Application of Derivatives · Lesson 1 of 7

Rates of Change in Real Situations

“Translate changing quantities into derivatives and interpret related rates with correct signs and units.”

Learning Objectives

• Distinguish a rate with respect to a variable from a rate with respect to time. • Use chain and product rules to connect changing geometric quantities. • Use geometric constraints and similar triangles to relate variables. • Interpret positive, negative, and zero rates with appropriate units. • Calculate and interpret marginal cost and marginal revenue.

A Derivative Tells Us How Fast Something Changes

A quantity can change because time passes, because a length changes, or because another quantity changes. For a circle, its area changes as its radius changes. For a business, total cost changes as output changes. A derivative describes the instantaneous change in one quantity compared with a small change in the quantity we choose as the input.

The phrase “with respect to” tells you the input variable. The rate of area with respect to radius and the rate of area with respect to time are different questions. Write the variables, their meanings, and their units before differentiating; this prevents many errors later.

Definition

Instantaneous rate of change

If $y = f (x)$ , then $f^{'} (x) = d y / d x$ is the instantaneous rate of change of $y$ with respect to $x$ . At $x = x_{0}$ , the rate is $f^{'} (x_{0})$ . Its units are the units of $y$ divided by the units of $x$ .

Rate at a particular inputLaTeX

\frac{d y}{d x}_{x = x_{0}} = f^{'} (x_{0})

First differentiate the relationship; then substitute the specified input.

For example, if distance $s$ is measured in metres and time $t$ in seconds, $d s / d t$ is measured in metres per second. If area $A$ is in square centimetres and radius $r$ in centimetres, $d A / d r$ is in square centimetres per centimetre. It is not an area-per-second rate unless time is actually the denominator.

Example — Area compared with radius

Problem
For a circle, find the rate of change of area with respect to radius when $r = 5$ cm.

1.The relationship is $A = π r^{2}$ , where $A$ is area and $r$ is radius.
2.Differentiate with respect to $r$ : $d A / d r = 2 π r$ .
3.At $r = 5$ , the rate is $10 π$ cm² per cm. A very small radius increase of $Δ r$ cm produces an area increase close to $10 π Δ r$ cm² near this radius.
4.No time rate was supplied, so we cannot report $d A / d t$ from this information alone.

Read the denominator

A derivative is a rate with respect to its denominator variable. Do not attach “per second” automatically. Also, a positive derivative means the output increases as the chosen input increases; a negative derivative means it decreases.

Linking Rates with the Chain Rule

Often the quantity we want depends on a length, while that length changes with time. The chain rule connects these two changes. Think of the changing length as the link between the geometric formula and the clock.

A related rate through one intermediate variableLaTeX

\frac{d y}{d t} = \frac{d y}{d x} \frac{d x}{d t}

Here

y

depends on

x

, and

x

depends on time

t

. Both derivatives on the right must be evaluated for the same instant.

Comparing two quantities that depend on timeLaTeX

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}, \frac{d x}{d t} \neq = 0

This form compares the two time rates where

x

can locally serve as the varying input.

The useful working order is to name the variables, write a relationship valid throughout the motion, differentiate that relationship, and only then substitute the values at the instant in question. Substituting a temporary radius or height too early can turn a changing quantity into a constant and destroy the relationship you need.

Example — Surface area of an expanding cube

Problem
A cube has $d V / d t = 9$ cm³/s. Find its surface-area rate when its edge is 10 cm.

1.Let $x (t)$ be the edge, $V = x^{3}$ , and $S = 6 x^{2}$ . Both $V$ and $S$ depend on the same changing edge.
2.Differentiating volume gives $d V / d t = 3 x^{2} (d x / d t)$ . Thus $d x / d t = 9/ (3 x^{2}) = 3/ x^{2}$ .
3.Differentiating surface area gives $d S / d t = 12 x (d x / d t) = 12 x (3/ x^{2}) = 36/ x$ .
4.At $x = 10$ , $d S / d t = 3.6$ cm²/s. The positive sign means the surface area is increasing.

Example — Circular ripples

Problem
A circular ripple expands at 4 cm/s. How fast does its enclosed area increase when its radius is 10 cm?

1.Write $A = π r^{2}$ and identify the given rate as $d r / d t = 4$ cm/s.
2.Differentiate with respect to time: $d A / d t = 2 π r (d r / d t)$ .
3.At $r = 10$ , $d A / d t = 2 π (10) (4) = 80 π$ cm²/s.
4.The radius rate is constant, but the area rate grows with $r$ : a larger circle has a longer boundary along which new area is added.

When Several Dimensions Change Together

A shape can change in more than one dimension. In that case, differentiate the full formula, allowing every changing dimension to vary. Product and chain rules are still the tools; the difference is that more than one rate may contribute to the final answer.

Example — A rectangle that changes shape

Problem
A rectangle has $d x / d t = - 3$ cm/min and $d y / d t = 2$ cm/min. At $x = 10$ cm and $y = 6$ cm, find the perimeter and area rates.

1.Since $P = 2 (x + y)$ , $d P / d t = 2 (d x / d t + d y / d t) = 2 (- 3 + 2) = - 2$ cm/min.
2.Since $A = x y$ , the product rule gives $d A / d t = y (d x / d t) + x (d y / d t)$ .
3.At this instant, $d A / d t = 6 (- 3) + 10 (2) = - 18 + 20 = 2$ cm²/min.
4.The perimeter decreases while the area increases. These are different quantities, so their rates need not have the same sign.

Example — A ladder sliding down a wall

Problem
A 5 m ladder has its foot moving away from a wall at 2 cm/s. Find the height rate when the foot is 4 m from the wall.

1.Let $x$ be the foot distance and $y$ the height, both in metres. The given horizontal rate is $d x / d t = 0.02$ m/s.
2.The ladder length is fixed, so $x^{2} + y^{2} = 25$ . At $x = 4$ , $y = 3$ .
3.Differentiate before substituting: $2 x (d x / d t) + 2 y (d y / d t) = 0$ . Hence $d y / d t = - (x / y) (d x / d t)$ .
4.Thus $d y / d t = - (4/3) (0.02) = - 2/75$ m/s, or $- 8/3$ cm/s. The negative sign describes downward motion.

Geometry Can Remove an Extra Variable

Related-rate problems sometimes seem to contain more unknown dimensions than available rates. A fixed shape, fixed angle, or similar triangles can supply the missing relationship. Use that relationship to express the required quantity in one changing variable before differentiating.

For a sphere, volume is $V = 4 π r^{3} /3$ , so $d V / d r = 4 π r^{2}$ and $d V / d t = 4 π r^{2} (d r / d t)$ . The first compares volume with radius; the second includes the radius time rate. If a problem supplies diameter $D$ instead, use $r = D /2$ before differentiating. For a cone, use $V = π r^{2} h /3$ and account for whichever dimensions vary.

Example — A sphere specified by its diameter

Problem
A spherical balloon has diameter $D (x) = 3 (2 x + 1) /2$ . Find $d V / d x$ wherever this diameter is positive.

1.The radius is $r (x) = 3 (2 x + 1) /4$ , and $d r / d x = 3/2$ . The positive-diameter domain is $x > - 1/2$ .
2.Using $V = 4 π r^{3} /3$ , the chain rule gives $d V / d x = 4 π r^{2} (d r / d x)$ .
3.Substitute the radius and its derivative: $d V / d x = 4 π [9 (2 x + 1)^{2} /16] (3/2) = 27 π (2 x + 1)^{2} /8$ .
4.This is a rate with respect to $x$ , not time. Its physical units depend on the units assigned to $x$ and the diameter.

Water rising in a conical tank— The water surface and tank have the same semi-vertical angle, so radius is proportional to depth.

Example — Water rising in an inverted cone

Problem
An inverted conical tank has semi-vertical angle $tan^{- 1} (0.5)$ . Water enters at 5 m³/h. Find the depth rate when $h = 4$ m.

1.The angle gives $r / h = tan α = 0.5$ , so $r = h /2$ . This is true throughout filling, not just at the specified instant.
2.Write $V = π r^{2} h /3 = π h^{3} /12$ . Differentiating gives $d V / d t = (π h^{2} /4) (d h / d t)$ .
3.At $h = 4$ , $5 = 4 π (d h / d t)$ , so $d h / d t = 5/ (4 π)$ m/h.
4.Using $π \approx 22/7$ gives $35/88$ m/h. Keep the exact expression until a numerical approximation is requested.

A moving person and a growing shadow— The small and large right triangles are similar. The whole ground distance is l + s.

Example — The length of a shadow

Problem
A 2 m tall person walks away from a 6 m lamp at 5 km/h. How fast does the shadow length increase?

1.Let $l$ be the distance from the lamp to the person and $s$ the shadow length. Similar triangles give $s / (l + s) = 2/6$ .
2.Thus $3 s = l + s$ , so $l = 2 s$ . Differentiating gives $d l / d t = 2 (d s / d t)$ .
3.Therefore $d s / d t = 5/2$ km/h. Converting both distances to kilometres does not change the ratio of the fixed heights.
4.This is the growth rate of the shadow length. The shadow tip is at $l + s$ , so its ground speed is $5 + 5/2 = 15/2$ km/h; these two questions must be distinguished.

Motion and the Meaning of Stopping

For motion along a line, differentiating the position or the stated distance function gives its time rate. A zero rate is an instant of rest. To select the relevant stopping time, use the physical interval and the information about the start of the motion.

Example — A car that starts and stops

Problem
A car travels $x = t^{2} (2 - t /3)$ metres from its start until it next stops. Find that stopping time and the distance travelled.

1.Differentiate: $v = d x / d t = 4 t - t^{2} = t (4 - t)$ .
2.Setting $v = 0$ gives $t = 0$ and $t = 4$ . The first is the starting instant, so the next stop is after 4 s.
3.For $0 < t < 4$ , $v > 0$ , so the car moves forward throughout this interval. Its distance travelled equals the increase in $x$ .
4.At $t = 4$ , $x = 16 (2 - 4/3) = 32/3$ m. Extending the formula beyond the stated journey would describe a different motion.

Example — Expansion of a heated disc

Problem
A disc has radius increasing at 0.05 cm/s. Find the area rate when the radius is 3.2 cm.

1.Use $A = π r^{2}$ , so $d A / d t = 2 π r (d r / d t)$ .
2.Substitute the current radius, not an earlier radius: $d A / d t = 2 π (3.2) (0.05) = 0.32 π$ cm²/s.
3.This is the instantaneous area rate. Over a short time $Δ t$ , the area increase is approximately $(0.32 π) Δ t$ cm² if the rate has not changed appreciably.
4.Do not confuse the finite change $Δ A$ with the derivative $d A / d t$ , or a radius increase with a radius rate.

Marginal Cost and Marginal Revenue

The input variable in a business model is often output rather than time. A marginal quantity describes the instantaneous rate at which a total changes as output changes. Because output counts items, a continuous differentiable model is being used to estimate behaviour near a production level.

Definition

Marginal cost and marginal revenue

For total cost $C (x)$ and total revenue $R (x)$ at output $x$ , marginal cost is $C^{'} (x)$ and marginal revenue is $R^{'} (x)$ . Their units are currency per unit of output.

Example — Marginal cost at a production level

Problem
A model gives $C (x) = 0.005 x^{3} - 0.02 x^{2} + 30 x + 5000$ rupees. Find marginal cost at $x = 3$ .

1.Differentiate each term: $C^{'} (x) = 0.015 x^{2} - 0.04 x + 30$ . The fixed cost 5000 contributes zero to the derivative.
2.At $x = 3$ , $C^{'} (3) = 0.015 (9) - 0.04 (3) + 30 = 30.015$ rupees per unit.
3.This rounds to 30.02 rupees per unit. It is the instantaneous model rate, not exactly the discrete difference $C (4) - C (3)$ .

Example — Marginal revenue

Problem
Given $R (x) = 3 x^{2} + 36 x + 5$ rupees, find marginal revenue when $x = 5$ .

1.Differentiate with respect to output: $R^{'} (x) = 6 x + 36$ .
2.At $x = 5$ , $R^{'} (5) = 30 + 36 = 66$ rupees per unit.
3.Revenue itself is $R (5)$ ; marginal revenue is its rate of change. A large total and a large marginal value describe different features of the model.

Question	Useful relationship	Interpretation
Rate with respect to radius	$d A / d r = 2 π r$	Area change per unit radius change
Rate with respect to time	$d A / d t = 2 π r (d r / d t)$	Area change per unit time
Coupled dimensions	$d (x y) / d t = x (d y / d t) + y (d x / d t)$	Both dimensions may contribute
Business rate	$M C = C^{'} (x)$ , $M R = R^{'} (x)$	Change in a total per unit output

Check Your Understanding

Use these questions to check both your calculations and your interpretation of the ideas. Read the conditions carefully before choosing an answer.

Quick check

For $A = π r^{2}$ with both lengths measured in centimetres, what are the units of $d A / d r$ ?

Quick check

A sphere has radius 3 cm and $d r / d t = 2$ cm/s. What is $d V / d t$ ?

Quick check

For $A = x y$ , which derivative is correct when both $x$ and $y$ change with time?

Quick check

A fixed-length ladder has its foot moving away from the wall. Before it reaches the ground, what is the sign of the height rate?

Quick check

If $C (x) = 2 x^{2} + 10 x + 100$ , what is marginal cost at $x = 4$ ?

Quick check

In a related-rate problem, when should an instantaneous radius normally be substituted?

Practice Problems

A circle has radius 6 cm. Find $d A / d r$ , and explain why its units differ from those of $d A / d t$ .
The radius of a spherical balloon grows at 0.2 cm/s. Find its volume rate at radius 5 cm.
A cube has edge growing at 3 cm/s. Find its volume rate and surface-area rate when the edge is 10 cm.
A rectangle has $x = 8$ cm, $y = 6$ cm, $d x / d t = - 5$ cm/min, and $d y / d t = 4$ cm/min. Find both rates and interpret their signs.
Sand falls at 12 cm³/s and forms a cone whose height is one-sixth of its radius. Find $d h / d t$ when $h = 4$ cm.
A particle moves on $6 y = x^{3} + 2$ . Find the points where $d y / d t = 8 (d x / d t)$ , assuming $d x / d t \neq = 0$ .
For $C (x) = 0.007 x^{3} - 0.003 x^{2} + 15 x + 4000$ and $R (x) = 13 x^{2} + 26 x + 15$ , calculate $C^{'} (17)$ and $R^{'} (7)$ .
An isosceles triangle has fixed base $b$ . Its equal sides shrink at 3 cm/s. Express its area in terms of one equal side, and find the area rate when that side equals $b$ .

Key Takeaways

• The denominator variable determines what a rate means and what units it has. • Differentiate a relationship before substituting dimensions valid at only one instant. • Chain and product rules connect rates of quantities that change together. • A geometric constraint can reduce several dimensions to one variable. • Marginal cost and revenue are derivatives with respect to output, not time.

Start of chapter

Next · Lesson 2

Increasing and Decreasing Functions

A Derivative Tells Us How Fast Something Changes

Linking Rates with the Chain Rule

When Several Dimensions Change Together

Geometry Can Remove an Extra Variable

Motion and the Meaning of Stopping

Marginal Cost and Marginal Revenue

Check Your Understanding

Quiz

Practice Problems

Key Takeaways