Quick questions on General rates of change: rate as a derivative, related rates, integrating a rate to recover a quantity, and interpreting rate graphs

If a quantity $Q$ depends on time $t$, the instantaneous rate of change of $Q$ is the derivative $\frac{dQ}{dt}$, the gradient of the tangent to the $Q$ against $t$ graph. The sign tells the story: $\frac{dQ}{dt} > 0$ means $Q$ is increasing, $\frac{dQ}{dt} < 0$ means $Q$ is decreasing, and $\frac{dQ}{dt} = 0$ means $Q$ is momentarily stationary. The average rate of change over an interval is the gradient of the chord, $\dfrac{Q(t_2) - Q(t_1)}{t_2 - t_1}$. Unless a question says "average", a rate always means the instantaneous one.

When two quantities are tied by an equation and both change with time, their rates are tied by the chain rule. This is the "related rates" idea developed on the applications of differentiation page; here is the recap you need. If $y$ depends on $x$ and $x$ depends on $t$, then

The genuinely new skill on this page is the reverse of related rates: you are handed the rate as a function of time and asked for the quantity. This is exactly the antiderivative-with-an-initial-condition method from rectilinear motion, where you integrate velocity to recover displacement, applied to any quantity at all.

Many HSC items give you the graph of the rate $\frac{dQ}{dt}$ and ask about the quantity $Q$ itself, with no equation in sight. Everything you need comes from the sign and shape of the rate curve, read like a first-derivative sign analysis:

When you do have a formula for the rate, the recovered quantity is a curve you can plot exactly. The filling reservoir from the first worked example, $V(t) = 600 + 80t - 2t^2$, is the parabola below: it climbs from $600$ kL and levels off at $1400$ kL when the inflow stops at $t = 20$ h, concave down throughout because the inflow is always easing. This is the visual meaning of "increasing at a decreasing rate".