b How long will it take the eggs temperature to reach 20 Solution a The

# B how long will it take the eggs temperature to reach

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(b) How long will it take the egg’s temperature to reach 20 ? Solution. (a) The differential equation for this model is T 0 ( t ) = - k ( T ( t ) - 18), and the general solution is T ( t ) = 18 + Ce - kt . We need both C and k . We know the initial temperature T (0) = 98, so T (0) = 18 + Ce 0 = 98 = C = 80 . To get k , we use the fact that T (5) = 38: 38 = 18 + 80 e - k · 5 = e - 5 k = 1 4 = ⇒ - 5 k = ln(1 / 4) = - ln(4) = k = ln(4) 5 . So putting it all together, we get T ( t ) = 18 + 80 e - ln(4) 5 t . (b) We want to know when T ( t ) = 20, so solve: 20 = 18 + 80 e - ln(4) 5 t = 2 = 80 e - ln(4) 5 t = e - ln(4) 5 t = 1 40 = - ln(4) 5 t = ln(1 / 40) = - ln(40) = t = 5 ln(40) ln(4) . So the answer is 5 ln(40) ln(4) minutes . Example 6.3. Suppose you have a cup of coffee with cooling constant k = . 09 min - 1 . It is placed in a room of temperature 20 C. (a) At what rate is the temperature changing when the temperature of the coffee is 80 C. (b) If the coffee is served at 90 C, how long will it take to reach 30 C. (c) At what time will the temperature reach 15 . Solution. (a) The differential equation is T 0 ( t ) = - k ( T ( t ) - T 0 ) = - . 09( T ( t ) - 20). So if T ( t ) = 80, then T 0 ( t ) = - . 09(80 - 20) = - . 09 · 60 C/min . 11
Math 31B Notes Sudesh Kalyanswamy (b) The general solution is T ( t ) = 20 + Ce - kt = 20 + Ce - . 09 t . First, since T (0) = 90, we can solve for C : 90 = 20 + Ce - . 09 · 0 = 90 = 20 + C = C = 70 , so T ( t ) = 20 + 70 e - . 09 t . We want to know when T ( t ) = 30, so 30 = 20 + 70 e - . 09 t = 1 7 = e - . 09 t = ⇒ - . 09 t = ln(1 / 7) = - ln(7) = t = ln(7) . 09 minutes . (c) In theory, you would set T ( t ) = 15, so 15 = 20 + 70 e - . 09 t = ⇒ - 5 70 = e - . 09 t . However, you cannot take logs because you cannot have the log of a negative number. That tells us there is no t which works. We should have seen this immediately because the temperature should never go below the temperature of the ambient space, which in this case is 20. Example 6.4. Suppose that a skydiver jumps out of an airplane. If the terminal velocity is - 98 m/s, find the velocity of the skydiver after 15 seconds. Assume k = 8 kg/s. Solution. Here, v ( t ) = - mg k + Ce - ( k/m ) t . We don’t know m yet. However, terminal velocity is - mg/k , so - 98 = - m · 9 . 8 8 = m = 98 · 8 9 . 8 = 80 . So the skydiver’s mass is 80 kg, and v ( t ) = - 80 g 8 + Ce - (8 / 80) t . We also need C , but we know v (0) = 0, so - 80 g 8 + Ce 0 = 0 = C = 10 g = 10(9 . 8) = 98 . Hence v ( t ) = - 10 g + 98 e - t/ 10 = - 98 + 98 e - t/ 10 . After 15 seconds, the velocity is v (15) = - 98 + 98 e - 15 / 10 m/s . Example 6.5. What is the minimum deposit P 0 necessary that will allow an annuity to pay out 1000 dollars/year indefinitely if it earns interest at a rate of 4% . Solution. We know P ( t ) = N r + Ce rt = 1000 . 04 + Ce . 04 t . The annuity will pay out indefinitely if C 0. To see why, notice that if C < 0, then P ( t ) is decreasing, and so will eventually hit 0, which is not what we want. However, if C 0, then at worst P ( t ) is constant, which is ok. If the initial amount is P 0 , then P 0 = 1000 . 04 + C = C = P 0 - 1000 . 04 . Since we want C 0, this means P 0 1000 . 04 dollars . Example 6.6. A skydiver jumps out of an airplane with zero initial velocity. Before the parachute opens, the skydiver’s velocity satisfies v 0 ( t ) = - g - 2 v ( t ) , and after the parachute opens, the skydiver’s velocity satisfies v 0 ( t ) = - g - v . If the parachute opens 10 seconds after the initial jump, find the velocity 20 seconds after the initial jump.