dT/dt = -0.2(T - 25), T(0) = 85, find T(5) Step by Step

Step by step

The question

Question: find T(5), if dT/dt = −0.2(T − 25) and T(0) = 85.

dTdt = −0.2(T − 25)

T(0) = 85

1 Read the equation

T is the temperature, t is the time, and 25 is the room temperature.

The bigger the gap T − 25, the faster it cools.

the rate = −0.2 × (T − 25): T − 25 is the gap to the room

2 Separate and integrate

This is a separable equation.

Divide by T − 25, times dt.

dTT − 25 = −0.2 dt

Integrate both sides, as in the separable lesson.

ln |T − 25| = −0.2t + C

Use e on both sides.

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

T − 25 = Ae^(−0.2t)

Add 25 to both sides.

T = 25 + Ae^(−0.2t)

3 Use the starting value

Put in t = 0 and T = 85.

e⁰ = 1.

85 = 25 + Ae^(0)

So A = 60: the gap at the start.

A = 85 − 25 = 60

The formula: T = 25 + 60e^(−0.2t).

T = 25 + 60e^(−0.2t)

Careful!

Don't forget the room temperature.

Only the gap, 60, shrinks; T gets close to 25, not to 0.

Careful: not T = 85e^(−0.2t) ✗

4 Answer the question

Put in t = 5.

T(5) = 25 + 60e^(−0.2 × 5)

The power: −1.

= 25 + 60e^(−1)

A calculator: 60 × e^(−1) ≈ 22.07.

≈ 25 + 22.07

So T(5) ≈ 47.07.

≈ 47.07

5 Check

Let's check.

At t = 0: 85, the start.

Right!

T(0) = 25 + 60 × 1 = 85

And the derivative of the formula is −0.2 times the gap T − 25: the equation again.

Right!

T′ = 60 × −0.2e^(−0.2t) = −0.2 × (T − 25)

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer