NAME____________________________________ SCHOOL__________________________________ TEST 4 Friday, December 12, 2008 _____________________________________________ / ___________________________________________ \ / / \ \ | |FAX your answers to: | | | | | | | |Irvin Roy Hentzel at Fax 515-294-5454 | | | | | | \ \___________________________________________/ / \_____________________________________________/ _____________________________________________ / ___________________________________________ \ / / \ \ | | Snail Mail your answers to: | | | | | | | | Irvin Roy Hentzel | | | | Department of Mathematics | | | | 432 Carver Hall | | | | Iowa State University | | | | Ames, Iowa 50011-2064 | | \ \___________________________________________/ / \_____________________________________________/ 1. (a) Give the definition of y = ln(x). ln(x) is the area under the curve y = 1/x from 1 to x. x (b) Give the definition of y = e x e is the inverse of the ln function. x (c) If (a,b) is on the curve, y = e , what is the ln(b)? ln(b) = a 2. Express the following as rational numbers. 2 ln(5) (a) e ans = 25 5/2 (b) ln( Sqrt[ e ] ) ans = 5/4 (c) Log (27/8) 2/3 ans = -3 7 (d) Log Sqrt[a ] 3 a ans = 7/6 3. Find the following derivatives. (a) y = ln(x) / ans: y = 1/x x (b) y = e / x ans: y = e (c) y = Sinh[x] / ans: y = Cosh[x] (d) y = Cosh[x] / ans: y = Sinh[x] (e) y = Tanh[x] 2 2 / Cosh [x] - Sinh [x] 1 ans: y = --------------------- = --------- 2 2 Cosh [x] Cosh [x] (f) y = ArcSin[x] / 1 ans: y = ---------------- _________ / 2 \/ 1 - x (g) y = ArcCos[x] / -1 ans: y = ---------------- _________ / 2 \/ 1 - x (h) y = ArcTan[x] / dx ans: y = ------------ 2 1 + x (i) y = ArcSec[x] / dx ans: y = ---------------------- _________ / 2 |x| \/ x - 1 / 4. Derive the solution of y = -a y + b (a >= 0) dy --------- = dt -a y + b -a dy --------- = -a dt -a y + b ln(-a y + b) = -a t + c -a t + c -a y + b = e -a t + c -a y = e - b -a t + c -a t e - b C e - b y = ------------------ or y = -------------- -a -a 5. A population started with 75 individuals and every 125 years the population triples. Write the equation: t/125 P(t) = 75 3 6. Room temperature is 71 degrees fahrenheit. A hot brick starts at 800 degrees fahrenheit and cools to 135 degrees fahrenheit in one hour. (a) How hot will it be after a total of 30 minutes? (b) When will it be 167 degrees fahrenheit? k t T = Ts + (To-Ts) e k t T = 71 + (800-71) e k t T = 71 + 729 e 60 k T(60) = 135 = 71 + 729 e 60 k 64 = 729 e 60 k 64/729 = e ln(64/729) = 60 k ln(64/729) k = ----------- 60 ln(64/729) ---------- t 60 T = 71 + 729 e t/60 T = 71 + 729 (64/729) 1/2 T(30) = 71 + 729 (64/729) (a) T(30) = 71 + 729 (8/27) = 287 degrees Fahrenheit <========== t/60 (b) 167 = 71 + 729 (64/729) t/60 96 = 729 (64/729) t/60 96/729 = (64/729) ln(96/729) = t/60 ln(64/729) ln(96/729) t = 60 ------------ ln(64/729) 60 Log[96/729]/Log[64/729] = 50 minutes <================ 7. A brick starts at 300 degrees at t = 0. At t = 10 it is 264 degrees. At t = 20 it is 234 degrees. What is the ambient temperature? k t T - Ts = (300-Ts) e 10 k 264 - Ts = (300-Ts) e 20 k 234 - Ts = (300-Ts) e 264 - Ts 10 k ------------ = e 300 - Ts 234 - Ts 20 k ------------ = e 300 - Ts _ _ 2 | | | 264 - Ts | 234 - Ts |------------ | = ------------ | 300 - Ts | 300 - Ts |_ _| (264 - Ts)(264 - Ts) = (234-Ts)(300-Ts) 2 2 264*264 - 528 Ts + Ts = 234*300 - 534 Ts + Ts 6 Ts = 234*300-264*264 = 504 Ts = 504/6 = 84 degrees <===== Answer / \/ (264-84)/(300-84) = e^(10 k) 5/6 = e^(10 k) ln(5/6) = 10 k k = (1/10) ln(5/6) (1/10)ln(5/6) t T(t) = 84 + (300-84) e t/10 T(t) = 84 + 216 (5/6) T[t_] := 84 + 216 (5/6)^(t/10) 8. Find the derivative of 5 6 (x-5) (x-6) x y = --------------- e 2 4 (x -29) 2 ln(y) = 5 ln(x-5) + 6 ln(x-6) - 4 ln(x - 29) + x / y 5 6 -8 x ------- = --------- + ------ + ----------- + 1 y x-5 x-6 2 x - 29 - _ _ _ _ | | | | | 5 6 | | | / | (x-5) (x-6) x| | 5 6 -8 x | y = | ------------- e | | ------- + ------ + ----------- + 1 | | 2 4 | | x-5 x-6 2 | |_ (x -29) _| |_ x - 29 _| 9. Find the derivative. Do not simplify. (x + 1/x ) (a) e _ _ 1 | | / (x + ---) | 1 | y = e x | 1 - ------ | | 2 | |_ x _| 2 (b) ln( 1/x + 3 x ) -1 ------ 2 + 6 x / x y = ---------------------- 2 1/x + 3 x _ _ | 3 | (c) ArcTan| ------- | |_ x+1 _| -3 ------ 2 (x+1) ------------------------------- _ _ 2 | 3 | 1 + | ----- | |_ x+1 _| (d Sqrt[ Cosh[x] ] -1/2 1/2 (Cosh[x]) Sinh[x] 10. Evaluate these integrals. 2x (a) INT -------- dx 4 1 + x 2 ArcTan[x ] + C Cosh[x] (b) INT ----------- dx Sinh[x] ln(Sinh[x]) + C 1/7 (c) INT --------------- dx _ _ | x 2 | Sqrt| 1-(---) | |_ 7 _| ArcSin(x/7) + C Tan[x] e (d) INT -------------- dx 2 Cos [x] Tan[x] e + C 11. Give the formula for interest at 2 % for n years for these compounding frequencies. (a) Yearly (b) Monthly (c) Daily (d) Instantaneously n (a) A (1.02) 12 n (b) A (1+0.02/12) 365 n (c) A (1+0.02/365) 0.02 n (d) A e 12. A tank has 50 gallons of water and contains 50 pounds of dissolved salt. Water with 2 pounds of salt per gallon flows into the tank at 3 gal per minute and the well mixed solution drains out of the tank at 3 gallons per minute. (a) Write the differential equation for the amount of salt in the tank at time t. (b) How much salt will be in the tank after 2 hours. S(0) = 50 dS/dt = 6 - 3/50 S dS/dt + 3/50 S = 6 3/50 t 3/50 t e ( dS/dt + 3/50 S) = 6 e 3/50 t 3/50 t d/dt( e S ) = 6 e 3/50 t 3/50 t e S = 100 e + C -3/50 t S = 100 + C e 50 = 100 + C so C = -50 -3/50 t S = 100 - 50 e -360/50 S(120) = 100 - 50 e = 99.9627 lbs of salt. / \/ f[t_] := 100 - 50 E^(-3/50 t)