# Suppose that represents the temperature of a cup of coffee set out in a room, where T is expressed in degrees Fahrenheit and t in minutes.

A physical principle known as Newton’s Law of Cooling tells us that
dT/dt = -1/15T+5
15T + 5.
a) Supposes that T(0) = 105. What does the differential equation give us for the
value of dT
dt |T=0? Explain in a complete sentence the meaning of these two
facts.
(b) Is T increasing or decreasing at t = 0?
(c) What is the approximate temperature at t = 1?
(d) On a graph, make a plot of dT/dt as a function of T.
(e)For which values of T does T increase?
(f) What do you think is the temperature of the room? Explain your thinking.
(g) Verify that T(t) = 75 + 30e^(-t/15) is the solution to the differential equation with initial value T(0) = 105. What happens to this solution after a long time?

## (a) To find the value of dT/dt when T = 0, we substitute T = 0 into the differential equation:

dT/dt = -1/15(0)+5 = 5.

This means that at T = 0, the rate of change of temperature with respect to time is 5 degrees Fahrenheit per minute.

(b) At t = 0, we need to evaluate the sign of dT/dt to determine whether T is increasing or decreasing. Since dT/dt is positive (5) at T = 0, this means that the temperature is increasing at t = 0.

(c) To find the approximate temperature at t = 1, we need to solve the differential equation numerically or use an approximation method. One way to do this is by using Euler's method:
- Use the equation dT/dt = -1/15T + 5 to find the rate of change at t = 0 and calculate T(1) approximately using the formula: T(1) = T(0) + (dT/dt)(∆t), where ∆t is the time step (in this case, 1 minute).
By substituting the values, we have:
T(1) = 105 + (5)(1) = 110.
So, the approximate temperature at t = 1 is 110 degrees Fahrenheit.

(d) To plot dT/dt as a function of T, we can choose different values for T and compute the corresponding values of dT/dt using the differential equation. We can create a table of values or use software like Excel or a graphing calculator. The resulting plot will show the relationship between T and dT/dt.

(e) To determine when T increases, we need to examine the sign of dT/dt. Since dT/dt = -1/15T + 5, we can see that T increases when dT/dt > 0. Simplifying the inequality, we have -1/15T + 5 > 0, which gives T < 75. Therefore, T increases when the temperature is less than 75 degrees Fahrenheit.

(f) The temperature of the room can be estimated by observing the behavior of T over time. As t approaches infinity, the term e^(-t/15) approaches zero, making T(t) approach the constant term in the equation, which is 75. This suggests that the temperature of the room is approximately 75 degrees Fahrenheit.

(g) To verify that T(t) = 75 + 30e^(-t/15) is the solution to the differential equation with initial value T(0) = 105, we substitute T(t) into the differential equation:

dT/dt = -1/15(75 + 30e^(-t/15)) + 5
= -5 + (2/5)e^(-t/15).

Taking the derivative of T(t) with respect to t, we have:

d/dt (75 + 30e^(-t/15)) = -30(1/15)e^(-t/15)
= -2e^(-t/15).

This derivative matches the expression -5 + (2/5)e^(-t/15), confirming that T(t) = 75 + 30e^(-t/15) is a solution to the differential equation.

As time (t) goes to infinity, e^(-t/15) goes to zero, making T approach 75. Thus, after a long time, the temperature of the cup of coffee will approach the temperature of the room, which is approximately 75 degrees Fahrenheit.