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MATH 100, Section 110 (CSP) Week 6: Marked Homework Solutions 2010 Oct 21 1. [3] After t seconds, the radius of the circular ripple is r = 50t cm, and the area within the circle is A(t) = πr 2 = π(50t)2 = 2500πt2 cm2 . The rate at which the area is increasing is A0 (t) = 5000πt cm2 /s, so (a) A0 (1) = 5000π cm2 /s, (b) A0 (3) = 15000π cm2 /s, (c) A0 (5) = 25000π cm2 /s. 2. [6] (a) When t = 0, the population is N = 4200 and the birth rate is B = b N = (0.025)(4200) = 105 individuals·yr−1 . (b) When t = 0, the death rate is D = d N = (0.016)(4200) = 67.2 individuals·yr−1 . (c) When t = 0, the (net) growth rate is dN = B − D = 105 − 67.2 = 37.8 individuals·yr−1 , dt or dN = r N = (b − d)N = (0.025 − 0.016)(4200) = (0.009)(4200) = 37.8 individuals·yr −1 . dt (d) By Theorem 2 (p. 234), the differential equation dN = r N, dt with initial condition N (0) = 4200 has the solution N (t) = 4200ert , where r = b − d = 0.025 − 0.016 = 0.009 yr−1 . 1 To find the time it takes for the population to quadruple, set N (t) = 4 · 4200 and solve for t: ert = 4, rt = ln 4, ln 4 ln 4 t= = ( ≈ 154 ) yr. r 0.009 (Notice that we don’t really need the specific information N (0) = 4200 to answer part (d). The initial population could be N (0) = N0 with N0 unspecified.) 3. [5] Let m(t) denote the amount of uranium-238 at time t years, and assume that the amount decays at a rate proportional to the amount present. Then the differential equation for m(t) is dm = k m, dt which by Theorem 2 (p. 234) has the solution m(t) = m(0)ekt , where m(0) is the amount of uranium-238 at time t = 0. To find k we use the half-life: when t = 4.5×109 years, m(4.5×109 ) = 21 m(0). On the other hand, the solution to the differential 9 9 equation gives m(4.5 × 109 ) = m(0)ek(4.5×10 ) . Therefore e4.5×10 k = ln(1/2) = − ln 2, k=− ln 2 4.5 × 109 yr−1 . 9 (a) When t = 109 years, m(109 ) = m(0)e10 k = m(0)e−(ln 2)/4.5 and the percentage remaining is m(109 )/m(0) × 100% = e−(ln 2)/4.5 × 100% (≈ 85.7%). (b) We solve for t when m(t) = 0.75m(0) = 43 m(0) (one quarter of the initial amount decayed means three quarters of the initial amount is present at this time). ekt = 34 gives t = (ln(3/4))/k = 4.5 × 109 (ln 4 − ln 3)/ ln 2 (≈ 1.87 × 109 ) years. 4. [10] Newton’s Law of Cooling (or warming, in this case) gives the differential equation dT = k(T − 200) dt and the initial condition is T (0) = 20, where T (t) is the temperature of the turkey in degrees Celcius, at time t in minutes. To solve this, make the change of variables y(t) = T (t) − 200, then (as shown in the lecture and textbook) y(t) must solve the differential equation dy = k y, dt 2 with initial condition y(0) = 20 − 200 = −180. Then by Theorem 2 (p. 234) the solution of the initial-value problem for y(t) is y(t) = −180ekt . We are also given that y(30) = 30 − 200 = −170, and we are asked to determine t such that y(t) = 80 − 200 = −120. To determine t, we need k, which we get by setting t = 30, −180ek(30) = −170, and solving: e30k = then k= 170 , 180 30k = ln 17 , 18 1 17 1 18 ln = − ln . 30 18 30 17 We set y(t) = −120 and solve for t: −180ekt = −120, and t= ekt = ln(2/3) 1 2 ln = 30 k 3 ln(17/18) 2 120 = , 180 3 (≈ 213) min. 5. [7] We solve this differential equation in the same way as we solve the differential equation for Newton’s Law of Cooling. Using the hint, write the differential equation as dy b =a y+ , dt a and make the change of variables b u(t) = y(t) + . a Differentiate with respect to t and use the differential equation for y: du dy b = = a y+ = au dt dt a so u(t) must solve the differential equation du = a u, dt with initial condition u(0) = y(0) + 3 b b =c+ . a a Then by Theorem 2 (p. 234) the solution of the initial value problem for u(t) is b u(t) = c + eat , a then y(t) = u(t) − (b/a), so y(t) = Then the solution to b c+ a dy = −3y + 10, dt b eat − . a y(0) = −99, is (a = −3, b = 10, c = −99) y(t) = 10 −99 − 3 4 e−3t + 10 . 3