Worked examples
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.
▶ Watch it solved
The question
Question: find the time t when P = 200, if dP/dt = 0.04P and P(0) = 100.
dPdt = 0.04P
P(0) = 100
1 Read the equation
The rate of change is 0.04 times the amount P.
The more there is, the faster it grows.
k > 0: growth.
the rate = 0.04 × P: the more P, the faster it changes
2 Separate and integrate
This is a separable equation.
Divide by P, times dt.
dPP = 0.04 dt
Integrate both sides, as in the separable lesson.
ln |P| = 0.04t + C
Use e on both sides.
e^C is just a number: call it A.
P = Ae^(0.04t)
3 Use the starting value
Put in t = 0 and P = 100.
e⁰ = 1.
100 = Ae^(0)
So A = 100: the amount at the start.
A = 100
The formula: P = 100e^(0.04t).
P = 100e^(0.04t)
Careful!
This is not a steady rate.
The rate grows with P, so the formula has e, not a plus.
Careful: not P = 100 + 0.04 × 100t ✗
4 Answer the question
Put in P = 200.
Now t is the unknown.
200 = 100e^(0.04t)
Divide both sides by 100.
2 = e^(0.04t)
Take ln of both sides.
ln 2 = 0.04t
Divide.
So t ≈ 17.33.
t = ln 20.04 ≈ 17.33
5 Check
Let's check.
At t = 0: 100, the start.
Right!
P(0) = 100 × 1 = 100
Put t ≈ 17.33 back in: about 200.01.
Right!
P(17.33) ≈ 200.01
Let's look back.
First: read the equation.
Then: separate and integrate.
Then: use the starting value.
Then: answer the question.
Last: check.
Answer t = ln 20.04 ≈ 17.33
▶ Watch it solved
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 t = ln(14)−0.0693 ≈ 20
▶ Watch it solved
