dT/dt = -k(T - 22), T(0) = 82, T(4) = 62, find T(12) Step by Step

Step by step

The question

Question: find T(12), if dT/dt = −k(T − 22) and T(0) = 82, T(4) = 62.

dTdt = −k(T − 22)

T(0) = 82, T(4) = 62

1 Read the equation

T is the temperature, t is the time, and 22 is the room temperature.

The bigger the gap T − 22, the faster it cools.

the rate = −k × (T − 22): T − 22 is the gap to the room

2 Separate and integrate

This is a separable equation.

Divide by T − 22, times dt.

dTT − 22 = −k dt

Integrate both sides, as in the separable lesson.

ln |T − 22| = −kt + C

Use e on both sides.

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

T − 22 = Ae^(−kt)

Add 22 to both sides.

T = 22 + Ae^(−kt)

3 Use the starting value

Put in t = 0 and T = 82.

e⁰ = 1.

82 = 22 + Ae^(0)

So A = 60: the gap at the start.

A = 82 − 22 = 60

The formula: T = 22 + 60e^(−kt).

T = 22 + 60e^(−kt)

Careful!

Don't forget the room temperature.

Only the gap, 60, shrinks; T gets close to 22, not to 0.

Careful: not T = 82e^(−kt) ✗

4 Find k

Now the second value: t = 4, T = 62.

62 = 22 + 60e^(−4k)

Take away 22 from both sides.

62 − 22 = 60e^(−4k)

Divide both sides by 60.

23 = e^(−4k)

Take ln of both sides: ln and e cancel, so the power comes down.

ln(23) = −4k

Divide by −4: k is about 0.1014.

k = ln(23)−4 ≈ 0.1014

So T ≈ 22 + 60e^(−0.1014t).

T ≈ 22 + 60e^(−0.1014t)

5 Answer the question

Put in t = 12.

T(12) = 22 + 60e^(−0.1014 × 12)

The power: −1.2164.

≈ 22 + 60e^(−1.2164)

A calculator: 60 × e^(−1.2164) ≈ 17.78.

≈ 22 + 17.78

So T(12) ≈ 39.78.

≈ 39.78

6 Check

Let's check.

At t = 0: 82, the start.

Right!

T(0) = 22 + 60 × 1 = 82

With this k, at t = 4 the formula gives 62: the second value.

Right!

T(4) ≈ 22 + 60e^(−0.1014 × 4) ≈ 62

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: find k.

Then: answer the question.

Last: check.

Answer