dT/dt = -0.15(T - 20), T(0) = 80, find T(8) Step by Step

Step by step

The question

Question: find T(8), if dT/dt = −0.15(T − 20) and T(0) = 80.

dTdt = −0.15(T − 20)

T(0) = 80

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 cools.

the rate = −0.15 × (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 = −0.15 dt

Integrate both sides, as in the separable lesson.

ln |T − 20| = −0.15t + C

Use e on both sides.

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

T − 20 = Ae^(−0.15t)

Add 20 to both sides.

T = 20 + Ae^(−0.15t)

3 Use the starting value

Put in t = 0 and T = 80.

e⁰ = 1.

80 = 20 + Ae^(0)

So A = 60: the gap at the start.

A = 80 − 20 = 60

The formula: T = 20 + 60e^(−0.15t).

T = 20 + 60e^(−0.15t)

Careful!

Don't forget the room temperature.

Only the gap, 60, shrinks; T gets close to 20, not to 0.

Careful: not T = 80e^(−0.15t) ✗

4 Answer the question

Put in t = 8.

T(8) = 20 + 60e^(−0.15 × 8)

The power: −1.2.

= 20 + 60e^(−1.2)

A calculator: 60 × e^(−1.2) ≈ 18.07.

≈ 20 + 18.07

So T(8) ≈ 38.07.

≈ 38.07

5 Check

Let's check.

At t = 0: 80, the start.

Right!

T(0) = 20 + 60 × 1 = 80

And the derivative of the formula is −0.15 times the gap T − 20: the equation again.

Right!

T′ = 60 × −0.15e^(−0.15t) = −0.15 × (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