Chapter 1.6a Modeling an oil spill

# Chapter 1.6a Modeling an oil spill - How precalculus can...

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How precalculus can save the dolphins At high noon, an oil tanker explosion releases an oil slick on the surface of the ocean. A dolphin located d meters (m) from the tanker immediately swims away from the tanker in a straight line at a speed of v meters per second (m/sec). The spilled oil volume grows at constant rate K cubic meters (m 3 ) per second and forms an expanding cylinder whose height is one meter at all times. For what minimum value of K will the oil overtake the dolphin? When and where will this happen? Solution: At time t , Let R ( t ) be how far the oil has spread (this is the radius of the oil slick), and let D ( t ) be the dolphin’s distance from the tanker. If R ( t ) < D ( t ) , the oil slick has not reached the dolphin. If R ( t ) = D ( t ) , the oil just reaches the dolphin. If R ( t ) > D ( t ) , the oil reaches beyond the dolphin. This is bad for the dolphin. Stanley Ocken Math 19500 Precalculus: Using functions to model real life

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How precalculus can save the dolphins To find R ( t ) , compute the oil slick’s volume V at time t in two ways: Oil slick volume is V = Kt , the oil spilled per second times the number of seconds. Cylinder volume is V = πr 2 h = πR ( t ) 2 since its height is 1 meter. Solving Kt = πR ( t ) 2 for R ( t ) yields R ( t ) = q Kt π . Next find D ( t ) , the dolphin’s distance from the tanker at time t . That distance starts out as d at time t = 0 and increases at the rate of v m/sec. Therefore D ( t ) is given by the simple linear function D ( t ) = vt + d . The oil first touches the dolphin at time t when R ( t ) = D ( t ) . At that time q Kt π = vt + d , and so Kt π = ( vt + d ) 2 . We can rewrite this equation (try it!) as v 2 t 2 + 2 vd - K π t + d 2 = 0 . Stanley Ocken Math 19500 Precalculus: Using functions to model real life
How precalculus can save the dolphins The quadratic equation v 2 t 2 + (2 vd - K π ) t + d 2 = 0 has general form at 2 + bt + c = 0 . Here a = v 2 , b = (2 vd - K π ) , and c = d 2 . The quadratic formula yields solution(s) t = - b ± D 2 a where the discriminant is D = b 2 - 4 ac . If D = 0 , there is one solution t = - b 2 a . There are no solutions if D < 0 and two solutions if D > 0 . Calculate D = b 2 - 4 ac = (2 vd - K π ) 2 - 4 v 2 d 2 = 4 v 2 d 2 - 4 vd K π + K 2 π 2 - 4 v 2 d 2 = K 2 π 2 - 4 vd K π = K 2 - 4 πvdK π 2 = K ( K - 4 πvd ) π 2 Since K/π 2 > 0 , both D and K - 4 πvd have the same sign. There are three cases: If K < 4 πvd then D < 0 and the equation has no solutions. If K > 4 πvd then D > 0 and the equation has two solutions. If K = 4 πvd then D = 0 and the equation has one solution. In this case the oil touches the dolphin at time t = - b 2 a = K π - 2 vd 2 v 2 = 4 vd - 2 vd 2 v 2 = d v seconds, when the dolphin’s distance from the tanker is vt + d meters. Stanley Ocken Math 19500 Precalculus: Using functions to model real life

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A precalculus version of the oil slick problem We rephrase the previous slide from the dolphin’s point of view as follows: Summary: If K < 4 πvd , the oil slick radius R ( t ) is less than the dolphin’s distance D ( t ) from the tanker at all times t , and the dolphin escapes.
• Spring '14
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