Connected rates of change

Connected rates of change describe how two or more quantities change together. For example, inflating a spherical balloon connects its radius, surface area and volume. If the volume changes with time, the radius must also change with time.

The key is to write an equation connecting the quantities before differentiating with respect to time.

Prerequisites

You should be comfortable with:

• the product, quotient and chain rules
• implicit differentiation
• rearranging formulae and substituting with units
• the area and volume formulae for common shapes

The central idea

Suppose a quantity $y$ depends on $x$, while $x$ depends on time $t$. Then $y$ also depends on $t$:

$$y=f(x), \qquad x=g(t).$$

The chain rule gives

$$\boxed{\frac{\mathrm dy}{\mathrm dt}=\frac{\mathrm dy}{\mathrm dx}\frac{\mathrm dx}{\mathrm dt}}.$$

This formula connects three rates:

• $\dfrac{\mathrm dx}{\mathrm dt}$ tells us how quickly $x$ changes with time
• $\dfrac{\mathrm dy}{\mathrm dx}$ tells us how sensitive $y$ is to a change in $x$
• $\dfrac{\mathrm dy}{\mathrm dt}$ tells us how quickly $y$ changes with time

For several variables, differentiate the whole relationship with respect to $t$. For example, if

$$x^2+y^2=25,$$

then

$$2x\frac{\mathrm dx}{\mathrm dt}+2y\frac{\mathrm dy}{\mathrm dt}=0.$$

Both $x$ and $y$ vary with time, so both produce a time derivative.

A reliable method

Use this sequence for almost every connected rates problem.

1. Define the changing quantities. Include their units.
2. Draw and label a diagram if geometry is involved.
3. Write one equation connecting the variables. Use geometry, volume, Pythagoras or another model.
4. Eliminate unwanted variables if a fixed geometric relationship allows it.
5. Differentiate with respect to time. Do this before substituting the instant values.
6. Substitute rates and measurements, keeping their signs.
7. Solve and state the answer with units and direction.
8. Check the sign, size and dimensions against the situation.

Do not substitute too early

If the radius is $6$ cm at one instant, do not replace $r$ by $6$ before differentiating. The radius is still changing. Substituting first turns a variable into a constant and destroys the rate information.

Example 1: area of an expanding circle

The radius of a circular ripple increases at $0.4\text{ cm s}^{-1}$. Find the rate at which its area increases when the radius is $6$ cm.

The area and radius are connected by

$$A=\pi r^2.$$

Differentiate with respect to $t$:

$$\frac{\mathrm dA}{\mathrm dt} =2\pi r\frac{\mathrm dr}{\mathrm dt}.$$

At the required instant,

$$r=6, \qquad \frac{\mathrm dr}{\mathrm dt}=0.4.$$

Therefore

$$\frac{\mathrm dA}{\mathrm dt} =2\pi(6)(0.4) =4.8\pi.$$

So the area increases at

$$\boxed{4.8\pi\text{ cm}^2\text{ s}^{-1}}$$

or approximately $15.1\text{ cm}^2\text{ s}^{-1}$.

Notice that a constant radial speed does not produce a constant rate of area growth. Since

$$\frac{\mathrm dA}{\mathrm dt}=2\pi r\frac{\mathrm dr}{\mathrm dt},$$

the same radial speed creates faster area growth when the circle is larger. The factor $2\pi r$ is the circumference, so this result also has a geometric interpretation: the new area is approximately circumference multiplied by a very small increase in radius.

Example 2: finding the radial speed of a sphere

Air is pumped into a spherical balloon at $80\text{ cm}^3\text{ s}^{-1}$. Find the rate at which its radius increases when $r=2$ cm.

For a sphere,

$$V=\frac{4}{3}\pi r^3.$$

Differentiate with respect to time:

$$\frac{\mathrm dV}{\mathrm dt} =4\pi r^2\frac{\mathrm dr}{\mathrm dt}.$$

Substitute $\dfrac{\mathrm dV}{\mathrm dt}=80$ and $r=2$:

$$80=4\pi(2)^2\frac{\mathrm dr}{\mathrm dt}.$$

Hence

$$\frac{\mathrm dr}{\mathrm dt} =\frac{80}{16\pi} =\boxed{\frac{5}{\pi}\text{ cm s}^{-1}}.$$

The units confirm the algebra. Dividing a volume rate, measured in $\text{cm}^3\text{ s}^{-1}$, by an area, measured in $\text{cm}^2$, leaves $\text{cm s}^{-1}$.

There is also a useful general identity here:

$$\frac{\mathrm dV}{\mathrm dt} =\underbrace{4\pi r^2}_{\text{surface area}} \frac{\mathrm dr}{\mathrm dt}.$$

For a fixed volume inflow, $\dfrac{\mathrm dr}{\mathrm dt}$ becomes smaller as the balloon grows because each further increase in radius requires a thicker shell of volume spread over a larger surface.

Example 3: water entering a conical tank

An inverted conical tank has height $6$ m and top radius $3$ m. Water enters at a constant rate of $3\pi\text{ m}^3\text{ min}^{-1}$. Find the rate at which the water depth increases when the water is $2$ m deep.

Let $h$ be the water depth and $r$ the radius of its surface. The water forms a smaller cone similar to the tank. Similar triangles give

$$\frac{r}{h}=\frac{3}{6}=\frac12,$$

so

$$r=\frac h2.$$

The cone volume formula contains two changing lengths:

$$V=\frac13\pi r^2h.$$

Use similarity to eliminate $r$ before differentiating:

$$V =\frac13\pi\left(\frac h2\right)^2h =\frac{\pi h^3}{12}.$$

Now differentiate with respect to $t$:

$$\frac{\mathrm dV}{\mathrm dt} =\frac{\pi h^2}{4}\frac{\mathrm dh}{\mathrm dt}.$$

At $h=2$ and $\dfrac{\mathrm dV}{\mathrm dt}=3\pi$,

$$3\pi =\frac{\pi(2)^2}{4}\frac{\mathrm dh}{\mathrm dt} =\pi\frac{\mathrm dh}{\mathrm dt}.$$

Therefore

$$\boxed{\frac{\mathrm dh}{\mathrm dt}=3\text{ m min}^{-1}}.$$

Why similarity matters

The original volume equation contains $V$, $r$ and $h$. The question supplies a volume rate and asks for a height rate, but supplies no radius rate. Similarity reduces the model to an equation containing only $V$ and $h$.

An alternative is to differentiate $V=\frac13\pi r^2h$ using the product rule, but this introduces $\dfrac{\mathrm dr}{\mathrm dt}$. You would then also need to differentiate $r=\frac h2$. Eliminating $r$ first is shorter and less error prone.

Example 4: a sliding ladder and a negative rate

A ladder of length $5$ m leans against a vertical wall. Its foot slides away from the wall at $0.6\text{ m s}^{-1}$. How quickly is the top moving when the foot is $3$ m from the wall?

Let $x$ be the distance from the wall to the foot and $y$ the height of the top. The fixed ladder length gives

$$x^2+y^2=25.$$

When $x=3$,

$$y=\sqrt{25-9}=4.$$

Differentiate the relationship with respect to time:

$$2x\frac{\mathrm dx}{\mathrm dt} +2y\frac{\mathrm dy}{\mathrm dt}=0.$$

Substitute $x=3$, $y=4$ and $\dfrac{\mathrm dx}{\mathrm dt}=0.6$:

$$2(3)(0.6)+2(4)\frac{\mathrm dy}{\mathrm dt}=0.$$

Thus

$$\frac{\mathrm dy}{\mathrm dt} =-\frac{3.6}{8} =\boxed{-0.45\text{ m s}^{-1}}.$$

The negative sign is part of the answer. It means the height is decreasing. In words, the top of the ladder moves downwards at $0.45\text{ m s}^{-1}$.

Understanding signs

A rate includes direction:

$$\frac{\mathrm dx}{\mathrm dt}>0 \quad\Longrightarrow\quad x\text{ is increasing},$$ $$\frac{\mathrm dx}{\mathrm dt}<0 \quad\Longrightarrow\quad x\text{ is decreasing}.$$

Translate words into signed rates before doing any algebra.

Wording	Mathematical sign
increasing, expanding, filling, moving away	positive
decreasing, shrinking, draining, moving towards	negative

If a cylindrical tank loses water at $12\text{ cm}^3\text{ s}^{-1}$, then

$$\frac{\mathrm dV}{\mathrm dt}=-12,$$

not $12$. If the question later asks for the speed at which the level falls, report the positive magnitude after interpreting the negative derivative.

Example 5: a draining cylinder

A vertical cylindrical tank has constant radius $5$ cm. Water drains at $20\pi\text{ cm}^3\text{ s}^{-1}$. Find the rate of change of the water depth.

Let $h$ be the depth. Then

$$V=\pi r^2h=25\pi h.$$

Since the tank is draining,

$$\frac{\mathrm dV}{\mathrm dt}=-20\pi.$$

Differentiate:

$$\frac{\mathrm dV}{\mathrm dt}=25\pi\frac{\mathrm dh}{\mathrm dt}.$$

Therefore

$$-20\pi=25\pi\frac{\mathrm dh}{\mathrm dt},$$

so

$$\boxed{\frac{\mathrm dh}{\mathrm dt}=-0.8\text{ cm s}^{-1}}.$$

The depth falls at $0.8\text{ cm s}^{-1}$. Because the cylinder has constant cross-sectional area, a constant volume outflow produces a constant rate of fall. This differs from the conical tank, whose cross-sectional area changes with depth.

Units are part of the reasoning

Rates use compound units. Match the quantity being differentiated:

Rate	Typical units
$\dfrac{\mathrm dr}{\mathrm dt}$	$\text{cm s}^{-1}$
$\dfrac{\mathrm dA}{\mathrm dt}$	$\text{cm}^2\text{ s}^{-1}$
$\dfrac{\mathrm dV}{\mathrm dt}$	$\text{cm}^3\text{ s}^{-1}$
$\dfrac{\mathrm d\theta}{\mathrm dt}$	$\text{rad s}^{-1}$

Convert all measurements to a consistent system before substitution. For example,

$$2\text{ litres min}^{-1} =2000\text{ cm}^3\text{ min}^{-1},$$

because $1$ litre is $1000\text{ cm}^3$. Do not mix metres with cubic centimetres or seconds with minutes in the same calculation.

Modelling assumptions

A connected rates calculation is only as good as the model connecting its variables. Typical assumptions include:

• a balloon remains perfectly spherical
• a tank has the stated geometric shape
• liquid enters or leaves at the quoted instantaneous rate
• the liquid surface remains horizontal
• a ladder remains in contact with both the floor and wall
• the floor and wall are perpendicular
• lengths, areas and volumes vary smoothly, so derivatives exist
• material thickness, splashing, evaporation and deformation are negligible

These assumptions do not make the mathematics weak. They identify exactly which real features have been simplified. In an exam, state an assumption when asked to criticise or refine a model.

Common mistakes

Treating a changing quantity as constant

In

$$A=\pi r^2,$$

$\pi$ is constant but $r$ changes. Therefore

$$\frac{\mathrm dA}{\mathrm dt} =2\pi r\frac{\mathrm dr}{\mathrm dt},$$

not merely $2\pi r$.

Differentiating with respect to the wrong variable

The derivative $\dfrac{\mathrm dA}{\mathrm dr}=2\pi r$ is not an area rate in time. It measures area change per unit radius. Multiply by $\dfrac{\mathrm dr}{\mathrm dt}$ to obtain $\dfrac{\mathrm dA}{\mathrm dt}$.

Missing a derivative term

If both $r$ and $h$ vary, then differentiating

$$V=\frac13\pi r^2h$$

directly requires the product rule:

$$\frac{\mathrm dV}{\mathrm dt} =\frac13\pi\left(2rh\frac{\mathrm dr}{\mathrm dt} +r^2\frac{\mathrm dh}{\mathrm dt}\right).$$

Writing only one term silently treats the other variable as constant.

Ignoring a negative rate

Draining, shrinking and falling normally correspond to negative derivatives. Decide the sign from the words, not from the answer you hope to obtain.

Confusing a rate with a percentage rate

A radius increasing at $5\%$ per second means

$$\frac{1}{r}\frac{\mathrm dr}{\mathrm dt}=0.05,$$

so $\dfrac{\mathrm dr}{\mathrm dt}=0.05r$. It does not mean $\dfrac{\mathrm dr}{\mathrm dt}=0.05$ unless the units and context say so.

Quick checks

Check 1

The side length $x$ of a square increases at $3\text{ cm s}^{-1}$. Find the rate of increase of its area when $x=4$ cm.

Answer

Since $A=x^2$,

$$\frac{\mathrm dA}{\mathrm dt}=2x\frac{\mathrm dx}{\mathrm dt} =2(4)(3)=24.$$

The area increases at $\boxed{24\text{ cm}^2\text{ s}^{-1}}$.

Check 2

The radius of a sphere decreases at $0.2\text{ m min}^{-1}$. Find $\dfrac{\mathrm dV}{\mathrm dt}$ when $r=3$ m.

Answer

Use $\dfrac{\mathrm dr}{\mathrm dt}=-0.2$:

$$\frac{\mathrm dV}{\mathrm dt} =4\pi r^2\frac{\mathrm dr}{\mathrm dt} =4\pi(3)^2(-0.2) =-7.2\pi.$$

Thus the volume decreases at $\boxed{7.2\pi\text{ m}^3\text{ min}^{-1}}$, or $\dfrac{\mathrm dV}{\mathrm dt}=-7.2\pi\text{ m}^3\text{ min}^{-1}$.

Check 3

Oil fills a vertical cylinder of radius $2$ m at $5\pi\text{ m}^3\text{ min}^{-1}$. How quickly does its depth rise?

Answer

Since $V=\pi r^2h=4\pi h$,

$$5\pi=4\pi\frac{\mathrm dh}{\mathrm dt}.$$

Therefore $\boxed{\dfrac{\mathrm dh}{\mathrm dt}=\dfrac54\text{ m min}^{-1}}$.

Exam strategy

Most marks are visible in the setup. Even if the final arithmetic fails, write:

1. the correct connecting equation
2. its derivative with respect to time
3. the known rates with signs and units
4. the relevant measurements at the stated instant

Always distinguish an equation true for the whole motion from values true only at one instant. For the ladder,

$$x^2+y^2=25$$

is always true, while $x=3$ and $y=4$ are true only at the instant in the question.

What to learn next

• Use rates as velocities and accelerations in calculus in kinematics.
• Apply derivatives to relationships defined indirectly in implicit differentiation.
• Study how rates appear in differential equation models.
• Explore the geometric relationship between radius and circumference in the interactive circle model.