#### Transcript Applications of Exponential Change

Find the exponential function y y0 e whose graph passes through the two points. Initial value: y y 0 1.1 kt 0 y 1.1e k 3 3 k Other point: 3 1.1e ln 3 ln 1.1e 1 ln3 ln1.1 3k k ln1.1 ln 3 3 ln1.1 ln 3t 3 0.3344 t Equation: Function: y 1.1e kt 1.1e Section 6.4b Suppose that you just took a delicious hot pocket out of the microwave. The tasty treat will gradually cool to the temperature of the surrounding air… As it turns out, the rate at which the hot pocket’s temperature is changing at any given time is proportional to the difference between its temperature and the temperature of the surrounding medium!!! This leads us to derive NEWTON’S LAW OF COOLING (which, incidentally, works for warming as well) Let T be the temperature of the object in question at time t, and Ts be the surrounding temperature. Then: dT k T TS dt Since dT d T TS , we can rewrite the equation: d T TS k T TS dt By the law of exponential change, the solution is T TS T0 TS e kt T TS T0 TS e kt where T 0 is the temperature at time t = 0. A hard-boiled egg at 98 C is put in a pan under running 18 C water to cool. After 5 minutes, the egg’s temperature is found to be 38 C. How much longer will it take the egg to reach 20 ? Define Variables: Law of Cooling: Substitute: T0 98 TS 18 T TS T0 TS e kt T 18 98 18 e T 18 80e kt kt A hard-boiled egg at 98 C is put in a pan under running 18 C water to cool. After 5 minutes, the egg’s temperature is found to be 38 C. How much longer will it take the egg to reach 20 ? 5 k To find k, use the point (5, 38): 38 18 80e e 5 k 0.25 1 5k ln 0.25 ln 4 k ln 4 5 0.2ln 4t The Final Equation: T 18 80e 18 80 4 0.2 t A hard-boiled egg at 98 C is put in a pan under running 18 C water to cool. After 5 minutes, the egg’s temperature is found to be 38 C. How much longer will it take the egg to reach 20 ? T 18 80 4 0.2 t 0.2 t 20 18 80 4 ln 0.025 0.2 t t 13.305 0.025 4 0.2 ln 4 Solve analytically: After about 13.305 minutes, the egg will reach 20 degrees C. In many situations, the resistance encountered by a moving object (i.e., from friction) is proportional to the object’s velocity… That is, the slower the object moves, the less its forward progress is resisted by the air through which it passes… To model such a situation, we’ll start with Newton’s Second Law of Motion… The resisting force opposing the motion: dv Force = mass x acceleration m dt If this force is proportional to the velocity, then: dv m kv dt or dv k v dt m k 0 This is a differential equation of exponential change… The solution: v v0e k m t For a 50-kg ice skater, the k in the previous equation is about 2.5 kg/sec. How long will it take the skater to coast from 7 m/sec to 1 m/sec? How far will the skater coast before coming to a complete stop? First, the general model: v 7e 2.5 50t 7e 0.05 t For a 50-kg ice skater, the k in the previous equation is about 2.5 kg/sec. How long will it take the skater to coast from 7 m/sec to 1 m/sec? How far will the skater coast before coming to a complete stop? v 7e 0.05t Now, we want the value of t when v = 1. 0.05 t 1 7e 0.05t e 1 7 0.05t ln 1 7 ln 7 t 38.918 0.05 The skater will reach 1 m/sec ln 7 from 7 m/sec after about 38.918 sec of coasting For a 50-kg ice skater, the k in the previous equation is about 2.5 kg/sec. How long will it take the skater to coast from 7 m/sec to 1 m/sec? How far will the skater coast before coming to a complete stop? v 7e 0.05t To find distance, we need the integral of velocity: s 7e 0.05t 0.05t 7e dt C 0.05 0.05t 140e C For a 50-kg ice skater, the k in the previous equation is about 2.5 kg/sec. How long will it take the skater to coast from 7 m/sec to 1 m/sec? How far will the skater coast before coming to a complete stop? v 7e s 140e 0.05t 0.05t C 140e Assuming s = 0 when t = 0, we have 0 140e C C 140 0 0.05t 140 For a 50-kg ice skater, the k in the previous equation is about 2.5 kg/sec. How long will it take the skater to coast from 7 m/sec to 1 m/sec? How far will the skater coast before coming to a complete stop? Finally, for distance: s 140 140e Find 0.05t lim s t 140 140 1 e 0.05 t t Mathematically, s never quite reaches 140. But for all practical purposes, the skater comes to a complete stop after traveling 140 m… The distance traveled by a moving object that encounters resistance proportional to its velocity: v0 m k m t s t 1 e k The total distance traveled by this object: v0 m s k Suppose a battleship has mass around 51,000 metric tons (51,000,000 kg) and a k value of about 59,000 kg/sec. Assume the ship loses power when it is moving at a speed of 9 m/sec. v t 9e 59t 51,000 459, 000 59 t 51,000 s t 1 e 59 (a) About how long will it take the ship’s speed to drop to 1 m/sec? 1 9e 59 t 51,000 51000 ln 1 9 t 59 t 1899.296 The ship will reach 1 m/sec in about 1899.296 seconds, or in about 31.655 minutes Suppose a battleship has mass around 51,000 metric tons (51,000,000 kg) and a k value of about 59,000 kg/sec. Assume the ship loses power when it is moving at a speed of 9 m/sec. v t 9e 59t 51,000 459, 000 59 t 51,000 s t 1 e 59 (b) About how far will the ship coast before it is dead in the water? 459, 000 459, 000 59 t 51,000 lim s t lim 1 e t t 59 59 The ship will coast for a distance of about 7779.661 meters, or 7.780 kilometers