dP/dt = 0.03P, P(0) = 500, find P(10) Step by Step

Step by step

The question

Question: find P(10), if dP/dt = 0.03P and P(0) = 500.

dPdt = 0.03P

P(0) = 500

1 Read the equation

The rate of change is 0.03 times the amount P.

The more there is, the faster it grows.

k > 0: growth.

the rate = 0.03 × P: the more P, the faster it changes

2 Separate and integrate

This is a separable equation.

Divide by P, times dt.

dPP = 0.03 dt

Integrate both sides, as in the separable lesson.

ln |P| = 0.03t + C

Use e on both sides.

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

P = Ae^(0.03t)

3 Use the starting value

Put in t = 0 and P = 500.

e⁰ = 1.

500 = Ae^(0)

So A = 500: the amount at the start.

A = 500

The formula: P = 500e^(0.03t).

P = 500e^(0.03t)

Careful!

This is not a steady rate.

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

Careful: not P = 500 + 0.03 × 500t ✗

4 Answer the question

Put in t = 10.

P(10) = 500e^(0.03 × 10)

The power: 0.3.

= 500e^(0.3)

So P(10) ≈ 674.93.

≈ 674.93

5 Check

Let's check.

At t = 0: 500, the start.

Right!

P(0) = 500 × 1 = 500

And the derivative of the formula is 0.03 times P: the equation again.

Right!

P′ = 500 × 0.03e^(0.03t) = 0.03 × P

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Last: check.

Answer