dP/dt = 0.04P, P(0) = 100, when is P = 200 Step by Step

Step by step

The question

Question: find the time t when P = 200, if dP/dt = 0.04P and P(0) = 100.

dPdt = 0.04P

P(0) = 100

1 Read the equation

The rate of change is 0.04 times the amount P.

The more there is, the faster it grows.

k > 0: growth.

the rate = 0.04 × P: the more P, the faster it changes

2 Separate and integrate

This is a separable equation.

Divide by P, times dt.

dPP = 0.04 dt

Integrate both sides, as in the separable lesson.

ln |P| = 0.04t + C

Use e on both sides.

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

P = Ae^(0.04t)

3 Use the starting value

Put in t = 0 and P = 100.

e⁰ = 1.

100 = Ae^(0)

So A = 100: the amount at the start.

A = 100

The formula: P = 100e^(0.04t).

P = 100e^(0.04t)

Careful!

This is not a steady rate.

The rate grows with P, so the formula has e, not a plus.

Careful: not P = 100 + 0.04 × 100t ✗

4 Answer the question

Put in P = 200.

Now t is the unknown.

200 = 100e^(0.04t)

Divide both sides by 100.

2 = e^(0.04t)

Take ln of both sides.

ln 2 = 0.04t

Divide.

So t ≈ 17.33.

t = ln 20.04 ≈ 17.33

5 Check

Let's check.

At t = 0: 100, the start.

Right!

P(0) = 100 × 1 = 100

Put t ≈ 17.33 back in: about 200.01.

Right!

P(17.33) ≈ 200.01

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer