A population P grows by dP/dt = 0.05P, with P(0) = 200. What is… Step by Step

Step by step

The question

Question: find P(10), if dP/dt = 0.05P and P(0) = 200.

dPdt = 0.05P

P(0) = 200

There are four choices.

First we solve it, the normal way.

Then we pick.

1 Read the equation

The rate of change is 0.05 times the amount P.

The more there is, the faster it grows.

k > 0: growth.

the rate = 0.05 × P: the more P, the faster it changes

2 Separate and integrate

This is a separable equation.

Divide by P, times dt.

dPP = 0.05 dt

Integrate both sides, as in the separable lesson.

ln |P| = 0.05t + C

Use e on both sides.

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

P = Ae^(0.05t)

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^(0.05t).

P = 200e^(0.05t)

Careful!

This is not a steady rate.

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

The right way: P = 200e^(0.05t).

Careful: not P = 200 + 0.05 × 200t ✗

right: P = 200e^(0.05t) ✓

4 Answer the question

Put in t = 10.

P(10) = 200e^(0.05 × 10)

The power: 0.05 times 10 is 0.5.

= 200e^(0.5)

So P(10) ≈ 329.74.

≈ 329.74

5 Check

Let's check.

At t = 0: 200, the start.

Right!

P(0) = 200 × 1 = 200

And the derivative of the formula is 0.05 times P: the equation again.

Right!

P′ = 200 × 0.05e^(0.05t) = 0.05 × P

6 Pick the choice

Our answer is P(10) ≈ 329.74.

That is choice C.

C) 329.74

7 Why not the others

A is 300.

Wrong: grew by 0.05 × 200 a year in a straight line.

That is the Careful mistake from before.

A) 300 ✗ grew by 0.05 × 200 a year in a straight line

B is 210.

Wrong: grew once, not for 10 years.

B) 210 ✗ grew once, not for 10 years

D is 543.66.

Wrong: used e¹, not e⁰.5.

D) 543.66 ✗ used e^1, not e^0.5

Let's look back.

First: read the equation.

Then: separate and integrate.

Then: use the starting value.

Then: answer the question.

Then: check.

Then: pick the choice.

Last: why not the others.

Answer