Take the Particle Away and the Sign Still Carries the Direction

You'll learnto write an outflow, a cooling or a depletion with its minus sign already inside the equation, to build a net rate as in minus out rather than in plus out, and to keep both halves of a unit when the quantity you are differentiating is itself a rate.

Topic 4.2 gave the derivative a particle to follow. Take the particle away and everything else survives: tanks fill and drain, coffee cools, cultures grow, costs climb. What changes is that the sign of a rate is no longer a direction on a line — it is a claim about whether the quantity is going up or going down. A tank can be losing water while water pours into it the whole time, because the derivative of the amount is the rate in minus the rate out, and only the comparison decides. Set that sign wrong in the first line and every line after it inherits the error, with no natural moment to notice.

8 STEPS · 6 QUICK CHECKS · AN OUTFLOW IS NEGATIVE FROM THE START · NET = IN − OUT · A RATE OF A RATE STACKS A UNIT · v1

R(t) = 20 + 4t in · S(t) = 3t² out ⇒ dA/dt = R − S, zero at t = 10/3

Before you start

What you're looking at

A tank, for 0 ≤ t ≤ 5 minutes. Water enters at R(t) = 20 + 4t gallons per minute and drains at S(t) = 3t² gallons per minute. The two bars beside the tank are those rates drawn to scale, and the dashed box around their overhang is the net rate — the derivative of the amount in the tank.

The question

The inflow rate climbs for the entire five minutes. So why does the tank start losing water partway through, and when exactly?

Watch for

The moment the out bar grows past the in bar. Nothing about the inflow changes there, and yet the water level turns around — and the strip for the amount itself never once leaves positive while the strip for its derivative flips.

Step 1 / 8