Connected Rates of Change and Small Changes

Link rates over time with the chain rule, find how fast a length, area or volume changes, and estimate small changes with δy ≈ (dy/dx)δx.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 37 real past-paper questions below

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Rates that are connected

You blow air into a balloon. You know how fast the air goes in, in $cm^{3}$ per second. But how fast is the balloon's radius growing? The two rates are connected, and calculus links them.

A rate of change is a derivative with respect to time t. For example, $\frac{d V}{d t}$ is how fast the volume V changes, in $cm^{3}$ per second.

Rule

Increasing means the rate is positive. Decreasing means it is negative: if V falls by 5 $cm^{3}$ each second, $\frac{d V}{d t} = - 5$ .

Formula

\frac{d A}{d t} = \frac{d A}{d r} \times \frac{d r}{d t}

$A$ — the quantity whose rate you want.
$r$ — the quantity that links it to time, for example a radius.
$\frac{d A}{d r}$ — found by differentiating the formula for A.

Rule

This is the chain rule. No formula sheet gives it: learn it.

Memory trick

The links look like fractions that cancel: the $d r$ on the bottom of one meets the $d r$ on the top of the next.

Worked example

1
The radius of a circle grows at 0.5 cm per second. Find how fast the area is growing when the radius is 6 cm.
2
$A = π r^{2}, \frac{d A}{d r} = 2 π r = 12 π$ — Differentiate the area formula, then put r = 6.
3
$\frac{d r}{d t} = 0.5$ — The rate you are given.
4
$\frac{d A}{d t} = 12 π \times 0.5 = 6 π cm^{2} per second$ — Multiply along the chain. Exact, with π left in.

Key idea

Write down the rate you know, the rate you want, and a formula that links the two letters. Then build the chain.

Find the missing link

Often the rate you know and the rate you want are both about a shape, but the formula is about a length. Go through the length: find its rate first, then use it.

Worked example

1
The surface area of a cube is increasing at 3 $cm^{2}$ per second. Find how fast its volume is increasing when each edge is 5 cm.
2
$S = 6 x^{2}, \frac{d S}{d x} = 12 x = 60$ — Surface area of a cube with edge x. Put x = 5.
3
$\frac{d x}{d t} = \frac{d S}{d t} \div \frac{d S}{d x} = \frac{3}{60} = \frac{1}{20}$ — From dS/dt = dS/dx × dx/dt, divide to find the edge's rate.
4
$V = x^{3}, \frac{d V}{d x} = 3 x^{2} = 75$ — Volume of the cube. Put x = 5.
5
$\frac{d V}{d t} = 75 \times \frac{1}{20} = 3.75 cm^{3} per second$ — Multiply along the chain.

Common mistake

Multiplying when you should divide. Check with units: $\frac{cm ^{2} / s}{cm ^{2} / cm}$ gives cm/s, which is the rate of a length.

Mark schemes give a mark for a correct chain written out, even before any numbers. So write the chain in letters first.

Key idea

Write the chain in letters. Put in the numbers you know. Solve for the one you do not.

Containers: one letter first

Water pouring into a cone is a classic. The volume formula has two letters, r and h. Before you differentiate, write V with one letter only, using a fact about the shape.

Rule

0606 and 4037 list the volume of a cone and of a sphere, and the surface areas. The 4PM1 sheet lists the sphere formulas and the curved surface of a cone, but NOT the volume of a cone. 4PM1 students must learn $V = \frac{1}{3} π r^{2} h$ .

Worked example

1
An upside-down cone holds water. When the depth is h cm, the radius of the surface is $\frac{h}{2}$ cm. Water goes in at $4 π cm^{3}$ per second. Find how fast the depth is rising when $h = 6$ .
2
$V = \frac{1}{3} π r^{2} h = \frac{1}{3} π (\frac{h}{2})^{2} h = \frac{π h ^{3}}{12}$ — Replace r with $\frac{h}{2}$ so that V has one letter.
3
$\frac{d V}{d h} = \frac{3 π h ^{2}}{12} = \frac{π h ^{2}}{4} = 9 π$ — Differentiate, then put h = 6.
4
$\frac{d h}{d t} = \frac{d V}{d t} \div \frac{d V}{d h} = \frac{4 π}{9 π}$ — From dV/dt = dV/dh × dh/dt.
5
$\frac{d h}{d t} = \frac{4}{9} cm per second$ — The π cancels. No calculator needed.

Common mistake

Treating r as a fixed number. In a cone, r grows with h. Replace it before you differentiate.

Common mistake

Putting h = 6 into V before differentiating. Differentiate first, then substitute.

Key idea

One letter, then differentiate, then substitute. In that order.

Small changes

Zoom in close to a point on a curve and it looks almost straight, just like its tangent. So for a small step in x, the change in y is almost the tangent's gradient times the step.

Formula

δ y \approx \frac{d y}{d x} δ x

$δ x$ — a small change in x. Negative if x decreases.
$δ y$ — the change in y it causes, approximately.
$\frac{d y}{d x}$ — the gradient at the starting value of x.

Rule

Learn it. 4PM1 expects you to know it, and 0606 and 4037 do not give it. It only works when $δ x$ is small.

Worked example

1
$y = 4 x + \frac{8}{x}$ . Find the approximate change in y when x increases from 4 to $4 + h$ , where h is small.
2
$\frac{d y}{d x} = 2 x^{- \frac{1}{2}} - 8 x^{- 2} = \frac{2}{x} - \frac{8}{x ^{2}}$ — Write as powers of x, then differentiate.
3
$x = 4 : \frac{2}{2} - \frac{8}{16} = \frac{1}{2}$ — The gradient at the start value.
4
$δ y \approx \frac{1}{2} \times h = \frac{1}{2} h$ — Gradient times the small change, δx = h.

Worked example

1
The radius of a circle is 10 cm. It increases by 1%. Estimate the increase in the area.
2
$δ r = 1% of 10 = 0.1$ — Turn the percentage into an actual change in r.
3
$\frac{d A}{d r} = 2 π r = 20 π$ — $A = π r^{2}$ . Put r = 10.
4
$δ A \approx 20 π \times 0.1 = 2 π \approx 6.28 cm^{2}$ — Gradient times the small change.

Common mistake

Using the percentage as the change: $δ r = 1$ from “1%”. First find 1% of the value: here 0.1.

Common mistake

Leaving out $δ x$ . The answer is a change, so it must have the h (or the small number) in it.

Key idea

Gradient at the start, times the small step. That is the whole method.

Estimating a value

The same idea estimates a number you cannot work out easily. Start from a nearby value you know exactly, then add the small change.

Worked example

1
Use calculus to estimate $9.06$ .
2
$y = x, x = 9, δ x = 0.06$ — 9 is close to 9.06 and has an exact root, 3.
3
$\frac{d y}{d x} = \frac{1}{2 x} = \frac{1}{6}$ — Differentiate and put x = 9.
4
$δ y \approx \frac{1}{6} \times 0.06 = 0.01$ — Gradient times the small step.
5
$9.06 \approx 3 + 0.01 = 3.01$ — Start value plus the change. A calculator gives 3.00998..., very close.

Key idea

Pick a start value with an easy exact answer. Then the estimate is start value plus gradient times step.

Which method? Read the question

Look for the words “rate” and “small”. They tell you which tool you need.

You see	You do
“Rate”, “per second”, “how fast”	A derivative with respect to time, like $\frac{d V}{d t}$
“Increasing” or “decreasing”	Positive or negative rate
Know one rate, want another	Chain: $\frac{d A}{d t} = \frac{d A}{d r} \times \frac{d r}{d t}$
A formula with two letters	Use the shape to replace one letter first
“Small”, “approximate change”, from a to a + h	$δ y \approx \frac{d y}{d x} δ x$ , with the gradient at a
“Increases by p%”	Change the percentage into an actual δx first
“Estimate” a value like a root	Nearby exact value, plus gradient times the step

Key idea

Engineers use $δ y \approx \frac{d y}{d x} δ x$ all the time. It tells them how much a small error in measuring a radius or a length will change an area or a volume.

Common mistake

Substituting the numbers before differentiating. Differentiate the formula first, then put in the values.

Key idea

Rates: build a chain. Small changes: gradient times step.

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