This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ct’s temperature T as a function of time t. With specific time and temperature information we can derive a particular solution to be used in solving a problem related to that given information. dT
dT
dT
= K (T ! A ) "
= K # dt " $
= K $ dt dt
T!A
T!A and " dT
= K " dt
T!A
# ln T ! A = Kt + C, with C the constant of integration
#e ln T ! A = e Kt +C , and with T $ A, we have T ! A = e Kt +C
# T = e Kt +C + A
In general, then, T = e Kt +C + A is the solution to the differential equation dT
= K (T ! A) . dt Suppose, further, we know that the temperature of the object is 95° C, that the ambient temperature is 20° C, and that exactly 20 minutes after the object began to cool its temperature was 70° C. We have A = 20° C and (0,95) and (20,70) as known conditions. Wi...
View Full
Document
 Summer '09
 Calculus, Equations

Click to edit the document details