According to Newton's law of cooling, the rate at which an object cools is directly proportional to the difference in temperature between the object and the surrounding medium. The face of a household iron cools from 130° to 100° in 30 minutes in a room that remains at a constant temperature of 70°. From calculus, the temperature f(t) of the face after t hours of cooling is given by
f(x) = 60(2)^−2t + 70.
(a) Assuming t = 0 corresponds to 1:00 P.M., approximate to the nearest tenth of a degree the temperature of the face at 2:00 P.M., 3:30 P.M., and 4:00 P.M. (Round your answers to one decimal place.)
2:00 P.M. °
3:30 P.M. °
4:00 P.M. °
so just do it for t = 1 , 2.5 and 3
To approximate the temperature of the face at a given time using the equation f(t) = 60(2)^(-2t) + 70, where t represents the hours of cooling, we need to convert the given times to hours since 1:00 P.M.
- 2:00 P.M. corresponds to t = 2 - 1 = 1 hour.
Substituting t = 1 into the equation, we have:
f(1) = 60(2)^(-2*1) + 70
f(1) = 60(2)^(-2) + 70
- 3:30 P.M. corresponds to t = 3.5 - 1 = 2.5 hours.
Substituting t = 2.5 into the equation, we have:
f(2.5) = 60(2)^(-2*2.5) + 70
f(2.5) = 60(2)^(-5) + 70
- 4:00 P.M. corresponds to t = 4 - 1 = 3 hours.
Substituting t = 3 into the equation, we have:
f(3) = 60(2)^(-2*3) + 70
f(3) = 60(2)^(-6) + 70
Now, we can calculate these values using a calculator or by performing the necessary calculations step by step in the equation.
For 2:00 P.M.:
f(1) = 60(2)^(-2) + 70 ≈ 60(0.25) + 70 ≈ 15 + 70 ≈ 85
For 3:30 P.M.:
f(2.5) = 60(2)^(-5) + 70 ≈ 60(0.03125) + 70 ≈ 1.875 + 70 ≈ 71.9
For 4:00 P.M.:
f(3) = 60(2)^(-6) + 70 ≈ 60(0.015625) + 70 ≈ 0.9375 + 70 ≈ 70.9
Therefore, the approximation of the temperature of the face at 2:00 P.M., 3:30 P.M., and 4:00 P.M. are as follows:
2:00 P.M. ≈ 85°C
3:30 P.M. ≈ 71.9°C
4:00 P.M. ≈ 70.9°C