] f[x] - f[a] x-a ----- = ----------- = ---------- ---> --------- g[x] g[x] - g[a] g[x]-g[a] ----------- g'(a) x-a By graphing: | / . | /f'(a) x . ' | / . ' | / . ' g'(a) x |/. ' -----------+--------------------- | | f(b)-f(a) By algebra: s(x) = f(x)-f(a) - ------------ (g(x)-g(a)) g(b)-g(a) s(a) = s(b) = 0 f(b)-f(a) s'(x) = f'(x) - ---------- g'(x) g(b)-g(a) So when s'(c) = 0, we get f'(c) f(b)-f(a) --------- = -------------- g'(c) g(b)-g(a) Then when f(a) = g(a) = 0 f(b) f'(c) --------- = ------- g(b) g'(c) f'(c) f(b) And if ------ has a limit, then so much -------. g'(c) g(b)