dP/dt = kP, P(0) = 400, P(10) = 600, when is P = 1000 Step by Step

Step by step

The question

Question: find the time t when P = 1000, if dP/dt = kP and P(0) = 400, P(10) = 600.

dPdt = kP

P(0) = 400, P(10) = 600

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 = 400.

e⁰ = 1.

400 = Ae^(0)

So A = 400: the amount at the start.

A = 400

The formula: P = 400e^(kt).

P = 400e^(kt)

4 Find k

Now the second value: t = 10, P = 600.

600 = 400e^(10k)

Divide both sides by 400.

32 = e^(10k)

Careful!

k is inside a power.

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

Careful: not k = 600400 × 10 ✗

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

ln(32) = 10k

Divide by 10: k is about 0.0405.

k = ln(32)10 ≈ 0.0405

So P ≈ 400e^(0.0405t).

P ≈ 400e^(0.0405t)

5 Answer the question

Put in P = 1000.

Now t is the unknown.

1000 ≈ 400e^(0.0405t)

Divide both sides by 400.

52 ≈ e^(0.0405t)

Take ln of both sides.

ln(52) ≈ 0.0405t

Divide.

So t ≈ 22.6.

t = ln(52)0.0405 ≈ 22.6

6 Check

Let's check.

At t = 0: 400, the start.

Right!

P(0) = 400 × 1 = 400

Put t ≈ 22.6 back in: about 1000.06.

Right!

P(22.6) ≈ 1000.06

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