MGU Mathematics Series — Calculus

Part II: Derivatives and Differentiation

Implicit Differentiation · Related Rates · Inverse Derivatives

Chapter 6 of 14

MGU Mathematics Series · Calculus · Library Textbook Edition

Chapter 6:
Implicit Differentiation
and Related Rates

Differentiating relationships that are not solved explicitly, finding slopes of implicit curves, and connecting quantities that change together over time

■ Math Level: Calculus ■

■ 14 Sections • Part II of 5 ■

Follows Chapter 5: Transcendental Derivatives · Prepares students for Chapter 7: Graph Analysis with Derivatives

Chapter 6 — Front Matter

Purpose and Overview

Chapter 5 extended differentiation to the full family of functions from precalculus: trigonometric, exponential, and logarithmic. In each case, the function was given explicitly — the output was written as a clear formula in terms of the input. You were given \( f(x) \), and you differentiated it.

But many of the most important mathematical relationships in science, engineering, economics, and geometry are not written that way. A circle, a sphere, a cooling object, a filling tank, an expanding ripple, or an economic equilibrium may be described by an equation relating two or more variables — without solving cleanly for one in terms of another.

This chapter introduces implicit differentiation, a method that extends the chain rule to equations where two variables are related but not explicitly solved. It then introduces related rates, one of the most powerful and practical early applications of derivatives: the analysis of quantities that change together over time.

By the end of this chapter, you should understand how to differentiate relationships that are not in explicit form, how to find slopes and tangent lines on implicit curves, how to connect changing quantities through their rates, and how to read, set up, and solve related rate problems in geometry, motion, science, and engineering.

Opening Question

A ladder 10 feet long leans against a vertical wall. The base of the ladder is being pulled away from the wall at a rate of 2 feet per second. How fast is the top of the ladder sliding down the wall at the moment when the base is 6 feet from the wall?

This is not a differentiation problem in the usual sense — there is no single function \( f(x) \) to differentiate. Instead, there is a geometric relationship between the height of the top of the ladder and the distance of the base from the wall. Both are changing. Calculus lets you find how fast one changes given how fast the other changes. This is a related rates problem, and this chapter builds everything you need to solve it.

14Sections

10Objectives

8Key Terms

15+Worked Examples

Why This Chapter Matters

The ideas in this chapter appear throughout applied mathematics. A satellite's position, velocity, and orbital path are described by equations relating multiple changing variables. The volume, pressure, and temperature of a gas are related by equations that must be differentiated implicitly when any of them changes over time. In economics, the demand for a product depends on its price, and the price depends on production levels — quantities that change together. In machine learning and optimization, gradient computations often involve implicit relationships between parameters, predictions, and loss functions.

Calculus without implicit differentiation is limited to functions written explicitly as \( y = f(x) \). With implicit differentiation, the full power of derivatives extends to relationships described in any form.

Learning Objectives

1Distinguish between explicit and implicit representations of a relationship.
2Apply implicit differentiation to find \( \frac{dy}{dx} \) for equations in \( x \) and \( y \).
3Use the chain rule correctly when differentiating \( y \) as a function of \( x \).
4Find the slope of a curve at a given point using implicit differentiation.
5Write equations of tangent lines to implicitly defined curves.
6Use implicit differentiation to find derivatives of inverse functions.
7Explain what a related rate measures and what information is required to find it.
8Set up and solve related rate problems in geometry, motion, science, and engineering.
9Identify and correct common mistakes in implicit differentiation and related rate problems.
10Interpret rates of change in context using correct units.

Chapter 6 — Contents

Sections at a Glance

6.1Explicit and Implicit RelationshipsWhat explicit means; what implicit means; examples of implicit curves; when solving for \( y \) is impractical
6.2Why Implicit Differentiation Is NeededLimits of explicit methods; equations that cannot be solved for \( y \); the chain rule as the key mechanism
6.3Differentiating Both SidesThe core method; differentiating expressions in \( y \) with respect to \( x \); attaching \( \frac{dy}{dx} \)
6.4Treating \( y \) as a Function of \( x \)Why the chain rule applies to \( y \); implicit function theorem concept; notation \( \frac{dy}{dx} \) and \( y' \)
6.5Finding Slopes of Implicit CurvesFull procedure; circle, folium, trigonometric examples; solving for \( \frac{dy}{dx} \); interpreting the slope formula
6.6Tangent Lines to Implicit CurvesPoint-slope form; evaluating at a specific point; perpendicularity of tangent and radius
6.7Derivatives of Inverse RelationshipsInverse function derivative rule; deriving arcsin, arctan, and the full inverse trig family
6.8What Related Rates MeasureBig idea; notation \( \frac{dx}{dt} \), \( \frac{dV}{dt} \); what is given, what is asked; units of rates
6.9Setting Up Related Rate ProblemsFive-step method; critical warning about substituting before differentiating; expanding circle example
6.10Geometry-Based Related RatesLadder, draining cone, expanding ripple; Pythagorean theorem; similar triangles; volume formulas
6.11Motion-Based Related RatesTwo moving objects; distances changing; angle of approach; observer tracking a moving car
6.12Applied Related Rates in Science and EngineeringBoyle's Law; shadow length; filling tanks; physical interpretation; MGU library connections
6.13Common Implicit and Related Rate MistakesEight recurring errors; missing \( \frac{dy}{dx} \); substituting early; sign errors; product rule omissions
6.14Preparing for Derivative ApplicationsSummary of methods; connection to graph analysis; preview of Chapter 7

Section 6.1

Explicit and Implicit Relationships

What Does Explicit Mean?

When a function is written explicitly, one variable is isolated on one side of the equation and defined as a clear formula in terms of the other. Every function you have differentiated so far in this course has been explicit:

Examples of Explicit Functions

y = x³ − 4x + 1

f(x) = sin(2x) · eˣ

g(x) = (x² + 3) / (x − 1)

In each case, \( y \) or \( f(x) \) is on the left. The right side is a formula involving only \( x \).

What Does Implicit Mean?

A relationship is implicit when both variables appear together in an equation without one being solved in terms of the other. The equation defines a relationship between \( x \) and \( y \), but does not spell out \( y \) as a formula in \( x \).

Implicit Equation	Name / Description	Can be solved explicitly?
\( x^2 + y^2 = 25 \)	Circle, radius 5	Only as two separate halves: \( y = \pm\sqrt{25-x^2} \)
\( x^3 + y^3 = 6xy \)	Folium of Descartes	No clean explicit formula exists
\( \sin(xy) = x + y \)	Transcendental curve	Not possible explicitly
\( x^2 - 3xy + y^2 = 7 \)	Conic section	Possible but messy

Why Not Always Solve Explicitly?

Consider \( x^2 + y^2 = 25 \). Solving gives \( y = \pm\sqrt{25-x^2} \) — two functions, not one. The circle is not a single function. Keeping the equation in implicit form and differentiating implicitly handles the entire curve with one unified calculation.

Graph Instruction

Draw the circle \( x^2 + y^2 = 25 \) centered at the origin with radius 5. Mark the point \( (3, 4) \) on the upper half and \( (3, -4) \) on the lower half. Label both. Note that neither half alone is the whole circle, but both points satisfy the same implicit equation.

Section 6.2

Why Implicit Differentiation Is Needed

Suppose you want the slope of \( x^2 + y^2 = 25 \) at \( (3, 4) \). You could solve for \( y = \sqrt{25-x^2} \) and differentiate to get \( \frac{dy}{dx} = \frac{-x}{\sqrt{25-x^2}} \), then substitute \( x = 3 \) to get \( -3/4 \). This works — for the upper semicircle only. For \( (3,-4) \) you would need a second calculation using the lower semicircle formula.

For equations where \( y \) cannot be solved explicitly at all — like \( x^3 + y^3 = 6xy \) or \( \sin(xy) = x + y \) — there is no choice. Implicit differentiation handles every point on every such curve through one systematic method.

The Chain Rule Is the Key

Implicit differentiation is not a new rule. It is the chain rule, applied carefully to a variable that depends on another. When you differentiate \( y^2 \) with respect to \( x \), you treat \( y \) as a function of \( x \):

\( \frac{d}{dx}[y^2] = 2y \cdot \frac{dy}{dx} \)

This is the central idea. Once you see why this step is necessary, the rest follows naturally.

Section 6.3

Differentiating Both Sides

The Five-Step Method for Implicit Differentiation

1. Start with an equation relating \( x \) and \( y \).

2. Differentiate both sides of the equation with respect to \( x \).

3. Apply the chain rule to every term involving \( y \) — attach \( \frac{dy}{dx} \) as a factor.

4. Collect all terms with \( \frac{dy}{dx} \) on one side.

5. Factor and solve for \( \frac{dy}{dx} \).

Example 6.3.1

Differentiating expressions involving \( y \) with respect to \( x \)

ProblemFind \( \frac{d}{dx} \) of each: (a) \( y^3 \) (b) \( 5y \) (c) \( y^2 - 3y \) (d) \( \sin(y) \)
ThinkEach involves \( y \), which depends on \( x \). The chain rule requires multiplying by the derivative of the inner function \( y \) with respect to \( x \).
Calculate (a) d/dx [y³] = 3y² · dy/dx (b) d/dx [5y] = 5 · dy/dx (c) d/dx [y² − 3y] = 2y · dy/dx − 3 · dy/dx = (2y − 3) dy/dx (d) d/dx [sin(y)] = cos(y) · dy/dx
InterpretEvery expression in \( y \) differentiated with respect to \( x \) picks up a factor of \( \frac{dy}{dx} \). This is the chain rule at work — \( y \) is the inner function, and its derivative with respect to \( x \) is \( \frac{dy}{dx} \).

Section 6.4

Treating \( y \) as a Function of \( x \)

When you differentiate \( y^2 \) with respect to \( x \), you are differentiating the composition \( [y(x)]^2 \). By the chain rule: \( \frac{d}{dx}\left[(y(x))^2\right] = 2 \cdot y(x) \cdot y'(x) = 2y \cdot \frac{dy}{dx} \). You are not assuming \( y \) has an explicit formula. You are only assuming that \( y \) is a function of \( x \) — which is guaranteed by the equation relating them, at least locally near a given point.

This assumption is supported by the implicit function theorem: under mild conditions, an implicit equation \( F(x, y) = 0 \) defines \( y \) as a function of \( x \) near any point where the partial derivatives are not both zero — even if we cannot write an explicit formula.

Notation Reminder

\( \frac{dy}{dx} \) and \( y' \) both mean the same thing: the derivative of \( y \) with respect to \( x \). In implicit differentiation, \( \frac{dy}{dx} \) appears naturally in the calculation and is the preferred form because it reminds you that \( y \) is changing with respect to \( x \).

Section 6.5

Finding Slopes of Implicit Curves

Example 6.5.1

Implicit differentiation — circle

ProblemFind \( \frac{dy}{dx} \) for the circle \( x^2 + y^2 = 25 \).
ThinkDifferentiate both sides with respect to \( x \). The term \( x^2 \) differentiates normally. The term \( y^2 \) requires the chain rule.
Calculate d/dx [x² + y²] = d/dx [25] 2x + 2y · dy/dx = 0 2y · dy/dx = −2x dy/dx = −x / y
CheckAt \( (3, 4) \): slope = \( -3/4 \). At \( (3, -4) \): slope = \( -3/(-4) = 3/4 \). One formula handles both points.
InterpretThe slope of the circle at any point \( (x, y) \) is \( -x/y \). At the top and bottom (\( x = 0 \)), the slope is zero — horizontal tangents. At the far sides (\( y = 0 \)), the slope is undefined — vertical tangents.

Answer: \( \displaystyle\frac{dy}{dx} = \frac{-x}{y} \)

Example 6.5.2

Implicit differentiation — folium of Descartes

ProblemFind \( \frac{dy}{dx} \) for \( x^3 + y^3 = 6xy \).
ThinkThis curve has no clean explicit formula for \( y \). The right side \( 6xy \) is a product of \( x \) and \( y \), so the product rule applies there.
Calculate Left side: d/dx [x³ + y³] = 3x² + 3y² · dy/dx Right side: d/dx [6xy] = 6(x · dy/dx + y · 1) = 6x dy/dx + 6y Setting equal: 3x² + 3y² dy/dx = 6x dy/dx + 6y Collect: 3y² dy/dx − 6x dy/dx = 6y − 3x² Factor: (3y² − 6x) dy/dx = 6y − 3x² Solve: dy/dx = (2y − x²) / (y² − 2x)
CheckThe formula requires \( y^2 \neq 2x \). Where this denominator is zero, the tangent line is vertical.
InterpretEven without an explicit formula for \( y \), implicit differentiation gives a usable slope expression for any point on the curve.

Answer: \( \displaystyle\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x} \)

Example 6.5.3

Implicit differentiation — trigonometric equation

ProblemFind \( \frac{dy}{dx} \) for \( \sin(x + y) = y^2 \cos x \).
ThinkThe left side uses chain rule for \( \sin(x+y) \). The right side requires the product rule on \( y^2 \cos x \).
Calculate Left: cos(x+y) · (1 + dy/dx) Right: 2y dy/dx · cos x + y²(−sin x) Expand left: cos(x+y) + cos(x+y) dy/dx = 2y cos x · dy/dx − y² sin x Collect: [cos(x+y) − 2y cos x] dy/dx = −y² sin x − cos(x+y) dy/dx = (−y² sin x − cos(x+y)) / (cos(x+y) − 2y cos x)
InterpretThe result is complex in appearance but was obtained by systematic application of the chain rule and product rule — no new technique was required.

Answer: \( \displaystyle\frac{dy}{dx} = \frac{-y^2\sin x - \cos(x+y)}{\cos(x+y) - 2y\cos x} \)

Section 6.6

Tangent Lines to Implicit Curves

Once you have \( \frac{dy}{dx} \) from implicit differentiation, finding the equation of a tangent line follows the same three steps as in Chapter 3: find \( \frac{dy}{dx} \) in terms of \( x \) and \( y \), evaluate it at the given point, then apply point-slope form \( y - y_0 = m(x - x_0) \).

Example 6.6.1

Tangent line to a circle at a given point

ProblemFind the equation of the tangent line to \( x^2 + y^2 = 25 \) at the point \( (3, 4) \).
ThinkFrom Example 6.5.1, \( \frac{dy}{dx} = -x/y \). Evaluate at \( (3, 4) \) to get the slope.
Calculate m = −3/4 y − 4 = −(3/4)(x − 3) y = −(3/4)x + 9/4 + 4 = −(3/4)x + 25/4
CheckAt \( x = 3 \): \( y = -9/4 + 25/4 = 16/4 = 4 \). ✓ The radius to \( (3,4) \) has slope \( 4/3 \). Product of slopes: \( (4/3)(-3/4) = -1 \). ✓ Tangent is perpendicular to radius.
InterpretThe tangent to a circle at any point is perpendicular to the radius at that point. Implicit differentiation confirms this geometric fact algebraically.

Answer: \( y = -\frac{3}{4}x + \frac{25}{4} \)

Example 6.6.2

Tangent line to a mixed implicit curve

ProblemFind the slope of the tangent line to \( x^2 y + y^3 = 10 \) at the point \( (1, 2) \).
ThinkThe term \( x^2 y \) requires the product rule because both \( x^2 \) and \( y \) depend on \( x \). The term \( y^3 \) requires the chain rule.
Calculate d/dx [x²y + y³] = d/dx [10] (2xy + x² dy/dx) + 3y² dy/dx = 0 2xy + dy/dx(x² + 3y²) = 0 dy/dx = −2xy / (x² + 3y²) At (1, 2): dy/dx = −2(1)(2) / (1 + 12) = −4/13
CheckVerify \( (1,2) \) is on the curve: \( (1)^2(2) + (2)^3 = 2 + 8 = 10 \). ✓
InterpretAt \( (1, 2) \), the curve is decreasing with slope \( -4/13 \). Tangent line: \( y - 2 = -\frac{4}{13}(x-1) \).

Answer: slope \( = -\frac{4}{13} \)

Section 6.7

Derivatives of Inverse Relationships

Implicit differentiation provides an elegant path to derivatives of inverse functions. If \( y = f^{-1}(x) \), then \( f(y) = x \). Differentiating both sides with respect to \( x \):

Inverse Function Derivative Rule

If \( y = f^{-1}(x) \), differentiate \( f(y) = x \):

f'(y) · dy/dx = 1

dy/dx = 1 / f'(y)

In other words: \( \frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))} \)

Example 6.7.1

Deriving \( \frac{d}{dx}[\arcsin x] \)

ProblemFind \( \frac{d}{dx}[\arcsin x] \) using implicit differentiation.
ThinkLet \( y = \arcsin x \), so \( \sin y = x \). Differentiate both sides with respect to \( x \).
Calculate cos y · dy/dx = 1 dy/dx = 1 / cos y Use sin²y + cos²y = 1: cos y = √(1 − sin²y) = √(1 − x²) dy/dx = 1 / √(1 − x²), −1 < x < 1
CheckThe derivative is always positive: arcsine is increasing. The slope grows larger as \( x \to \pm 1 \), consistent with how the arcsine curve rises sharply near its endpoints.
InterpretImplicit differentiation derives the arcsine formula cleanly, without needing an explicit formula for the inverse function.

Answer: \( \displaystyle\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1 - x^2}} \)

Example 6.7.2

Deriving \( \frac{d}{dx}[\arctan x] \)

ProblemFind \( \frac{d}{dx}[\arctan x] \) using implicit differentiation.
ThinkLet \( y = \arctan x \), so \( \tan y = x \).
Calculate sec²y · dy/dx = 1 dy/dx = 1 / sec²y = cos²y Use sec²y = 1 + tan²y = 1 + x²: dy/dx = 1 / (1 + x²)
InterpretAlways positive, always between 0 and 1. As \( x \to \pm\infty \), the slope approaches zero — consistent with arctangent leveling off toward \( \pm\pi/2 \).

Answer: \( \displaystyle\frac{d}{dx}[\arctan x] = \frac{1}{1 + x^2} \)

Summary: Inverse Trigonometric Derivatives

Function	Derivative	Domain
\( \arcsin x \)	\( \dfrac{1}{\sqrt{1-x^2}} \)	\( -1 < x < 1 \)
\( \arccos x \)	\( \dfrac{-1}{\sqrt{1-x^2}} \)	\( -1 < x < 1 \)
\( \arctan x \)	\( \dfrac{1}{1+x^2} \)	All real numbers
\( \text{arccot}\, x \)	\( \dfrac{-1}{1+x^2} \)	All real numbers
\( \text{arcsec}\, x \)	\( \dfrac{1}{\|x\|\sqrt{x^2-1}} \)	\( \|x\| > 1 \)
\( \text{arccsc}\, x \)	\( \dfrac{-1}{\|x\|\sqrt{x^2-1}} \)	\( \|x\| > 1 \)

Section 6.8

What Related Rates Measure

A related rate is the rate of change of one quantity that is connected to the rate of change of another quantity through a known equation. Suppose two variables \( A \) and \( r \) are related by \( A = \pi r^2 \) — the area of a circle. Both can change over time. If the radius is increasing at a known rate, calculus lets you find how fast the area is increasing. Differentiate both sides of \( A = \pi r^2 \) with respect to time \( t \):

The Related Rate for a Circle

\( A = \pi r^2 \)

\( \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} \)

This says: the rate at which area changes equals \( 2\pi r \) times the rate at which the radius changes. If you know \( r \) and \( \frac{dr}{dt} \) at a moment, you can compute \( \frac{dA}{dt} \) at that moment.

Notation for Rates of Change

Notation	Meaning	Typical Units
\( \frac{dx}{dt} \)	Rate of change of \( x \) with respect to time	m/s, ft/min, …
\( \frac{dV}{dt} \)	Rate of change of volume	m³/min, L/s, …
\( \frac{dA}{dt} \)	Rate of change of area	cm²/s, ft²/hr, …
\( \frac{dr}{dt} \)	Rate of change of radius	cm/s, m/min, …
\( \frac{dh}{dt} \)	Rate of change of height	m/min, ft/s, …

Section 6.9

Setting Up Related Rate Problems

Five-Step Method for Related Rates

Step 1. Draw and label. Sketch the situation. Label all changing quantities as variables. Do not assign numerical values to changing quantities yet.

Step 2. Identify the given rate and the unknown rate. Write down what you know and what you need.

Step 3. Write the relationship equation. Use geometry, physics, or a formula connecting the variables whose rates appear.

Step 4. Differentiate both sides with respect to \( t \). Apply the chain rule. Every variable picks up a rate factor.

Step 5. Substitute and solve. Plug in values at the specific moment. Solve for the unknown rate. Include units.

Critical Warning — Never Substitute Before Differentiating

If you substitute the specific numerical value of a changing variable before differentiating, you turn it into a constant. Differentiating a constant gives zero — which is wrong. The equation before differentiation is a general relationship. After differentiation, it is a rate equation. Only then do you substitute the values at the specific moment in question.

Example 6.9.1

Expanding circle — area changing

ProblemThe radius of a circle increases at 3 cm/s. How fast is the area increasing when the radius is 7 cm?
Given\( \frac{dr}{dt} = 3 \) cm/s. Find \( \frac{dA}{dt} \) when \( r = 7 \) cm.
Relate\( A = \pi r^2 \)
Differentiate dA/dt = 2πr · dr/dt
Substitute dA/dt = 2π(7)(3) = 42π ≈ 131.9 cm²/s
InterpretWhen the radius is 7 cm and growing at 3 cm/s, the area is growing at approximately 132 cm²/s. The unit cm²/s confirms this is an area rate.

Answer: \( \frac{dA}{dt} = 42\pi \approx 131.9 \) cm²/s

Section 6.10

Geometry-Based Related Rates

Example 6.10.1

The sliding ladder

ProblemA 10-foot ladder leans against a vertical wall. The base slides away at 2 ft/s. How fast is the top sliding down when the base is 6 feet from the wall?
LabelLet \( x \) = horizontal distance of base from wall, \( y \) = height of top on wall. Both change with time. Hypotenuse = 10.
Given\( \frac{dx}{dt} = 2 \) ft/s. Find \( \frac{dy}{dt} \) when \( x = 6 \) ft.
Relate\( x^2 + y^2 = 100 \)
Differentiate 2x dx/dt + 2y dy/dt = 0
Find yWhen \( x = 6 \): \( y = \sqrt{100 - 36} = \sqrt{64} = 8 \) ft.
Substitute 2(6)(2) + 2(8) dy/dt = 0 24 + 16 dy/dt = 0 dy/dt = −24/16 = −1.5 ft/s
InterpretThe negative sign means the top is sliding down at 1.5 ft/s. This is physically correct: as the base moves out, the top falls.

Answer: \( \frac{dy}{dt} = -1.5 \) ft/s (top slides down at 1.5 ft/s)

Example 6.10.2

Water draining from a conical tank

ProblemWater drains from an inverted conical tank (vertex at bottom, height 5 m, base radius 2 m) at 2 m³/min. How fast is the water level dropping when the water is 3 m deep?
Given\( \frac{dV}{dt} = -2 \) m³/min. Find \( \frac{dh}{dt} \) when \( h = 3 \) m.
RelateUse similar triangles to relate \( r \) and \( h \): full dimensions give \( r/h = 2/5 \), so \( r = 2h/5 \). Substitute into volume formula:
Equation V = (1/3)π r² h = (1/3)π (2h/5)² h = 4πh³/75
Differentiate dV/dt = (4π/75) · 3h² · dh/dt = (4πh²/25) dh/dt
Substitute −2 = (4π · 9 / 25) dh/dt = (36π/25) dh/dt dh/dt = −50 / (36π) = −25/(18π) ≈ −0.442 m/min
InterpretThe water level drops at approximately 0.44 m/min when the depth is 3 m. The negative sign confirms the level is falling.

Answer: \( \frac{dh}{dt} = \frac{-25}{18\pi} \approx -0.442 \) m/min

Graph Instruction — Conical Tank

Draw an inverted cone with vertex at the bottom. Mark the full dimensions (height 5 m, radius 2 m) at the top rim. Show the water line at height \( h = 3 \) m with water surface radius \( r = 6/5 \) m. Label \( h \) and \( r \). Show the similar triangles relating the water dimensions to the full tank dimensions.

Example 6.10.3

Expanding ripple on a pond

ProblemThe radius of a circular ripple increases at 4 m/s. How fast is the enclosed area increasing when the radius is 10 m?
Calculate A = πr² → dA/dt = 2πr · dr/dt = 2π(10)(4) = 80π ≈ 251.3 m²/s
InterpretThe area grows at approximately 251 m²/s — far faster than the 4 m/s radius growth, because area scales as the square of the radius.

Answer: \( \frac{dA}{dt} = 80\pi \approx 251.3 \) m²/s

Section 6.11

Motion-Based Related Rates

Example 6.11.1

Two cars moving at a right angle

ProblemCar A moves north at 60 mph, Car B moves east at 80 mph. Both start from the same intersection. How fast is the distance between them increasing after 1 hour?
Label\( y \) = northward distance of A, \( x \) = eastward distance of B, \( z \) = distance between them.
At t = 1\( y = 60 \), \( x = 80 \), \( z = \sqrt{60^2 + 80^2} = \sqrt{10000} = 100 \) miles.
Relate\( z^2 = x^2 + y^2 \)
Differentiate z dz/dt = x dx/dt + y dy/dt 100 · dz/dt = 80(80) + 60(60) = 6400 + 3600 = 10000 dz/dt = 100 mph
InterpretAfter 1 hour, the cars are separating at exactly 100 mph — a 3-4-5 triangle scaled by 20, giving a clean result.

Answer: \( \frac{dz}{dt} = 100 \) mph

Example 6.11.2

Observer tracking a moving car

ProblemA person stands 30 m from a straight road. A car passes at 20 m/s. How fast is the distance from the person to the car increasing 2 seconds after the car passes the closest point?
Label\( x \) = distance of car from closest point on road, \( z \) = distance from person to car. Fixed perpendicular distance = 30 m.
At t = 2 s\( x = 40 \) m, \( z = \sqrt{40^2 + 900} = \sqrt{2500} = 50 \) m.
Calculate z² = x² + 900 → 2z dz/dt = 2x dx/dt 50 dz/dt = 40 · 20 = 800 dz/dt = 16 m/s
Interpret16 m/s — less than the car's 20 m/s, because part of the car's motion is across the field of view rather than purely away from the observer.

Answer: \( \frac{dz}{dt} = 16 \) m/s

Section 6.12

Applied Related Rates in Science and Engineering

Example 6.12.1

Boyle's Law: pressure and volume

ProblemBoyle's Law: \( PV = k \) (constant temperature). When \( P = 50 \) lb/in² and \( V = 100 \) in³, the volume is increasing at 5 in³/min. How fast is the pressure changing?
Differentiate P dV/dt + V dP/dt = 0 50(5) + 100 dP/dt = 0 dP/dt = −2.5 lb/in²/min
InterpretPressure drops at 2.5 lb/in² per minute as the volume expands — exactly what Boyle's Law predicts: pressure and volume change in opposite directions at constant temperature.

Answer: \( \frac{dP}{dt} = -2.5 \) lb/in²/min

Example 6.12.2

Shadow length

ProblemA 2 m person walks away from a 6 m lamppost at 1.5 m/s. How fast is the shadow lengthening?
Label\( x \) = distance of person from lamppost, \( s \) = length of shadow. Tip of shadow is at \( x + s \) from the post.
Similar triangles 6 / (x + s) = 2 / s → 6s = 2x + 2s → 4s = 2x → s = x/2
Differentiate ds/dt = (1/2) dx/dt = (1/2)(1.5) = 0.75 m/s
InterpretThe shadow always grows at half the person's walking speed — at any distance from the post. The linear relationship \( s = x/2 \) means the rate is constant throughout the walk.

Answer: \( \frac{ds}{dt} = 0.75 \) m/s

Example 6.12.3

Filling a cylindrical tank

ProblemWater is pumped into a cylindrical tank (radius 3 m) at 5 m³/min. How fast is the water level rising?
Calculate V = π(9)h = 9πh → dV/dt = 9π dh/dt 5 = 9π dh/dt → dh/dt = 5/(9π) ≈ 0.177 m/min
InterpretThe level rises at about 0.177 m/min regardless of the current depth — the constant cross-section of a cylinder means the inflow rate produces a constant rise rate.

Answer: \( \frac{dh}{dt} = \frac{5}{9\pi} \approx 0.177 \) m/min

MGU Library Connection

The shadow problem appears in physics optics units, engineering drawing courses, and computer graphics ray-tracing algorithms. See the MGU Physics Library for discussion of light, angles, and projection.
Boyle's Law and related gas law differentiation appear in MGU Chemistry and Thermodynamics pages.
Related rates in fluid dynamics and pipe flow appear in MGU Engineering Mathematics chapters.
Implicit curves and gradient computation connect to MGU Machine Learning Mathematics: gradient descent, implicit regularization, and manifold learning.

Section 6.13

Common Implicit and Related Rate Mistakes

Mistake 1 — Forgetting dy/dx on y-terms

d/dx [y³] = 3y²

Correct: \( \frac{d}{dx}[y^3] = 3y^2 \cdot \frac{dy}{dx} \). The chain rule requires multiplying by the derivative of the inner function. Differentiating any expression in \( y \) with respect to \( x \) must include a \( \frac{dy}{dx} \) factor.

Mistake 2 — Substituting numbers before differentiating

r = 5, so V = (4/3)π(125) = constant; dV/dt = 0

Correct: Differentiate the general equation \( V = \frac{4}{3}\pi r^3 \) first to get \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \). Only then substitute \( r = 5 \). Substituting first replaces a variable with a constant; constants have zero derivative and all rate information is destroyed.

Mistake 3 — Missing the product rule in implicit differentiation

d/dx [x²y] = 2xy'

Correct: \( \frac{d}{dx}[x^2y] = 2x \cdot y + x^2 \cdot \frac{dy}{dx} \). The expression \( x^2y \) is a product of two functions of \( x \). The product rule applies: differentiate the first times the second, plus the first times the derivative of the second.

Mistake 4 — Assuming rates of related variables are equal

"The ladder is 10 feet, so dx/dt and dy/dt are the same."

Correct: The relationship \( x^2 + y^2 = 100 \) gives \( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \), so \( \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} \). The rates depend on the current values of \( x \) and \( y \), which change over time. They are almost never equal.

Mistake 5 — Getting the sign wrong

dh/dt = +0.44 m/min when the tank is draining

Correct: A decreasing quantity has a negative rate. If water is draining, \( \frac{dV}{dt} < 0 \). If the water level is dropping, \( \frac{dh}{dt} < 0 \). Always check whether the sign of your answer matches the physical direction of change.

Mistake 6 — Using the wrong geometric formula

Using the sphere volume formula for a cone problem

Correct: Always identify the shape before writing the formula. Verify: Sphere \( V = \frac{4}{3}\pi r^3 \). Cone \( V = \frac{1}{3}\pi r^2 h \). Cylinder \( V = \pi r^2 h \). Using the wrong formula gives a wrong rate equation and a wrong answer.

Mistake 7 — Failing to use similar triangles to reduce variables

Differentiating V = (1/3)πr²h without eliminating r

Correct: In cone problems, there are two changing spatial variables, \( r \) and \( h \). You can only differentiate one equation with respect to \( t \). Use similar triangles to express \( r \) in terms of \( h \) before differentiating, reducing to a single variable.

Mistake 8 — Omitting units from the answer

dy/dt = −1.5

Correct: A rate without units is incomplete. State the units: \( -1.5 \) ft/s, or m/min, etc. Units confirm the meaning of the derivative and catch unit-conversion errors. If the length is in feet and time in seconds, the rate is in ft/s.

Section 6.14

Preparing for Derivative Applications

This chapter extended differentiation in two important directions. Implicit differentiation lets you find \( \frac{dy}{dx} \) for any curve or relationship — even when you cannot solve for \( y \) explicitly. Differentiate both sides with respect to \( x \), attach \( \frac{dy}{dx} \) to every \( y \)-term by the chain rule, and solve. Related rates use derivatives to analyze how quantities change together over time. Label clearly, write the relationship equation, differentiate with respect to \( t \), and substitute only after differentiating.

Both methods rest on the same foundation: the chain rule. Implicit differentiation is the chain rule applied to \( y \) as a function of \( x \). Related rates is the chain rule applied to all variables as functions of \( t \). You have now differentiated everything from polynomial functions to implicit curves to changing geometric quantities.

Looking Ahead

Chapter 7: Graph Analysis with Derivatives

Now that you can differentiate nearly any function — explicitly or implicitly — the next step is to use derivatives to understand the shapes of graphs. Chapter 7 uses first and second derivatives to identify where functions increase and decrease, where they have local maxima and minima, and where they change concavity. You will build sign charts, apply the first and second derivative tests, and sketch curves using derivative information alone. Every technique from Chapters 3 through 6 carries into this analysis.

Chapter 6 — Back Matter

Chapter Summary

Sec 6.1An explicit function isolates \( y \) as a formula in \( x \). An implicit relationship gives an equation in both \( x \) and \( y \) without solving for either. Many important curves cannot be expressed explicitly, or require two separate explicit formulas.
Sec 6.2Implicit differentiation is needed when solving for \( y \) is impractical or impossible. The method works by differentiating both sides of an equation with respect to \( x \), treating \( y \) as a function of \( x \) and applying the chain rule.
Sec 6.3The five-step method: differentiate both sides → apply chain rule to \( y \)-terms (each picks up \( \frac{dy}{dx} \)) → collect \( \frac{dy}{dx} \) terms → factor → solve. The chain rule is the entire mechanism.
Sec 6.4Treating \( y \) as a function of \( x \) is justified by the implicit function theorem. The notation \( \frac{dy}{dx} \) (or \( y' \)) represents this derivative throughout implicit calculations.
Sec 6.5Implicit differentiation gives \( \frac{dy}{dx} \) as a formula in both \( x \) and \( y \). This formula gives the slope of the curve at any point where the denominator is nonzero. Examples include circles, the folium of Descartes, and trigonometric implicit equations.
Sec 6.6To find a tangent line: differentiate implicitly to get \( \frac{dy}{dx} \), evaluate at the given point, apply point-slope form. The tangent to a circle is perpendicular to the radius at that point, as confirmed algebraically.
Sec 6.7If \( y = f^{-1}(x) \), then \( f(y) = x \). Differentiating gives \( f'(y) \cdot \frac{dy}{dx} = 1 \), so \( \frac{dy}{dx} = 1/f'(y) \). This procedure derives all six inverse trigonometric derivative formulas.
Sec 6.8A related rate connects the rate of change of one quantity to the rate of another through a shared equation. Rates are written as derivatives with respect to time \( t \). Units of rates must be stated explicitly.
Sec 6.9The five-step method: draw and label → identify given and unknown rates → write the relationship equation → differentiate with respect to \( t \) → substitute and solve. Never substitute specific values of changing variables before differentiating.
Sec 6.10Geometry problems use area, volume, and the Pythagorean theorem. Cone problems require similar triangles to eliminate one variable before differentiating. The sign of the rate reflects the direction of change.
Sec 6.11Motion problems with two moving objects use the Pythagorean theorem to relate positions and the separating distance. The resulting rate of separation depends on both positions and both velocities at the moment in question.
Sec 6.12Physical applications include Boyle's Law, shadow length (via similar triangles), and tank filling (where constant cross-section gives constant rate). Units and context guide interpretation of every answer.
Sec 6.13Eight recurring errors: missing \( \frac{dy}{dx} \) on \( y \)-terms; substituting before differentiating; missing the product rule; assuming equal rates; sign errors; wrong formulas; failing to use similar triangles; omitting units.
Sec 6.14Both methods in this chapter rest on the chain rule. Chapter 7 uses the first and second derivatives to analyze graph behavior: increasing/decreasing intervals, extrema, concavity, and inflection points.

Key Rate Equations Reference

Situation	Equation	Rate Equation (after differentiating w.r.t. \( t \))
Circle area	\( A = \pi r^2 \)	\( \dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt} \)
Sphere volume	\( V = \frac{4}{3}\pi r^3 \)	\( \dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt} \)
Cylinder volume (fixed radius)	\( V = \pi r^2 h \)	\( \dfrac{dV}{dt} = \pi r^2\dfrac{dh}{dt} \)
Cone volume (similar triangles required)	\( V = \frac{1}{3}\pi r^2 h \)	Express \( r \) in terms of \( h \) first, then differentiate
Right triangle hypotenuse	\( x^2 + y^2 = z^2 \)	\( 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 2z\dfrac{dz}{dt} \)
Boyle's Law	\( PV = k \)	\( P\dfrac{dV}{dt} + V\dfrac{dP}{dt} = 0 \)

Key Terms Review

explicit function

A relationship written as \( y = f(x) \), where one variable is solved as a clear formula in terms of the other.

implicit relationship

An equation involving two or more variables where one is not solved explicitly in terms of the other. Examples: \( x^2 + y^2 = 25 \), \( x^3 + y^3 = 6xy \).

implicit differentiation

Differentiating both sides of an equation with respect to \( x \), treating \( y \) as a function of \( x \) and applying the chain rule to every \( y \)-term.

chain rule

The rule \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \). The engine behind implicit differentiation: \( y \) is the inner function, and every \( y \)-term picks up a \( \frac{dy}{dx} \) factor.

related rates

Using derivatives to connect the rates of change of two or more variables that are related by an equation. Both variables are functions of time \( t \).

implicit curve

A curve defined by an implicit equation such as \( x^2 + y^2 = r^2 \). Not necessarily the graph of a single function.

tangent line

A line touching a curve at a single point with slope equal to \( \frac{dy}{dx} \) at that point. Found using point-slope form after evaluating the derivative at the given point.

similar triangles

Triangles with the same angles and proportional sides. Used in cone and shadow problems to express one variable in terms of another before differentiating.

Concept Review Questions

What does it mean for a function to be defined implicitly? Give an example and explain why it cannot always be solved for \( y \) explicitly.
When you differentiate \( y^2 \) with respect to \( x \), why does \( \frac{dy}{dx} \) appear in the result? Which differentiation rule is responsible?
A curve is defined by \( x^2 + 3y^2 = 12 \). Without solving for \( y \), describe the complete procedure you would use to find the slope of this ellipse at a specific point.
Explain in your own words why you must differentiate before substituting specific values in a related rate problem.
What does the sign of a related rate tell you? What does it mean when \( \frac{dh}{dt} \) is negative?
Two quantities are related by the Pythagorean theorem \( x^2 + y^2 = z^2 \). Write the equation that governs their rates of change with respect to time.
In the conical tank problem, why is it necessary to use similar triangles before differentiating? What goes wrong if you skip this step?

Implicit Differentiation Practice

Part AFinding dy/dx

Find \( \frac{dy}{dx} \) by implicit differentiation. Show all steps.

\( x^2 + y^2 = 36 \)
\( x^3 - y^3 = 8 \)
\( x^2y + y^3 = 5 \)
\( 2x^2 - 3xy + y^2 = 1 \)
\( \sqrt{x} + \sqrt{y} = 9 \)
\( e^y = x^2 + y \)
\( \ln(xy) = x^2 \)
\( \sin x + \cos y = 1 \)
\( \tan(x + y) = x \)
\( x^2\sin y = y\cos x \)

Part BTangent Lines

Find the slope at the given point, then write the tangent line equation.

\( x^2 + y^2 = 25 \) at \( (4, 3) \)
\( x^2 + 4y^2 = 8 \) at \( (2, 1) \)
\( xy = 6 \) at \( (2, 3) \)
\( x^3 + y^3 = 9 \) at \( (1, 2) \)
\( x^2y^2 = 16 \) at \( (2, 2) \)
\( e^x + e^y = e \) at \( (0, 1) \)

Extension: Verify that \( (3, 4) \) and \( (3, -4) \) both lie on \( x^2 + y^2 = 25 \). Use implicit differentiation to find the tangent slope at each. Are the tangent lines parallel? Perpendicular? Explain.

Part CRate Equation Setup

Write the relationship equation and differentiate with respect to \( t \). Do not solve — just produce the rate equation.

Area of a square with side \( s \)
Volume of a cube with side \( s \)
Volume of a sphere with radius \( r \)
Hypotenuse \( z \) of a right triangle with legs \( x \) and \( y \)
Perimeter of a rectangle with length \( l \) and width \( w \)
Boyle's Law: \( PV = k \)

Application Problems

For each problem: draw and label, write the relationship equation, differentiate with respect to \( t \), substitute, solve, and interpret with units.

Expanding Balloon. A spherical balloon is inflated so the radius increases at 2 cm/s. How fast is the volume increasing when the radius is 5 cm?
Draining Tank. A cylindrical tank with radius 4 m drains at 3 m³/min. How fast is the water level dropping?
Longer Ladder. A 13-foot ladder leans against a wall. The base slides away at 3 ft/s. How fast is the top sliding down when the base is 5 feet from the wall?
Shadow Length. A 1.8 m person walks away from a 4 m lamppost at 1 m/s. How fast is the shadow lengthening?
Two Ships. Ship A moves south at 15 km/h. Ship B moves east at 20 km/h. They start from the same port at the same time. How fast is the distance between them increasing after 2 hours?
Conical Pile. Gravel is dumped forming a cone where the radius always equals the height. Volume increases at 10 ft³/min. How fast is the height increasing when the height is 5 ft?
Angle of Elevation. A balloon rises vertically from a point 200 m from an observer at 5 m/s. How fast is the angle of elevation increasing when the balloon is 150 m high? (Hint: \( \tan\theta = h/200 \).)
Profit and Production. The profit \( P \) (thousands of dollars) and production level \( q \) (units) satisfy \( P^2 - 4Pq + 3q^2 = 0 \). Production increases at 100 units/month. When \( q = 5 \) and \( P = 5 \), how fast is profit changing?

Error Analysis

Each problem contains a mistake. Identify the error and write the correct solution.

Problem 1

d/dx [x² + y²] = 9 → 2x + 2y = 0 → y = −x

This is a student's attempt at finding \( \frac{dy}{dx} \) for \( x^2 + y^2 = 9 \). What step is wrong? What is the correct \( \frac{dy}{dx} \)?

Problem 2

r = 4, so V = (4/3)π(64) = 256π/3. Differentiating: dV/dt = 0.

A student wants \( \frac{dV}{dt} \) for a sphere when \( r = 4 \) and \( \frac{dr}{dt} = 3 \). What went wrong? Give the correct calculation.

Problem 3

d/dx [x³y + y²] = 3x²y' + 2yy'

The student differentiated \( x^3y + y^2 = 7 \). What is missing? Write the complete correct differentiation.

Problem 4

In the ladder problem, dy/dt = dx/dt = 2 ft/s since both come from the same ladder.

Explain why the rates \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are not equal. Use the rate equation to explain the relationship between them.

Chapter 6 Checkpoint

Complete without assistance — assesses all key skills and concepts from Chapter 6

Part A — Implicit Differentiation

Find \( \frac{dy}{dx} \) for \( 3x^2 + 2xy - y^2 = 8 \). Show all steps.
Find the slope of the tangent line to \( x^2 + y^3 = 9 \) at the point \( (0, \sqrt[3]{9}) \).
Write the equation of the tangent line to \( x^2y - y^3 = 2 \) at the point \( (1, -1) \).
Find \( \frac{d^2y}{dx^2} \) for \( x^2 + y^2 = r^2 \). (Hint: differentiate \( \frac{dy}{dx} = -x/y \) implicitly.)

Part B — Inverse Function Derivatives

Use implicit differentiation to derive \( \frac{d}{dx}[\arccos x] = \frac{-1}{\sqrt{1-x^2}} \). Let \( y = \arccos x \) so \( \cos y = x \). Show all steps.

Part C — Related Rates

A 15-foot ladder leans against a wall. The base slides out at 4 ft/s. How fast is the top sliding down when the base is 9 feet from the wall?
A spherical snowball melts so its surface area decreases at 2 cm²/min. How fast is the radius decreasing when the radius is 3 cm? (Surface area: \( SA = 4\pi r^2 \).)
A conical water tank (vertex down) has height 12 m and base radius 4 m. Water drains at 2 m³/min. How fast is the water level dropping when the water is 6 m deep?
Two people start walking from the same corner. Person A walks north at 4 ft/s. Person B walks west at 3 ft/s. How fast is the distance between them increasing after 10 seconds?

Part D — Challenge

The volume of a box is \( V = lwh \) where \( l \), \( w \), and \( h \) are all functions of time. At a moment when \( l = 2 \), \( w = 3 \), \( h = 4 \), \( \frac{dl}{dt} = 1 \), \( \frac{dw}{dt} = -1 \), and \( \frac{dh}{dt} = 2 \), find \( \frac{dV}{dt} \).

Part E — Interpretation

In the conical draining tank (Problem 8 above), explain in plain language what \( \frac{dh}{dt} \) represents. Why is the value negative? What does the magnitude of the rate tell you about how quickly the water level is falling?
In Problem 9 above, the two people are walking away from each other. Explain why \( \frac{dz}{dt} \) is less than the sum of their individual speeds (4 + 3 = 7 ft/s). What geometric feature of their paths explains this?

Answers may be placed in the answer key.