# 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) Given that T(0) = 105, the differential equation gives us the value of dT/dt at t=0 as:

dT/dt |T=0 = -1/15(105) + 5 = -7 + 5 = -2.

This means that at the initial time (t=0), the rate of change of temperature is -2 degrees Fahrenheit per minute.

b) Since dT/dt at t=0 is negative (-2), the temperature T is decreasing at t=0.

c) To find the approximate temperature at t=1, we can solve the differential equation using separation of variables. Rearranging the equation, we have:

dT/dt = -1/15T + 5

1/(T/15 - 5)dT = dt

Integrating both sides, we get:

15ln|T/15 - 5| = t + C

Since T(0) = 105, we can substitute these values into the equation and solve for C:

15ln|105/15 - 5| = 0 + C

15ln(7) = C

Substituting C back into the equation:

15ln|T/15 - 5| = t + 15ln(7)

ln|T/15 - 5| = (1/15)t + ln(7)

|T/15 - 5| = e^[(1/15)t + ln(7)]

Since the temperature cannot be negative, we take the positive value:

(T/15 - 5) = e^[(1/15)t + ln(7)]

T/15 = 5 + e^[(1/15)t + ln(7)]

T = 75 + 15e^[(1/15)t + ln(7)]

Approximate temperature at t=1 can be found by substituting t=1:

T = 75 + 15e^[(1/15)(1) + ln(7)]

T ≈ 91.373 degrees Fahrenheit.

So, the approximate temperature at t=1 is approximately 91.373 degrees Fahrenheit.

d) To plot the graph of dT/dt as a function of T, we can substitute the given equation into the differential equation:

dT/dt = -1/15T + 5

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

Simplifying, we have dT/dt = -1/3e^(-t/15).

We can then plot dT/dt on the y-axis and T on the x-axis.

e) T increases when dT/dt is positive. By observing the graph, we can see that T increases for T values greater than approximately 60 degrees Fahrenheit.

f) The temperature of the room can be estimated by observing the behavior of the equation T(t) = 75 + 30e^(-t/15). As time approaches infinity (t → ∞), the term e^(-t/15) approaches 0. Therefore, T(t) approaches 75, suggesting 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 these values into the differential equation:

d/dt (75 + 30e^(-t/15)) = -1/15(75 + 30e^(-t/15)) + 5

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

Multiplying both sides by 15:

-30e^(-t/15) = -(75 + 30e^(-t/15)) + 75

-30e^(-t/15) + 75 + 30e^(-t/15) = 75

The left-hand side simplifies to:

45

Therefore, the equation is satisfied.

After a long time, as t approaches infinity, e^(-t/15) approaches 0. As a result, T(t) approaches 75, which confirms the earlier observation that the temperature of the room is approximately 75 degrees Fahrenheit.

## (a) To find the value of dT/dt at T=0, we substitute T=0 into the differential equation dT/dt = -1/15T + 5:

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

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

(b) To determine if T is increasing or decreasing at t=0, we can examine the sign of dT/dt at that point. Given that dT/dt | T=0 = 5 (positive), we can conclude that the temperature T is increasing at t=0.

(c) To find the approximate temperature at t=1, we can solve the differential equation:

dT/dt = -1/15T + 5

Separating variables, we get:

dT/(-1/15T + 5) = dt
-15T/(T-75) = dt

Integrating both sides, we have:

-15 ln|T-75| = t + C

Applying the initial condition T(0) = 105, we can solve for the constant C:

-15 ln|105-75| = 0 + C
C = -15 ln(30)

Substituting t=1, we can solve for T:

-15 ln|T-75| = 1 -15 ln(30)
ln|T-75| = -1 + ln(30)
|T-75| = e^(-1 + ln(30))

Since we are considering Fahrenheit temperatures, we take only the positive value:

T - 75 = e^(-1 + ln(30))
T = 75 + e^(-1 + ln(30))

Approximately, T ≈ 76.31 degrees Fahrenheit at t=1.

(d) To make a plot of dT/dt as a function of T, we can substitute different values of T into the equation dT/dt = -1/15T + 5.

(e) To determine for which values of T the temperature T increases, we look at the sign of dT/dt. When dT/dt is positive, T is increasing. From the differential equation dT/dt = -1/15T + 5, we know that T will increase when -1/15T + 5 > 0. Solving this inequality, we find that T > 75. Therefore, for temperatures above 75 degrees Fahrenheit, the temperature T increases.

(f) Based on the given information, we cannot determine the exact temperature of the room. However, when T approaches the temperature of the room, the rate of cooling, represented by dT/dt, should approach zero. Therefore, the temperature of the room can be approximated as the value of T where dT/dt ≈ 0.

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

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

This matches the right-hand side of the differential equation, so T(t) is a solution.

As t approaches infinity, e^(-t/15) approaches 0, and T(t) approaches 75. This means that the temperature of the cup of coffee will approach the temperature of the room over time, which is expected according to Newton's Law of Cooling.