Problems Warm-up - April 9, 2007 Solutions
1. Suppose you flip a coin 2 times. What are the possible outcomes? Solution: HH, HT, TH, TT 2. Suppose you roll a six-side die 600 times. How many times do you expect to roll a 6? Solution: You expect
1 6
of the outcomes to be a 6, so you expect to roll a six 100 times.
3. Suppose you roll a six-side die 10 times. How many times do you expect to roll a 6? Solution: You expect
1 6
of the outcomes to be a 6, so you expect to roll a six 1.67 times.
4. Suppose you roll two six side dice and add the numbers together? What are the possible outcomes? Solution: Outcomes: 2,3,4,5,6,7,8,9,10,11,12
Lecture Problems
5. Solve the initial value problem t2 y + ty = 2, y(1) = 1
Solution: Put into standard form, integrating factor is t. y(t) = 2 ln t + C t y(t) = 2 ln t + 1 t
6. You take out a car loan for $25,000 with interest rate 5% per year. You make payments at a rate of $4800 per year. Set up a differential equation modeling this situation. How long does it take to pay off the loan? Solution: y =0.05y - 4800, y(0) = 25, 000 (the differential eq) 0.05t y =96, 000 - 71, 000e (the solution) Solve y = 0 to get t = 6.03 years to pay off the loan. 7. We start with a tank containing 50 gallons of salt water with the salt concentration being 2 pounds per gallon. Salt water with a salt concentration of 3 pounds per gallon is then poured into the top of the tank at the rate of 5 gal/min. Salt water is at the same time drained from the bottom of the tank at the rate of 4 gal/min. How much salt will be in the tank after an hour? Let y be the amount of salt in the tank. Set up the differential equation, solve it, etc. Solution: y =(rate in) - (rate out) y(0) = 100 y y =(3)(5) - (4) 50 + t 4y y =15 - 50 + t C y =150 + 3t - (general solution) (50 + t)4 312, 500, 000 y =150 + 3t - (particular solution) (50 + t)4 y(60) 327.9
8. Let f (x) = ex . Using the 5th degree Taylor polynomial make the approximations. Estimate the error in each case. Solution: x2 x3 x4 x5 + + + p5 (x) = 1 + x + 2! 3! 4! 5! (a) e-1 Solution: 11 0.3667 30 M M 16 = |R5 (-1)| 6! 6! p5 (-1) = M is the maximum of f (6) (x) = ex on the interval [-1, 0], which is e0 = 1, so M = 1. This gives us |R5 (-1)| (b) e-1/2 Solution: 2329 0.60651 3840 M (1/2)6 M |R5 = (-1/2)| 6! 64 6! p5 (-1/2) = M is the maximum of f (6) (x) = ex on the interval [-1/2, 0], which is e0 = 1, so M = 1. This gives us |R5 (-1)| 1 1 = 0.0000217 64 6! 46080 M 1 1 = = 0.00139 6! 6! 720
(c) e1/2 (find something useful for M , e1/2 isn't good) Solution: 6331 1.6487 3840 M (1/2)6 M |R5 (1/2)| = 6! 64 6! p5 (1/2) = M is the maximum of f (6) (x) = ex on the interval [0, 1/2], which is e1/2 . This is not a useful choice for M . Instead note that e < 4 so e1/2 < 41/2 = 2, so we choose M = 2. This gives us |R5 (-1)| 2 1 = 0.0000434 64 6! 23040
9. Find a Taylor series for Solution: Note that
1 . (1 - x)2 1 1-x = 1 : (1 - x)2
1 =1 + x + x2 + x3 + x4 + = 1-x
xn
n=0
1 =1 + 2x + 3x2 + 4x3 + 5x4 + = (1 - x)2 10. Find a Taylor series for Solution: Note that 1 . (1 - x)3 1 1-x = 2 : (1 - x)3
(n + 1)xn
n=0
1 =1 + x + x2 + x3 + x4 + = 1-x
xn
n=0
1 =1 + 2x + 3x2 + 4x3 + 5x4 + = (1 - x)2
(n + 1)xn
n=0
2 =2 + 2 3x + 3 4x2 + 4 5x3 + = 3 (1 - x) 1 =1 + 3x + 6x2 + 10x3 + = (1 - x)3
(n + 1)(n + 2)xn
n=0
n=0
1 (n + 1)(n + 2)xn 2
11. Using the second degree taylor polynomials in the previous problems, find an approximation for the integrals: (a)
1/2 0
7 1 dx 2 (1 - x) 8
(b)
1/2 0
1 9 dx 3 (1 - x) 8
12. Find a Taylor series for
1 . (1 + x)3 Solution: Use a previous problem: 1 =1 + 3x + 6x2 + ...