dT/dt = -0.3(T - 18), T(0) = 68, find T(2) Step by Step

Step by step

The question

Question: find T(2), if dT/dt = −0.3(T − 18) and T(0) = 68.

dTdt = −0.3(T − 18)

T(0) = 68

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 = −0.3 × (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 = −0.3 dt

Integrate both sides, as in the separable lesson.

ln |T − 18| = −0.3t + C

Use e on both sides.

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

T − 18 = Ae^(−0.3t)

Add 18 to both sides.

T = 18 + Ae^(−0.3t)

3 Use the starting value

Put in t = 0 and T = 68.

e⁰ = 1.

68 = 18 + Ae^(0)

So A = 50: the gap at the start.

A = 68 − 18 = 50

The formula: T = 18 + 50e^(−0.3t).

T = 18 + 50e^(−0.3t)

Careful!

Don't forget the room temperature.

Only the gap, 50, shrinks; T gets close to 18, not to 0.

Careful: not T = 68e^(−0.3t) ✗

4 Answer the question

Put in t = 2.

T(2) = 18 + 50e^(−0.3 × 2)

The power: −0.6.

= 18 + 50e^(−0.6)

A calculator: 50 × e^(−0.6) ≈ 27.44.

≈ 18 + 27.44

So T(2) ≈ 45.44.

≈ 45.44

5 Check

Let's check.

At t = 0: 68, the start.

Right!

T(0) = 18 + 50 × 1 = 68

And the derivative of the formula is −0.3 times the gap T − 18: the equation again.

Right!

T′ = 50 × −0.3e^(−0.3t) = −0.3 × (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