dN/dt = kN, N(0) = 1000, N(3) = 1500, find N(6) Step by Step

Step by step

The question

Question: find N(6), if dN/dt = kN and N(0) = 1000, N(3) = 1500.

dNdt = kN

N(0) = 1000, N(3) = 1500

1 Read the equation

The rate of change is a number k times the amount N.

The more there is, the faster it changes.

the rate = k × N: the more N, the faster it changes

2 Separate and integrate

This is a separable equation.

Divide by N, times dt.

dNN = k dt

Integrate both sides, as in the separable lesson.

ln |N| = kt + C

Use e on both sides.

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

N = Ae^(kt)

3 Use the starting value

Put in t = 0 and N = 1000.

e⁰ = 1.

1000 = Ae^(0)

So A = 1000: the amount at the start.

A = 1000

The formula: N = 1000e^(kt).

N = 1000e^(kt)

4 Find k

Now the second value: t = 3, N = 1500.

1500 = 1000e^(3k)

Divide both sides by 1000.

32 = e^(3k)

Careful!

k is inside a power.

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

Careful: not k = 15001000 × 3 ✗

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

ln(32) = 3k

Divide by 3: k is about 0.1352.

k = ln(32)3 ≈ 0.1352

So N ≈ 1000e^(0.1352t).

N ≈ 1000e^(0.1352t)

5 Answer the question

Put in t = 6.

N(6) = 1000e^(0.1352 × 6)

The power: 0.8109.

≈ 1000e^(0.8109)

So N(6) ≈ 2250.

≈ 2250

6 Check

Let's check.

At t = 0: 1000, the start.

Right!

N(0) = 1000 × 1 = 1000

With this k, at t = 3 the formula gives 1500: the second value.

Right!

N(3) ≈ 1000e^(0.1352 × 3) ≈ 1500

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