116_pdfsam_math 54 differential equation solutions odd

116_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 3 and, hence, t= 1 ln(15) ≈ 1.354 (years). 2 To determine the fishing rate required to keep the fish mass constant we solve the general problem dm = 2m − r, dt with r as the fishing rate. Thus we obtain m(t) = Ke2t + r . 2 m(0) = 7, The initial mass was given to be 7 million tons. Substituting this into the above equation we can find the arbitrary constant K : m(0) = 7 = K + Thus m(t) is given by m(t) = 7 − r 2t r e+. 2 2 r 2 ⇒ K =7− r . 2 A fishing rate of r = 14 million tons/year will give a constant mass of fish by canceling out the coefficient of the e2t term. 21. Let D = D (t), S (t), and V (t) denote the diameter, surface area, and volume of the snowball at time t, respectively. From geometry, we know that V = πD 3 /6 and S = πD 2 . Since we are given that V (t) is proportional to S (t), the equation describing the melting process is dπ3 dV = kS ⇒ D = k πD2 dt dt 6 dD π 2 dD D = kπD 2 = 2k = const. ⇒ ⇒ 2 dt dt Solving, we get D = 2kt + C . Initially, D (0) = 4, and we also know that D (30) = 3. These data allow us to find k and C . 4 = D (0) = 2k · 0 + C ⇒ C = 4; ⇒ 2k = − 1 . 30 3 = D (30) = 2k · 30 + C = 2k · 30 + 4 112 ...
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