dT/dt = -0.2(T - 20), T(0) = 4, find T(3) Step by Step

Step by step

The question

Question: find T(3), if dT/dt = −0.2(T − 20) and T(0) = 4.

dTdt = −0.2(T − 20)

T(0) = 4

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 = −0.2 × (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 = −0.2 dt

Integrate both sides, as in the separable lesson.

ln |T − 20| = −0.2t + C

Use e on both sides.

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

T − 20 = Ae^(−0.2t)

Add 20 to both sides.

T = 20 + Ae^(−0.2t)

3 Use the starting value

Put in t = 0 and T = 4.

e⁰ = 1.

4 = 20 + Ae^(0)

So A = −16: the gap at the start.

A = 4 − 20 = −16

The formula: T = 20 − 16e^(−0.2t).

T = 20 − 16e^(−0.2t)

Careful!

Don't forget the room temperature.

Only the gap, -16, shrinks; T gets close to 20, not to 0.

Careful: not T = 4e^(−0.2t) ✗

4 Answer the question

Put in t = 3.

T(3) = 20 − 16e^(−0.2 × 3)

The power: −0.6.

= 20 − 16e^(−0.6)

A calculator: -16 × e^(−0.6) ≈ −8.78.

≈ 20 + −8.78

So T(3) ≈ 11.22.

≈ 11.22

5 Check

Let's check.

At t = 0: 4, the start.

Right!

T(0) = 20 − 16 × 1 = 4

And the derivative of the formula is −0.2 times the gap T − 20: the equation again.

Right!

T′ = -16 × −0.2e^(−0.2t) = −0.2 × (T − 20)

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer