dP/dt = kP, P(0) = 200, P(5) = 400, find P(8) Step by Step

Step by step

The question

Question: find P(8), if dP/dt = kP and P(0) = 200, P(5) = 400.

dPdt = kP

P(0) = 200, P(5) = 400

1 Read the equation

The rate of change is a number k times the amount P.

The more there is, the faster it changes.

the rate = k × P: the more P, the faster it changes

2 Separate and integrate

This is a separable equation.

Divide by P, times dt.

dPP = k dt

Integrate both sides, as in the separable lesson.

ln |P| = kt + C

Use e on both sides.

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

P = Ae^(kt)

3 Use the starting value

Put in t = 0 and P = 200.

e⁰ = 1.

200 = Ae^(0)

So A = 200: the amount at the start.

A = 200

The formula: P = 200e^(kt).

P = 200e^(kt)

4 Find k

Now the second value: t = 5, P = 400.

400 = 200e^(5k)

Divide both sides by 200.

2 = e^(5k)

Careful!

k is inside a power.

To bring it down, take ln of both sides first.

Careful: not k = 400200 × 5 ✗

Take ln of both sides: ln and e cancel, so the power comes down.

ln 2 = 5k

Divide by 5: k is about 0.1386.

k = ln 25 ≈ 0.1386

So P ≈ 200e^(0.1386t).

P ≈ 200e^(0.1386t)

5 Answer the question

Put in t = 8.

P(8) = 200e^(0.1386 × 8)

The power: 1.109.

≈ 200e^(1.109)

So P(8) ≈ 606.29.

≈ 606.29

6 Check

Let's check.

At t = 0: 200, the start.

Right!

P(0) = 200 × 1 = 200

With this k, at t = 5 the formula gives 400: the second value.

Right!

P(5) ≈ 200e^(0.1386 × 5) ≈ 400

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: find k.

Then: answer the question.

Last: check.

Answer