# 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?

## Hard to see just what you wrote, but I'm guessing

dT/dt = -1/(15T+5)
(15T+5) dT = -dt
No, that can't be right, if you want an exponential function.

dT/dt = 15T+5
then we have
dT/(3T+1) = 5dt
ln(3T+1) = 5t+k
3T+1 = ce^5t
T = (ce^5t - 1)/3

Hmm. Not that either. Let's do some answer analysis.

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

Well, if you fix up your T function, things should be pretty straightforward. Where do you get stuck?

## Sorry for the confusion, it's (-1/15)T+5.

And the last part is 75+30e^(-t/15) the fraction is all part of the e and is not divided by each other.

I am confused on a through e mainly and g at the end. Should I solve the equation and just plug in 105 and graph the function to solve the answers to the problems being asked? I just don't get what I'm supposed to do.

## so. you have

dT/dt = -T/15 + 5 = -1/15 (T-75)
dT/(T-75) = -1/15 dt
ln(T-75) = -t/15 + k
T-75 = e^(-t/15 + k)
T-75 = ce^(-t/15)
T = 75 + ce^(-t/15)

Now, you are told that T(0) = 105, so
105 = 75 + ce^0
c = 30

T(t) = 75 + 30e^(-t/15)
as desired in part (g)
So, the coffee starts at 105 and decreases from there, ever more slowly, as the temperature difference becomes less and less.

You can see that as t grows large, e^(-t/15) -> 0, so the coffee approaches 75° as a limit. That must be the temperature of the room.

As for the increasing and decreasing stuff, you know that T(t) is increasing where dT/dt is positive.

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

Substituting T=0 into the equation, we get:
dT/dt |T=0 = -1/15(0) + 5
= 5

The value of dT/dt |T=0 is 5. This means that at time t=0, the rate of change of temperature is increasing at a rate of 5 degrees Fahrenheit per minute.

(b) To determine whether T is increasing or decreasing at t=0, we need to consider the sign of dT/dt at t=0. Since dT/dt |T=0 is positive (5 > 0), it indicates that the temperature T is increasing at t=0.

(c) To find the approximate temperature at t=1, we can use the given solution to the differential equation, T(t) = 75 + 30e^(-t/15).

Substituting t=1 in the equation, we get:
T(1) = 75 + 30e^(-1/15)
≈ 87.36

Therefore, the approximate temperature at t=1 is 87.36 degrees Fahrenheit.

(d) To plot dT/dt as a function of T, we can choose different values of T and compute the corresponding values of dT/dt using the given differential equation.

Let's choose a range of values for T. For example, let T vary from 0 to 100 degrees Fahrenheit. We can then compute dT/dt using the equation dT/dt = -1/15T+5 for each T value. Plotting the resulting dT/dt values against T will give the graph.

(e) To determine for which values of T does T increase, we need to consider the sign of dT/dt. If dT/dt is positive, it indicates that T is increasing.

From the given differential equation dT/dt = -1/15T+5, we see that dT/dt is positive when -1/15T+5 > 0. Solving this inequality:

-1/15T + 5 > 0
-1/15T > -5
T < 75

Therefore, T will increase for values of T less than 75 degrees Fahrenheit.

(f) To estimate the temperature of the room, we can analyze the behavior of the solution T(t) = 75 + 30e^(-t/15) as t approaches infinity.

As t approaches infinity, the exponential term e^(-t/15) approaches 0. This means that the temperature T(t) approaches the constant term 75. Hence, we can infer 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 the values T(0) = 105 and T(t) = 75 + 30e^(-t/15) into the differential equation and see if it holds true.

Substituting T(0) = 105 and T(t) = 75 + 30e^(-t/15) into the differential equation dT/dt = -1/15T+5, we get:

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

Simplifying both sides of the equation, we find that they are equal, which verifies that T(t) = 75 + 30e^(-t/15) is the correct solution with the initial value T(0) = 105.

After a long time (as t approaches infinity), the term e^(-t/15) approaches 0 and T(t) approaches the constant term 75. This means that the temperature of the coffee eventually reaches the temperature of the room, which we estimated to be approximately 75 degrees Fahrenheit.