dT/dt = 0.2(18 - T), T(0) = 90, find T(5) Step by Step

Step by step

The question

Question: find T(5), if dT/dt = −0.2(T − 18) and T(0) = 90.

dTdt = −0.2(T − 18)

T(0) = 90

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 = −0.2 × (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 = −0.2 dt

Integrate both sides, as in the separable lesson.

ln |T − 18| = −0.2t + C

Use e on both sides.

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

T − 18 = Ae^(−0.2t)

Add 18 to both sides.

T = 18 + Ae^(−0.2t)

3 Use the starting value

Put in t = 0 and T = 90.

e⁰ = 1.

90 = 18 + Ae^(0)

So A = 72: the gap at the start.

A = 90 − 18 = 72

The formula: T = 18 + 72e^(−0.2t).

T = 18 + 72e^(−0.2t)

Careful!

Don't forget the room temperature.

Only the gap, 72, shrinks; T gets close to 18, not to 0.

Careful: not T = 90e^(−0.2t) ✗

4 Answer the question

Put in t = 5.

T(5) = 18 + 72e^(−0.2 × 5)

The power: −1.

= 18 + 72e^(−1)

A calculator: 72 × e^(−1) ≈ 26.49.

≈ 18 + 26.49

So T(5) ≈ 44.49.

≈ 44.49

5 Check

Let's check.

At t = 0: 90, the start.

Right!

T(0) = 18 + 72 × 1 = 90

And the derivative of the formula is −0.2 times the gap T − 18: the equation again.

Right!

T′ = 72 × −0.2e^(−0.2t) = −0.2 × (T − 18)

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer