121_pdfsam_math 54 differential equation solutions odd

# 121_pdfsam_math 54 - Exercises 3.3 where M(t = 32 is the temperature of ice This equation is linear and is rewritten in the standard form as dT

This preview shows page 1. Sign up to view the full content.

Exercises 3.3 where M ( t ) = 32 is the temperature of ice. This equation is linear and is rewritten in the standard form as dT dt + KT ( t )=32 K. We Fnd that the integrating factor is e Kt . Multiplying both sides by e Kt and integrating gives e Kt dT dt + e Kt KT ( t )=32 Ke Kt e Kt T ( t )= Z 32 Ke Kt dt e Kt T ( t )=32 e Kt + C T ( t )=32+ Ce Kt . By setting t = 0 and using the initial temperature 70 ±, we Fnd the constant C . 70 = 32 + C C =38 . Knowing that it takes 15 minutes for the wine to chill to 60 ±, we can Fnd the constant, K : 60 = 32 + 38 e K (15) . Solving for K yields K = 1 15 ln ± 60 32 38 ² 0 . 02035 . Therefore, T ( t )=32+38 e 0 . 02035 t . We can now determine how long it will take for the wine to reach 56
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

Ask a homework question - tutors are online