Written Homework 10 Solutions MATH 1310 - CSM Assignment: pp 217-210; problems: 1(abcd), 3, 4(ab), 5, 7. pp 238-240; problems: 3, 4, 5, 10. pp 48; problem 27. 1. Solutions: 1 (a) y = 1 = ( 5 − t)2 0 1 1 ( 5 − t) 2 = y2. 3 1 1 = y3. = q 3/2 1 − 2t) 25 − 2t 4 1 1 = y4. (c) y 0 = 1 = q 4/3 1 3 ( 125 − 3t) − 3t (b) y 0 = 1 ( 25 125 (d) So the general case should look like: y(t) = 1 r n−1 1 5n−1 3. Solutions: (a) x0 = − sin(t) = −(sin t) = −y. y 0 = cos t = x. (b) x0 = − sin(t + π/2) = −(sin(t + π/2)) = −y. y 0 = cos(t + π/2) = x. 4. Solutions: (a) y 0 = ln(10) · 5 · 10t ≈ 2.3 · 5 · 10t = 2.3y. (b) y 0 = ln(10) · C · 10t ≈ 2.3 · C · 10t = 2.3y. 5. Solutions: (a) C = 286. (b) C = 1/5. (c) C = −7/3. 1 − (n − 1)t . 7. Solutions: 1 ln 2 t/0.69 (a) dy/dt = ln 2 · 2 · = ·2 ≈ 1 · 2t/0.69 = y. 0.69 0.69 k ln 2 t/0.69 k kt/0.69 = (b) dy/dt = ln 2 · 2 · ·2 ≈ k · 2t/0.69 = ky. 0.69 0.69 k ln 2 k =A 2t/0.69 ≈ A·k·2t/0.69 = k·A·2t/0.69 = kP . (c) dP/dt = A·ln 2·2kt/0.69 0.69 0.69 t/0.69 PP 238-240: 3. Solutions: (a) (b) (c) (d) (e) (f) d 3x 7e = 7e3x · 3 = 21e3x . dx d Cekx = Cekx · k = kCekx . dx d 1.5et = 1.5et . dt d 1.5e2t = 3e2t . dt d 3x 2e − 3e2x = 6e3x − 6e2x = 6e2x (ex − 1). dx d cos t e = − sin t · ecos t . dt 4. Solutions: (a) fx = y · exy . fy = x · exy . (b) fx = 6x · e2y . fy = 3x2 · e2y · 2. (c) fu = eu sin v. fv = eu cos v. (d) fu = eu sin v · sin v. fv = eu sin v · 3 sin v · 3 cos v. 5. Solutions: (a) exp(2x + 3) = e2x+3 = e2x · e3 . 2 (b) (exp(x))2 = ex · ex = e2x 6= ex . (c) exp(17x)/exp(5x) = e17x = e17x−5x = e12x . e5x 2 27. Radioactivity. In radioactive decay, radium slowly changes into lead. If one sample of radium is twice the size of a second lump, then the larger sample will produce twice as much lead as the second in any given time. In other words, the rate of decay is proportional to the amount of radium present. Measurements show that 1 gram of radium decays into lead at the rate of 1/2337 grams per year. Write an equation that links the decay rate to the size of the radium sample. How does your equation indicate that the process involves decay rather than growth? Solution: Since the rate of decay is proportional to the among of radium present and 1 gram of radium decays into lead at the rate of 1/2337 grams per year, we can write the following equation where R represents the size of the radium sample and R0 gives the decay rate: R0 = −R , 2337 R(t) = R0 · e−t/2337 (1) 10. Radioactivity. (a) Assuming that when we begin the sample of radium weighs 1 gram, write out the initial value problem that summarizes the information about the weight R of the sample. Solution: R(t) = 1 · e−t/2337 . (b) How much did the sample weigh 20 years ago? How much will it weigh 200 years hence? Solution: R(−20) = 1 · e20/2337 ≈ 1.01 and R(200) = 1 · e−200/2337 ≈ 0.92. 3

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