dT/dt = -k(T - 20), T(0) = 5, T(10) = 12, find T(30) Step by Step

Step by step

The question

Question: find T(30), if dT/dt = −k(T − 20) and T(0) = 5, T(10) = 12.

dTdt = −k(T − 20)

T(0) = 5, T(10) = 12

1 Read the equation

T is the temperature, t is the time, and 20 is the room temperature.

The bigger the gap T − 20, the faster it warms.

the rate = −k × (T − 20): T − 20 is the gap to the room

2 Separate and integrate

This is a separable equation.

Divide by T − 20, times dt.

dTT − 20 = −k dt

Integrate both sides, as in the separable lesson.

ln |T − 20| = −kt + C

Use e on both sides.

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

T − 20 = Ae^(−kt)

Add 20 to both sides.

T = 20 + Ae^(−kt)

3 Use the starting value

Put in t = 0 and T = 5.

e⁰ = 1.

5 = 20 + Ae^(0)

So A = −15: the gap at the start.

A = 5 − 20 = −15

The formula: T = 20 − 15e^(−kt).

T = 20 − 15e^(−kt)

Careful!

Don't forget the room temperature.

Only the gap, -15, shrinks; T gets close to 20, not to 0.

Careful: not T = 5e^(−kt) ✗

4 Find k

Now the second value: t = 10, T = 12.

12 = 20 − 15e^(−10k)

Take away 20 from both sides.

12 − 20 = -15e^(−10k)

Divide both sides by −15.

815 = e^(−10k)

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

ln(815) = −10k

Divide by −10: k is about 0.0629.

k = ln(815)−10 ≈ 0.0629

So T ≈ 20 − 15e^(−0.0629t).

T ≈ 20 − 15e^(−0.0629t)

5 Answer the question

Put in t = 30.

T(30) = 20 − 15e^(−0.0629 × 30)

The power: −1.8858.

≈ 20 − 15e^(−1.8858)

A calculator: -15 × e^(−1.8858) ≈ −2.28.

≈ 20 + −2.28

So T(30) ≈ 17.72.

≈ 17.72

6 Check

Let's check.

At t = 0: 5, the start.

Right!

T(0) = 20 − 15 × 1 = 5

With this k, at t = 10 the formula gives 12: the second value.

Right!

T(10) ≈ 20 + -15e^(−0.0629 × 10) ≈ 12

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