dT/dt = -0.05(T - 18), T(0) = 98, when is T = 38 Step by Step

Step by step

The question

Question: find the time t when T = 38, if dT/dt = −0.05(T − 18) and T(0) = 98.

dTdt = −0.05(T − 18)

T(0) = 98

1 Read the equation

T is the temperature, t is the time, and 18 is the room temperature.

The bigger the gap T − 18, the faster it cools.

the rate = −0.05 × (T − 18): T − 18 is the gap to the room

2 Separate and integrate

This is a separable equation.

Divide by T − 18, times dt.

dTT − 18 = −0.05 dt

Integrate both sides, as in the separable lesson.

ln |T − 18| = −0.05t + C

Use e on both sides.

e^C is just a number: call it A.

T − 18 = Ae^(−0.05t)

Add 18 to both sides.

T = 18 + Ae^(−0.05t)

3 Use the starting value

Put in t = 0 and T = 98.

e⁰ = 1.

98 = 18 + Ae^(0)

So A = 80: the gap at the start.

A = 98 − 18 = 80

The formula: T = 18 + 80e^(−0.05t).

T = 18 + 80e^(−0.05t)

Careful!

Don't forget the room temperature.

Only the gap, 80, shrinks; T gets close to 18, not to 0.

Careful: not T = 98e^(−0.05t) ✗

4 Answer the question

Put in T = 38.

Now t is the unknown.

38 = 18 + 80e^(−0.05t)

Take away 18 from both sides.

38 − 18 = 80e^(−0.05t)

Divide both sides by 80.

14 = e^(−0.05t)

Take ln of both sides.

ln(14) = −0.05t

Divide.

So t ≈ 27.73.

t = ln(14)−0.05 ≈ 27.73

5 Check

Let's check.

At t = 0: 98, the start.

Right!

T(0) = 18 + 80 × 1 = 98

Put t ≈ 27.73 back in: about 38.

Right!

T(27.73) ≈ 38

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer