dP/dt = 0.06P, P(0) = 250, when is P = 1000 Step by Step

Step by step

The question

Question: find the time t when P = 1000, if dP/dt = 0.06P and P(0) = 250.

dPdt = 0.06P

P(0) = 250

1 Read the equation

The rate of change is 0.06 times the amount P.

The more there is, the faster it grows.

k > 0: growth.

the rate = 0.06 × P: the more P, the faster it changes

2 Separate and integrate

This is a separable equation.

Divide by P, times dt.

dPP = 0.06 dt

Integrate both sides, as in the separable lesson.

ln |P| = 0.06t + C

Use e on both sides.

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

P = Ae^(0.06t)

3 Use the starting value

Put in t = 0 and P = 250.

e⁰ = 1.

250 = Ae^(0)

So A = 250: the amount at the start.

A = 250

The formula: P = 250e^(0.06t).

P = 250e^(0.06t)

Careful!

This is not a steady rate.

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

Careful: not P = 250 + 0.06 × 250t ✗

4 Answer the question

Put in P = 1000.

Now t is the unknown.

1000 = 250e^(0.06t)

Divide both sides by 250.

4 = e^(0.06t)

Take ln of both sides.

ln 4 = 0.06t

Divide.

So t ≈ 23.1.

t = ln 40.06 ≈ 23.1

5 Check

Let's check.

At t = 0: 250, the start.

Right!

P(0) = 250 × 1 = 250

Put t ≈ 23.1 back in: about 999.71.

Right!

P(23.1) ≈ 999.71

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer