Newton's Law of Cooling

# 0202732554 20 3 t 2 ln t t 75e 20 3 20

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Unformatted text preview: th this we can determine a specific solution to the differential equation. A = 20 ! T (t ) = e Kt +C + 20 (0, 95) ! 95 = e K (0 )+C + 20 ! e K = 75 ! K = ln 75 " 4.3175 ! T (t ) = e Kt +ln 75 + 20 = e Kt e ln 75 + 20 = 75e Kt + 20 (20, 70) ! 70 = 75e K (20 ) + 20 ! 50 = 75e K (20 ) 2 ! = e20 K 3 #2& ! ln % ( = 20 K \$ 3' 1 #2& ! K = ln % ( " )0.0202732554 20 \$ 3 ' t #2& ln% ( ! T (t ) = 75e 20 \$ 3 ' + 20 t !2\$ ln# & The particular solution T (t ) = 75e 20 " 3 % + 20 , with T in degrees centigrade and t in minutes for the given conditions, can now be used to answer a wide variety of questions about the situation. (a) Determine the object’s temperature 45 min...
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## This document was uploaded on 03/05/2014 for the course MATH 146 at Illinois State.

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