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Calculus BC · Unit 4 · 4.3
AP Calculus · Contextual Applications of Differentiation
Rates of Change in Applied Contexts
In minus out. The sign of the net rate answers most of the question before any arithmetic.
key idea$\text{Rate in}-\text{Rate out}=\text{Net rate of change}$
Setting up the model
Outside motion, the commonest AP set-up is a quantity with something flowing in and something flowing out. The net rate is the difference, and its sign answers most of the question before any arithmetic.
The in-minus-out model
- Name the quantity and its units — litres of water, grams of salt, number of people.
- Identify the in-rate and the out-rate, both in quantity-units per time-unit.
- Net rate $=$ in $-$ out. This is the derivative of the quantity.
- Positive net rate ⇒ the quantity is growing. Negative ⇒ shrinking. Zero ⇒ a maximum, a minimum, or a steady state.
| Context | Quantity | In-rate | Out-rate |
|---|
| tank | litres of water | pumped in | drained out |
| population | number of people | births + immigration | deaths + emigration |
| bank account | dollars | deposits + interest | withdrawals + fees |
| reservoir | cubic metres | rainfall inflow | evaporation + release |
| inventory | units in stock | deliveries | sales |
| chemical | grams dissolved | added | reacted away |
The physical story changes; the structure never does. Find the two rates, subtract, read the sign.
Worked examples
EX 1
Net rate and its sign
tankWater enters a tank at $R_{\text{in}}(t)=12\sqrt{t}$ L/min and leaves at $R_{\text{out}}(t)=3t$ L/min. Is the amount of water increasing or decreasing at $t=9$?
1
Write the net rate
$W'(t)=R_{\text{in}}(t)-R_{\text{out}}(t)=12\sqrt{t}-3t$.
2
Evaluate at $t=9$
$12\sqrt{9}-3(9)=36-27=9$.
3
Read the sign
Positive, so water is entering faster than it leaves.
4
State it with units
At $t=9$ minutes the amount of water is increasing at $9$ litres per minute.
5
Note what was not needed
No formula for $W(t)$ itself — the question only asked about the rate.
EX 2
When is the amount greatest?
net rate zeroFor the same tank, when is the volume of water at a maximum on $[0,25]$?
1
A maximum needs the net rate to be zero
Set $12\sqrt{t}-3t=0$.
2
Solve
$3\sqrt{t}\left(4-\sqrt{t}\right)=0$, so $\sqrt{t}=0$ or $\sqrt{t}=4$, giving $t=0$ and $t=16$.
3
Test the sign around $t=16$
At $t=9$: $+9$. At $t=25$: $12(5)-75=-15$. So $W'$ goes $+\to-$.
4
Conclude
A sign change from positive to negative means a maximum at $t=16$ minutes.
5
Note the reasoning
This is the First Derivative Test wearing an applied costume — the net rate is the derivative.
EX 3
Interpret a zero net rate
steady stateA population model gives $P'(t)=0$ at $t=40$ with $P''(40)\gt 0$. Interpret both facts.
1
Read the first derivative
Births and deaths are exactly balanced — the population is momentarily not changing.
2
Read the second
$P''\gt 0$ means the net rate is increasing through zero, i.e. going from negative to positive.
3
Combine
The population was falling, has bottomed out, and is about to start growing.
4
Name it
$t=40$ is a minimum of the population.
5
Common misreading
“$P'=0$” does not mean the population is zero. It means the population is not changing.
EX 4
Two rates with different units
unit checkA pool gains water at $0.5$ m$^{3}$/min and loses $200$ litres/min. Find the net rate.
1
Spot the mismatch
One rate is in cubic metres, the other in litres. They cannot be subtracted as they stand.
2
Convert
$1$ m$^{3}=1000$ litres, so $0.5$ m$^{3}$/min $=500$ litres/min.
3
Subtract
$500-200=300$ litres per minute.
4
State the answer
The pool is gaining water at $300$ L/min, or equivalently $0.3$ m$^{3}$/min.
5
The habit
Check that both rates share units before subtracting. A unit mismatch produces a confidently wrong number.
Units & interpretation
| Statement | Means | Does NOT mean |
|---|
| $Q'(t)\gt 0$ | the quantity is growing | the quantity is large |
| $Q'(t)=0$ | momentarily not changing | the quantity is zero |
| $Q'(t)\lt 0$ | the quantity is shrinking | the quantity is negative |
| $Q''(t)\lt 0$ with $Q'\gt 0$ | growing, but more slowly | shrinking |
| in-rate $=$ out-rate | a steady state or an extremum | nothing is flowing |
Every row confuses the quantity with its rate. Naming which one you are describing is most of the interpretation mark.
Watch out
Rate zero is not quantity zero
$Q'=0$ says the quantity has stopped changing. The quantity itself is usually at its largest or smallest right there.
Match units before subtracting
Litres against cubic metres, minutes against hours — convert first. Example 4 is a factor-of-1000 trap.
The net rate is the derivative
Once you write $Q'=\text{in}-\text{out}$, everything from the derivative toolkit applies: sign charts, extrema, concavity.
A balanced flow is still a flow
In-rate equal to out-rate means the amount is steady, not that nothing is happening. Water may still be pouring through.