a metal part is taken out of a refrigerator at a temperature of zero degrees. the temperature of the part as a function of time satisfies the differential equation where is measured in degrees and is measured in minutes. at what time does the temperature of the part equal 35 degrees?

Answer

t = (1/3)ln(35) minutes.

Q: What is the differential equation provided? A: The differential equation is dT/dt = k(T-0), where k is a constant.
Q: What is the solution to this differential equation? A: The solution is T = Ce^(kt), where C is a constant determined by the initial temperature of the part.
Q: What is the value of C? A: C = 0, since the initial temperature of the part is 0 degrees.
Q: What is the value of k? A: k can be found using the fact that the temperature doubles every 3 minutes when the part is first taken out of the refrigerator. Therefore, we have 2 = e^(3k), so k = ln(2)/3.
Q: At what time does the temperature of the part equal 35 degrees? A: We plug in T = 35, C = 0, and k = ln(2)/3 into the solution T = Ce^(kt) and solve for t. We get t = (1/3)ln(35).