dm/dt = -km, m(0) = 60, half-life 10, when is m = 15 Step by Step

Step by step

The question

Question: find the time t when m = 15, if dm/dt = −km and m(0) = 60, with a half-life of 10.

dmdt = −km

Half-life 10 means: after 10, half is left.

So m(10) = 30.

m(0) = 60, half-life 10: m(10) = 30

1 Read the equation

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

The more there is, the faster it changes.

the rate = −k × m: the more m, the faster it changes

2 Separate and integrate

This is a separable equation.

Divide by m, times dt.

dmm = −k dt

Integrate both sides, as in the separable lesson.

ln |m| = −kt + C

Use e on both sides.

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

m = Ae^(−kt)

3 Use the starting value

Put in t = 0 and m = 60.

e⁰ = 1.

60 = Ae^(0)

So A = 60: the amount at the start.

A = 60

The formula: m = 60e^(−kt).

m = 60e^(−kt)

4 Find k

Now the second value: t = 10, m = 30.

30 = 60e^(−10k)

Divide both sides by 60.

12 = e^(−10k)

Careful!

k is inside a power.

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

Careful: not k = 3060 × 10 ✗

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

ln(12) = −10k

Divide by −10: k is about 0.0693.

k = ln(12)−10 ≈ 0.0693

So m ≈ 60e^(−0.0693t).

m ≈ 60e^(−0.0693t)

5 Answer the question

Put in m = 15.

Now t is the unknown.

15 ≈ 60e^(−0.0693t)

Divide both sides by 60.

14 ≈ e^(−0.0693t)

Take ln of both sides.

ln(14) ≈ −0.0693t

Divide.

So t ≈ 20.

t = ln(14)−0.0693 ≈ 20

6 Check

Let's check.

At t = 0: 60, the start.

Right!

m(0) = 60 × 1 = 60

Put t ≈ 20 back in: about 15.

Right!

m(20) ≈ 15

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