Everything else that changes AB & BC

Pick the quantity being tracked and name it. Everything follows from that choice.

If $V(t)$ is the volume of water in a tank, then $V'(t)$ is positive exactly when the tank is gaining and negative exactly when it is losing. Not sometimes. Always. So a tank losing water at 5 gallons per minute is written

$$\frac{dV}{dt} = -5,$$

with the minus sign in the equation, not saved for the end. Writing $\frac{dV}{dt} = 5$ and remembering later that it is really a loss is how sign errors survive to the final line.

The same convention covers every context:

1. Cooling is $\frac{dT}{dt} < 0$; heating is $\frac{dT}{dt} > 0$.
2. Depletion of fuel, inventory, or a drug in the bloodstream is a negative derivative of the amount remaining.
3. A cost that is falling has $C'(q) < 0$, even though the cost itself stays positive.

The sign of the derivative and the sign of the quantity have nothing to do with each other. A tank holding 400 gallons and draining fast is $V = 400$ with $V' = -18$.

When a quantity has something entering and something leaving, both described as positive rates, the derivative of the amount is the difference:

$$\frac{dA}{dt} = (\text{rate in}) - (\text{rate out}).$$

Subtract, never add. Adding two positive rates describes a tank with two inflows, which is a different tank. If water enters at 12 gallons per minute and leaves at 15, then $\frac{dV}{dt} = 12 - 15 = -3$, and the tank is losing 3 gallons every minute despite water pouring in the whole time.

The comparison, not either rate alone, decides everything:

1. In $>$ out: the amount is increasing.
2. In $<$ out: the amount is decreasing.
3. In $=$ out: the amount is momentarily unchanging, and this is where a maximum or minimum can occur.

That last line is the standard tank question. The amount is largest at the instant the inflow curve drops below the outflow curve, which is a crossing of the two rate graphs, not a peak of either one.

Applied contexts often track something that is itself measured per unit time: water use in millions of gallons per day, ice loss in cubic kilometers per year, a hiring rate in employees per month. Differentiating one of those does not cancel anything. It stacks another unit underneath.

If $W(t)$ is a city's water use in millions of gallons per day, and $t$ is in years, then

$$W'(t) \text{ is in millions of gallons per day, per year.}$$

Read out loud: how fast the daily usage is climbing, year over year. Dropping either half of that unit produces a claim that is off by a factor of 365 or is simply meaningless.

A related quantity worth naming is the relative rate of change, $\dfrac{f'(t)}{f(t)}$, which reports growth as a fraction of current size and is what a percentage growth rate means. For $P(t) = P_0 e^{kt}$ the relative rate is exactly the constant $k$, which is why exponential models get quoted as percentages. The relative rate and the rate itself are different numbers with different units, and 3.1 percent per year is not 3.1 people per year.

Topic 4.1's four slots carry over verbatim, and the sign is the slot that goes wrong most often out here.

1. When, with its units.
2. What quantity, in words.
3. Which direction, read off the sign and stated as increasing or decreasing.
4. How fast, with the full compound unit.

So for a melting sculpture with $V'(20) = -14$, where $V$ is in cubic centimeters and $t$ in minutes: twenty minutes in, the volume of the sculpture is decreasing at 14 cubic centimeters per minute.

Two ways that sentence gets damaged. Saying the volume is "changing at 14 cubic centimeters per minute" throws away the direction the minus sign was carrying. Saying it is "increasing at 14" reverses it. Both are one word away from correct and both are fully wrong.