dT/dt = -0.1(T - 20), T(0) = 90, find T(10) Step by Step

Step by step

The question

Question: find T(10), if dT/dt = −0.1(T − 20) and T(0) = 90.

dTdt = −0.1(T − 20)

T(0) = 90

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 = −0.1 × (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 = −0.1 dt

Integrate both sides, as in the separable lesson.

ln |T − 20| = −0.1t + C

Use e on both sides.

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

T − 20 = Ae^(−0.1t)

Add 20 to both sides.

T = 20 + Ae^(−0.1t)

3 Use the starting value

Put in t = 0 and T = 90.

e⁰ = 1.

90 = 20 + Ae^(0)

So A = 70: the gap at the start.

A = 90 − 20 = 70

The formula: T = 20 + 70e^(−0.1t).

T = 20 + 70e^(−0.1t)

Careful!

Don't forget the room temperature.

Only the gap, 70, shrinks; T gets close to 20, not to 0.

Careful: not T = 90e^(−0.1t) ✗

4 Answer the question

Put in t = 10.

T(10) = 20 + 70e^(−0.1 × 10)

The power: −1.

= 20 + 70e^(−1)

A calculator: 70 × e^(−1) ≈ 25.75.

≈ 20 + 25.75

So T(10) ≈ 45.75.

≈ 45.75

5 Check

Let's check.

At t = 0: 90, the start.

Right!

T(0) = 20 + 70 × 1 = 90

And the derivative of the formula is −0.1 times the gap T − 20: the equation again.

Right!

T′ = 70 × −0.1e^(−0.1t) = −0.1 × (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