Solve problems involving related and general rates of change, including integrating a given rate dQ/dt to recover a quantity with an initial condition, and interpreting rate graphs

What this dot point is asking

NESA wants you to treat a rate of change as a derivative and to move freely in both directions between a quantity and its rate. Going forwards, differentiate a quantity $Q(t)$ to get its rate $\frac{dQ}{dt}$, and use the chain rule when two changing quantities are linked (related rates). Going backwards, integrate a given rate $\frac{dQ}{dt} = g(t)$, using an initial condition to fix the constant, to recover the quantity itself. You must also be able to read a rate graph: describe behaviour as "increasing at a decreasing rate", locate where the quantity is greatest or least from the sign of the rate, and read net change as the signed area under the rate graph.

The answer

A rate of change is a derivative

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.

Read the units off the derivative: a volume in litres changing per minute has rate in L/min, a population per year has rate in (people)/year. Carrying units is not decoration, it is how you check that you differentiated or integrated the right way.

Related rates: the chain rule links two changing quantities

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

For example, a spherical weather balloon is inflated so that its volume grows at $\frac{dV}{dt} = 50$ cm$^3$/s. Since $V = \frac{4}{3}\pi r^3$, we have $\frac{dV}{dr} = 4\pi r^2$, and the chain rule gives $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. When $r = 5$ cm,

The radius is growing slowly even though the volume is pouring in, because the same volume is spread over an ever larger surface. The disciplined habit is to write down the rate you are given and the rate you want, differentiate the link with respect to $t$, and only substitute the instantaneous values at the very end. Substituting numbers too early freezes a variable that is actually changing and is the classic way to lose the method marks.

Integrating a given rate to recover the quantity

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.

and a single known value of $Q$ (often $Q$ at $t = 0$) pins down $C$. The recipe is always the same three steps: integrate, write $+ C$, then use the initial condition. This is the workhorse of "water flows into / out of a tank", "a dam fills", and "a population grows" questions.

A subtlety worth naming: when the rate can be negative (a tank draining), the quantity decreases while the rate is negative, and a question often asks "when does the flow stop?" That is when $\frac{dQ}{dt} = 0$, which is a turning point of $Q$, not when $Q = 0$. Treat "when does it stop flowing" and "when is the tank empty" as two different equations, the same way motion questions separate "at rest" ($v = 0$) from "at the origin" ($x = 0$).

Reading a rate graph

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:

• Where the rate is positive, $Q$ is increasing; where it is negative, $Q$ is decreasing.
• $Q$ has a turning point where the rate is zero and changes sign. Rate going $+$ to $-$ means $Q$ has a local maximum; rate going $-$ to $+$ means a local minimum.
• $Q$ changes fastest where the rate is largest in magnitude (the highest or lowest point of the rate graph), which is a point of inflection on the $Q$ curve.
• The net change in $Q$ over an interval is the signed area between the rate graph and the $t$-axis, because $\displaystyle\int_a^b \frac{dQ}{dt}\,dt = Q(b) - Q(a)$. Area above the axis adds to $Q$; area below subtracts.

Below is a worked rate graph for a wetland bird colony whose growth rate $\frac{dP}{dt}$ rises, peaks, then turns negative during a dry spell. Read the colony's story straight off it.

The rate is zero at $t = 3$ and $t = 9$. At $t = 3$ the rate changes from negative to positive, so the population stops falling and starts rising: that is the minimum. At $t = 9$ the rate changes from positive to negative, so the population stops rising and starts falling: that is the maximum. The rate peaks at $t = 6$, so the colony grows fastest there. None of this needs the equation of $P$.

Increasing or decreasing, at an increasing or decreasing rate

A favourite "describe the graph" task uses four phrases that combine the direction of change (sign of $\frac{dQ}{dt}$) with the change in the rate itself (sign of $\frac{d^2Q}{dt^2}$, i.e. concavity). They confuse students because the words "increasing" and "decreasing" appear twice meaning different things. Keep the two ideas separate: the first word is about $Q$ going up or down; the phrase "at an increasing/decreasing rate" is about whether the steepness is growing or easing, which is concavity.

In symbols: "increasing at an increasing rate" is $\frac{dQ}{dt} > 0$ and concave up; "increasing at a decreasing rate" is $\frac{dQ}{dt} > 0$ but concave down (the rise is flattening, as a tank fills and the level creeps up ever more slowly). For a decreasing quantity the language tracks the size of the rate $\left|\frac{dQ}{dt}\right|$: "decreasing at an increasing rate" means it is falling and the fall is getting steeper (concave down), while "decreasing at a decreasing rate" means it is falling but the fall is easing off (concave up), like a hot drink cooling quickly at first and then more gently.

How exam questions ask about general rates of change

The wording maps straight onto "differentiate or integrate, and which graph feature":

• "Find the rate at which ... is changing." Differentiate the quantity, or read the value off the given rate function; give units (quantity per time).
• "At what rate is ... changing?" with two linked quantities. A related rates question: differentiate the link with respect to $t$ and substitute the instantaneous values last.
• "The rate is $\frac{dQ}{dt} = g(t)$. Find $Q$." Integrate, write $+ C$, and use the initial/boundary condition to evaluate $C$.
• "When does the water stop flowing / the population stop changing?" Solve $\frac{dQ}{dt} = 0$. This is a turning point of $Q$, not $Q = 0$.
• "How much flows out / in during ...?" Either evaluate $Q$ at the two times and subtract, or compute the signed area $\int_a^b \frac{dQ}{dt}\,dt$. They are the same number.
• "From the rate graph, when is $Q$ greatest / least?" Find where the rate changes sign: $+$ to $-$ is a maximum of $Q$, $-$ to $+$ is a minimum.
• "When is $Q$ changing fastest?" Find the largest magnitude of the rate (the peak or trough of the rate graph), an inflection on the $Q$ curve.
• "Sketch a possible graph of $Q$." Match $Q$'s gradients to the rate graph: rising where the rate is positive, turning where it is zero, and never letting a physical quantity (a population, a volume) go below zero.
• "Describe the behaviour ..." Use the four phrases: state direction (sign of the rate) and whether the rate is growing or easing (concavity).

The recurring decision is always which direction of calculus: you differentiate to get a rate, and integrate (with an initial condition) to get the quantity back.

Stage by stage: from a rate graph to the quantity graph

When a question gives a rate graph and asks you to sketch the quantity, the safest route is to read off the four pieces of information in order and build the $Q$ curve from them. Below the method is built up for the bird-colony rate graph above, with the colony starting at $P(0) = 200$ birds.

Stage 1, mark where the rate is zero. Read the $t$-axis crossings of the rate graph: $\frac{dP}{dt} = 0$ at $t = 3$ and $t = 9$. These are the only places the population can turn, so they will be the turning points of the $P$ curve. Mark vertical guide lines there.

Stage 2, read the sign of the rate on each interval and turn it into rising or falling. The rate is negative on $0 \le t < 3$, positive on $3 < t < 9$, and negative on $9 < t \le 12$. So $P$ is falling, then rising, then falling. Combined with stage 1, $t = 3$ (rate $-$ to $+$) is a minimum and $t = 9$ (rate $+$ to $-$) is a maximum.

Stage 3, use the size of the rate for the steepness. The $P$ curve is steepest where the rate is largest in size. The rate peaks at $t = 6$ (the fastest rise) and is most negative at the ends $t = 0$ and $t = 12$ (the steepest falls). At the turning points $t = 3$ and $t = 9$ the curve is momentarily flat. This tells you how to shape the curve between the turning points: shallow near the turns, steepest at $t = 6$.

Stage 4, draw the quantity curve through the start value. Anchor the curve at $P(0) = 200$ and join the features: fall to a minimum at $t = 3$, rise to a maximum at $t = 9$, then fall again, staying above zero. The vertical scale is set by the net changes, which are the signed areas under the rate graph; here the rise from the minimum to the maximum, $\int_3^9 \frac{dP}{dt}\,dt = 36$, lifts $P$ from $164$ back to $200$.

Notice how the quantity curve mirrors the rate graph: every zero of the rate is a turning point of $P$, the rate's peak at $t = 6$ is where $P$ rises most steeply, and the area under the positive hump is exactly the height that $P$ climbs from its minimum to its maximum.

The recovered-quantity curve, from a formula

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".