Unit 4Related Rates Problem-Solving Steps

Concept 4.153Built for May 2027 AP Exam

Related Rates Problem-Solving Steps

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The procedure has five steps and they are always the same five: draw and label, write the relation, differentiate with respect to time, substitute the named instant, then solve for the rate that is left. Each step produces something the next one needs, which is why skipping the first one is expensive rather than merely untidy. A labelled diagram is the equation written in pictures, and without it the relation has to be guessed from prose. The differentiation is one line, the substitution is arithmetic, and the whole difficulty of the problem has already been paid for by the time the derivative appears. Following the order turns an unfamiliar context into a familiar sequence.

On the AP Exam

What you’ll see: A context with two or more changing measurements, a formula supplied or expected from geometry, and one rate given with another requested at a named instant.
What you’ll do: Name the quantities, write the relation, differentiate with respect to time, substitute, solve, and label the answer with units.
Memory hook: Label, relate, differentiate, substitute, solve. Five verbs in a fixed order, and the order is the method.
Use it on the exam: For MCQs: The named instant is usually a number you can substitute before differentiating only by accident, so read the stem twice to see which values are constant and which are momentary. For FRQs: On 2022 Q4(d) the guideline separates the product-rule point from the chain-rule point, which means the differentiated line has to be written out. A correct final number with no differentiated equation in between loses both.

Concept Review

The Five Steps as a Flow

Figure 4 — The five steps in order, each with the failure it prevents. The diagram comes first because the relation is read off it, and the substitution comes after the derivative because a frozen variable has no rate.

Read the flow from top to bottom and note that no step can be moved. Substituting before differentiating sets every rate to zero, and differentiating before writing the relation leaves nothing to differentiate. The middle step is the only calculus in the chain; the rest is bookkeeping that protects it.

Five numbered rows listing the related rate procedure with a note on what each step protects

What Each Step Produces

Step	What you write	What it hands to the next step
1. Draw and label	a figure with every changing length named	the symbols used in step 2
2. Write the relation	one equation holding for all $t$	something differentiable
3. Differentiate	every variable picks up its own rate	an equation in the rates
4. Substitute	the values at the named instant	one unknown left
5. Solve	the requested rate with units	the answer

The second column is the part examiners read. A solution that jumps from the context to a number has no second column, and the scoring guidelines award points per line rather than per answer.

Constants and Variables in the Same Formula

A formula such as $V = \pi r^2 h$ may describe a tank whose radius never changes, and the same formula may describe a cone whose radius shrinks as it melts. The formula is identical; the differentiation is not. Step 1 exists to settle which symbols are functions of time, and the answer changes what step 3 looks like:

$\frac{dV}{dt} = \pi r^2\frac{dh}{dt} \quad \text{versus} \quad \frac{dV}{dt} = 2\pi rh\frac{dr}{dt} + \pi r^2\frac{dh}{dt}$

The left version has one term because the radius is fixed. The right version has two because both measurements move, and it also needs a product rule.

Choosing the Instant

The substitution in step 4 uses values that hold only at one moment. Writing those values at the top of the page and marking them "at $t = t_0$ " prevents them from being treated as constants later. On a draining tank the height at the instant is a number, while the height itself is a function; confusing the two produces a derivative of zero for a quantity that is visibly changing.

Checking the Answer

Three checks catch most errors before they cost a point. The units must be a unit of the requested quantity per unit of time. The sign must match the wording, so a decreasing quantity has a negative rate. The size must be plausible, and a radius that grows faster than the volume that feeds it is usually a sign that a factor was dropped.

Common Misconception Check

Myth 1: "The five steps are a suggestion and can be reordered." Truth: Steps 3 and 4 cannot be swapped, and step 2 cannot be skipped. The order encodes which quantities are momentary and which are not.
Myth 2: "A diagram is only needed for geometry problems." Truth: Any labelled relation does the same job, including a table of symbols for a rate model that has no picture at all.
Myth 3: "The final number is the answer." Truth: The answer is the number with its units and its sign, and on a contextual part an unlabelled number is worth less than a labelled one.

Topic-Specific Study Method

Work the Five Steps in Order

Borrowed idea: Work the five steps in order. Draw and label first, because a labelled diagram is what turns a sentence into an equation. Write the relation, differentiate it with respect to time, substitute the named instant, and solve. Which step do students skip most often? The diagram, because it feels like drawing rather than doing mathematics. How much does that skip cost? Usually the whole part, since the equation that follows is built from the labels. Five steps, in order, every time.

Step	The move	The sentence that earns it
1	name and label every changing quantity	"Let $r$ be the radius and $h$ the height, both in centimeters."
2	write the relation that holds for all $t$	"The water forms a cone similar to the tank, so $\frac{r}{h} = \frac{4}{8}$ ."
3	differentiate with respect to $t$	"Differentiating with respect to $t$ gives $\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$ ."
4	substitute the instant	"At the instant when $h = 5$ , the volume rate is $3$ ."
5	solve and label	"Therefore $\frac{dh}{dt} = \frac{12}{25\pi}$ meters per minute."

① Take any related-rate stem and write only step 1 and step 2, with no calculus at all. Doing that for three problems in a row shows that the equation, once found, does most of the work.

② For one problem, write the differentiated line twice: once treating a measurement as constant and once treating it as a function of time. The two lines differ by a term, and seeing the term appear is what makes step 1 feel necessary.

③ After solving, write the units and read them aloud as a sentence. "Cubic centimeters per second" either matches the quantity asked for or it does not, and the mismatch is visible before any arithmetic is rechecked.

Training & Error Diagnosis

Core Archetype (Simplest Form): Water is pumped into an inverted conical tank at $3$ cubic meters per minute. The tank has a radius of $4$ meters at the top and a height of $8$ meters. Find the rate at which the water depth is increasing when the depth is $5$ meters.

Standard Execution:

Label. Let $h$ be the depth and $r$ the radius of the water surface.

Relate. By similar triangles $\frac{r}{h} = \frac{4}{8}$ , so $r = \frac{h}{2}$ and $V = \frac{1}{3}\pi r^2 h = \frac{\pi h^3}{12}$ .

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

Substitute. $3 = \frac{25\pi}{4}\frac{dh}{dt}$ .

Solve. $\frac{dh}{dt} = \frac{12}{25\pi} \approx 0.153$ meters per minute.

Exam Variation (Context / Parameter Shift): A square plate of metal is heated, and its side length increases at $0.02$ centimeters per minute. Find the rate at which the diagonal is increasing when the side is $20$ centimeters.

Solution: With diagonal $d$ and side $s$ , $d^2 = 2s^2$ , so $2d\frac{dd}{dt} = 4s\frac{ds}{dt}$ and $\frac{dd}{dt} = \frac{2s}{d}\frac{ds}{dt} = \sqrt{2}\frac{ds}{dt}$ . Substituting $\frac{ds}{dt} = 0.02$ gives $\frac{dd}{dt} = 0.02\sqrt{2} \approx 0.028$ centimeters per minute. The diagonal grows at a constant multiple of the side, which is why the named instant does not affect the answer here.

Common Traps & Failure Signals:

The unlabelled start (Signal: a formula containing a symbol never defined): the relation was copied from memory without deciding what its letters mean in this problem.
The frozen derivative (Signal: a rate of zero for a quantity the stem says is changing): the substitution happened before the differentiation.
The unitless finish (Signal: a bare number on the answer line): step 5 was treated as optional and the units were never checked.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Label Skipped: You looked for a formula before naming anything.
Order Reversed: You substituted the named values into the geometry before differentiating, which turns every changing quantity into a constant at exactly the instant the question cares about, so the rates on your page are all zero.
Similar Triangles Missed: You kept two variables in a relation that the context had already reduced to one, so the differentiated line contained two unknown rates and the final division could not be finished even though every earlier step was correct.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Five released BC sets contain a part whose scoring depends on the last two steps of the procedure, and the geometric relation has to be built rather than recognised
10-Year Total Count: 5 of 9 Exams · 5 FRQ Parts; Counted where the relation must be constructed from the context before it can be differentiated
FRQ Years Appeared: 2017 (Q1d), 2018 (Q4d, Q5c), 2022 (Q4d), 2025 (Q2D)

Practice Examples

MCQ 1

Which Step Comes First

A conical pile of gravel keeps its shape as gravel is added, and the problem asks for the rate at which the height grows. Which step of the related-rate procedure has to be completed before any derivative can be written?

MCQ 2

The Constant That Is Not

A cylindrical tank with a fixed radius of $2$ meters is filled at $5$ cubic meters per minute. At what rate is the water level rising?

MCQ 3

A Shrinking Cube

A cube of ice melts so that each edge decreases at $0.5$ centimeters per minute. At the instant when the edge is $10$ centimeters, at what rate is the surface area decreasing?

FRQ 1 • A Conical Tank

Water is pumped into an inverted conical tank at a constant rate of $3$ cubic meters per minute. The tank has a radius of $4$ meters at the top and a height of $8$ meters, and the water forms a cone similar to the tank.

(a) Write an equation that gives the volume of water in the tank in terms of the depth $h$ alone, and explain how the similarity of the cones is used.

(b) Find the rate at which the depth is increasing at the instant when the depth is $5$ meters. Include units.

(c) Find the rate at which the area of the water surface is increasing at the same instant. Include units.

(d) Explain why the depth rises more slowly as the tank fills, even though water enters at a constant rate.

Cheatsheet

	Do	Sentence to write	Frequent error
1	draw and label	"Let $r$ be the radius at time $t$ ."	a symbol used but never defined
2	write the relation	"Since the cones are similar, $r = \frac{h}{2}$ ."	two variables left where the context allows one
3	differentiate with respect to $t$	"Differentiating gives $\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$ ."	a constant given a rate
4	substitute the instant	"At $h = 5$ and $\frac{dV}{dt} = 3$ ."	substitution before differentiating
5	solve and label	"Therefore $\frac{dh}{dt} \approx 0.153$ m/min."	a bare number

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Start Topic Practice Next Topic: 4.5 Implicit Differentiation with Respect to Time

Unit 4Related Rates Problem-Solving Steps

Concept 4.153Built for May 2027 AP Exam

Related Rates Problem-Solving Steps

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: A context with two or more changing measurements, a formula supplied or expected from geometry, and one rate given with another requested at a named instant.
What you’ll do: Name the quantities, write the relation, differentiate with respect to time, substitute, solve, and label the answer with units.
Memory hook: Label, relate, differentiate, substitute, solve. Five verbs in a fixed order, and the order is the method.
Use it on the exam: For MCQs: The named instant is usually a number you can substitute before differentiating only by accident, so read the stem twice to see which values are constant and which are momentary. For FRQs: On 2022 Q4(d) the guideline separates the product-rule point from the chain-rule point, which means the differentiated line has to be written out. A correct final number with no differentiated equation in between loses both.

Concept Review

The Five Steps as a Flow

Five numbered rows listing the related rate procedure with a note on what each step protects

What Each Step Produces

Step	What you write	What it hands to the next step
1. Draw and label	a figure with every changing length named	the symbols used in step 2
2. Write the relation	one equation holding for all $t$	something differentiable
3. Differentiate	every variable picks up its own rate	an equation in the rates
4. Substitute	the values at the named instant	one unknown left
5. Solve	the requested rate with units	the answer

The second column is the part examiners read. A solution that jumps from the context to a number has no second column, and the scoring guidelines award points per line rather than per answer.

Constants and Variables in the Same Formula

$\frac{dV}{dt} = \pi r^2\frac{dh}{dt} \quad \text{versus} \quad \frac{dV}{dt} = 2\pi rh\frac{dr}{dt} + \pi r^2\frac{dh}{dt}$

The left version has one term because the radius is fixed. The right version has two because both measurements move, and it also needs a product rule.

Choosing the Instant

Checking the Answer

Common Misconception Check

Myth 1: "The five steps are a suggestion and can be reordered." Truth: Steps 3 and 4 cannot be swapped, and step 2 cannot be skipped. The order encodes which quantities are momentary and which are not.
Myth 2: "A diagram is only needed for geometry problems." Truth: Any labelled relation does the same job, including a table of symbols for a rate model that has no picture at all.
Myth 3: "The final number is the answer." Truth: The answer is the number with its units and its sign, and on a contextual part an unlabelled number is worth less than a labelled one.

Topic-Specific Study Method

Work the Five Steps in Order

Step	The move	The sentence that earns it
1	name and label every changing quantity	"Let $r$ be the radius and $h$ the height, both in centimeters."
2	write the relation that holds for all $t$	"The water forms a cone similar to the tank, so $\frac{r}{h} = \frac{4}{8}$ ."
3	differentiate with respect to $t$	"Differentiating with respect to $t$ gives $\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$ ."
4	substitute the instant	"At the instant when $h = 5$ , the volume rate is $3$ ."
5	solve and label	"Therefore $\frac{dh}{dt} = \frac{12}{25\pi}$ meters per minute."

① Take any related-rate stem and write only step 1 and step 2, with no calculus at all. Doing that for three problems in a row shows that the equation, once found, does most of the work.

Training & Error Diagnosis

Standard Execution:

Label. Let $h$ be the depth and $r$ the radius of the water surface.

Relate. By similar triangles $\frac{r}{h} = \frac{4}{8}$ , so $r = \frac{h}{2}$ and $V = \frac{1}{3}\pi r^2 h = \frac{\pi h^3}{12}$ .

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

Substitute. $3 = \frac{25\pi}{4}\frac{dh}{dt}$ .

Solve. $\frac{dh}{dt} = \frac{12}{25\pi} \approx 0.153$ meters per minute.

Solution: With diagonal $d$ and side $s$ , $d^2 = 2s^2$ , so $2d\frac{dd}{dt} = 4s\frac{ds}{dt}$ and $\frac{dd}{dt} = \frac{2s}{d}\frac{ds}{dt} = \sqrt{2}\frac{ds}{dt}$ . Substituting $\frac{ds}{dt} = 0.02$ gives $\frac{dd}{dt} = 0.02\sqrt{2} \approx 0.028$ centimeters per minute. The diagonal grows at a constant multiple of the side, which is why the named instant does not affect the answer here.

Common Traps & Failure Signals:

The unlabelled start (Signal: a formula containing a symbol never defined): the relation was copied from memory without deciding what its letters mean in this problem.
The frozen derivative (Signal: a rate of zero for a quantity the stem says is changing): the substitution happened before the differentiation.
The unitless finish (Signal: a bare number on the answer line): step 5 was treated as optional and the units were never checked.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Label Skipped: You looked for a formula before naming anything.
Order Reversed: You substituted the named values into the geometry before differentiating, which turns every changing quantity into a constant at exactly the instant the question cares about, so the rates on your page are all zero.
Similar Triangles Missed: You kept two variables in a relation that the context had already reduced to one, so the differentiated line contained two unknown rates and the final division could not be finished even though every earlier step was correct.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Five released BC sets contain a part whose scoring depends on the last two steps of the procedure, and the geometric relation has to be built rather than recognised
10-Year Total Count: 5 of 9 Exams · 5 FRQ Parts; Counted where the relation must be constructed from the context before it can be differentiated
FRQ Years Appeared: 2017 (Q1d), 2018 (Q4d, Q5c), 2022 (Q4d), 2025 (Q2D)

Practice Examples

MCQ 1

Which Step Comes First

MCQ 2

The Constant That Is Not

A cylindrical tank with a fixed radius of $2$ meters is filled at $5$ cubic meters per minute. At what rate is the water level rising?

MCQ 3

A Shrinking Cube

A cube of ice melts so that each edge decreases at $0.5$ centimeters per minute. At the instant when the edge is $10$ centimeters, at what rate is the surface area decreasing?

FRQ 1 • A Conical Tank

(a) Write an equation that gives the volume of water in the tank in terms of the depth $h$ alone, and explain how the similarity of the cones is used.

(b) Find the rate at which the depth is increasing at the instant when the depth is $5$ meters. Include units.

(c) Find the rate at which the area of the water surface is increasing at the same instant. Include units.

(d) Explain why the depth rises more slowly as the tank fills, even though water enters at a constant rate.

Cheatsheet

	Do	Sentence to write	Frequent error
1	draw and label	"Let $r$ be the radius at time $t$ ."	a symbol used but never defined
2	write the relation	"Since the cones are similar, $r = \frac{h}{2}$ ."	two variables left where the context allows one
3	differentiate with respect to $t$	"Differentiating gives $\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$ ."	a constant given a rate
4	substitute the instant	"At $h = 5$ and $\frac{dV}{dt} = 3$ ."	substitution before differentiating
5	solve and label	"Therefore $\frac{dh}{dt} \approx 0.153$ m/min."	a bare number

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Start Topic Practice Next Topic: 4.5 Implicit Differentiation with Respect to Time

Unit 4Related Rates Problem-Solving Steps

Concept 4.153Built for May 2027 AP Exam

Related Rates Problem-Solving Steps

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

What you’ll see: A context with two or more changing measurements, a formula supplied or expected from geometry, and one rate given with another requested at a named instant.
What you’ll do: Name the quantities, write the relation, differentiate with respect to time, substitute, solve, and label the answer with units.
Memory hook: Label, relate, differentiate, substitute, solve. Five verbs in a fixed order, and the order is the method.
Use it on the exam: For MCQs: The named instant is usually a number you can substitute before differentiating only by accident, so read the stem twice to see which values are constant and which are momentary. For FRQs: On 2022 Q4(d) the guideline separates the product-rule point from the chain-rule point, which means the differentiated line has to be written out. A correct final number with no differentiated equation in between loses both.

Concept Review

The Five Steps as a Flow

Five numbered rows listing the related rate procedure with a note on what each step protects

What Each Step Produces

Step	What you write	What it hands to the next step
1. Draw and label	a figure with every changing length named	the symbols used in step 2
2. Write the relation	one equation holding for all $t$	something differentiable
3. Differentiate	every variable picks up its own rate	an equation in the rates
4. Substitute	the values at the named instant	one unknown left
5. Solve	the requested rate with units	the answer

The second column is the part examiners read. A solution that jumps from the context to a number has no second column, and the scoring guidelines award points per line rather than per answer.

Constants and Variables in the Same Formula

$\frac{dV}{dt} = \pi r^2\frac{dh}{dt} \quad \text{versus} \quad \frac{dV}{dt} = 2\pi rh\frac{dr}{dt} + \pi r^2\frac{dh}{dt}$

The left version has one term because the radius is fixed. The right version has two because both measurements move, and it also needs a product rule.

Choosing the Instant

Checking the Answer

Common Misconception Check

Myth 1: "The five steps are a suggestion and can be reordered." Truth: Steps 3 and 4 cannot be swapped, and step 2 cannot be skipped. The order encodes which quantities are momentary and which are not.
Myth 2: "A diagram is only needed for geometry problems." Truth: Any labelled relation does the same job, including a table of symbols for a rate model that has no picture at all.
Myth 3: "The final number is the answer." Truth: The answer is the number with its units and its sign, and on a contextual part an unlabelled number is worth less than a labelled one.

Topic-Specific Study Method

Work the Five Steps in Order

Step	The move	The sentence that earns it
1	name and label every changing quantity	"Let $r$ be the radius and $h$ the height, both in centimeters."
2	write the relation that holds for all $t$	"The water forms a cone similar to the tank, so $\frac{r}{h} = \frac{4}{8}$ ."
3	differentiate with respect to $t$	"Differentiating with respect to $t$ gives $\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$ ."
4	substitute the instant	"At the instant when $h = 5$ , the volume rate is $3$ ."
5	solve and label	"Therefore $\frac{dh}{dt} = \frac{12}{25\pi}$ meters per minute."

① Take any related-rate stem and write only step 1 and step 2, with no calculus at all. Doing that for three problems in a row shows that the equation, once found, does most of the work.

Training & Error Diagnosis

Standard Execution:

Label. Let $h$ be the depth and $r$ the radius of the water surface.

Relate. By similar triangles $\frac{r}{h} = \frac{4}{8}$ , so $r = \frac{h}{2}$ and $V = \frac{1}{3}\pi r^2 h = \frac{\pi h^3}{12}$ .

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

Substitute. $3 = \frac{25\pi}{4}\frac{dh}{dt}$ .

Solve. $\frac{dh}{dt} = \frac{12}{25\pi} \approx 0.153$ meters per minute.

Solution: With diagonal $d$ and side $s$ , $d^2 = 2s^2$ , so $2d\frac{dd}{dt} = 4s\frac{ds}{dt}$ and $\frac{dd}{dt} = \frac{2s}{d}\frac{ds}{dt} = \sqrt{2}\frac{ds}{dt}$ . Substituting $\frac{ds}{dt} = 0.02$ gives $\frac{dd}{dt} = 0.02\sqrt{2} \approx 0.028$ centimeters per minute. The diagonal grows at a constant multiple of the side, which is why the named instant does not affect the answer here.

Common Traps & Failure Signals:

The unlabelled start (Signal: a formula containing a symbol never defined): the relation was copied from memory without deciding what its letters mean in this problem.
The frozen derivative (Signal: a rate of zero for a quantity the stem says is changing): the substitution happened before the differentiation.
The unitless finish (Signal: a bare number on the answer line): step 5 was treated as optional and the units were never checked.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Label Skipped: You looked for a formula before naming anything.
Order Reversed: You substituted the named values into the geometry before differentiating, which turns every changing quantity into a constant at exactly the instant the question cares about, so the rates on your page are all zero.
Similar Triangles Missed: You kept two variables in a relation that the context had already reduced to one, so the differentiated line contained two unknown rates and the final division could not be finished even though every earlier step was correct.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Five released BC sets contain a part whose scoring depends on the last two steps of the procedure, and the geometric relation has to be built rather than recognised
10-Year Total Count: 5 of 9 Exams · 5 FRQ Parts; Counted where the relation must be constructed from the context before it can be differentiated
FRQ Years Appeared: 2017 (Q1d), 2018 (Q4d, Q5c), 2022 (Q4d), 2025 (Q2D)

Practice Examples

MCQ 1

Which Step Comes First

MCQ 2

The Constant That Is Not

A cylindrical tank with a fixed radius of $2$ meters is filled at $5$ cubic meters per minute. At what rate is the water level rising?

MCQ 3

A Shrinking Cube

A cube of ice melts so that each edge decreases at $0.5$ centimeters per minute. At the instant when the edge is $10$ centimeters, at what rate is the surface area decreasing?

FRQ 1 • A Conical Tank

(a) Write an equation that gives the volume of water in the tank in terms of the depth $h$ alone, and explain how the similarity of the cones is used.

(b) Find the rate at which the depth is increasing at the instant when the depth is $5$ meters. Include units.

(c) Find the rate at which the area of the water surface is increasing at the same instant. Include units.

(d) Explain why the depth rises more slowly as the tank fills, even though water enters at a constant rate.

Cheatsheet

	Do	Sentence to write	Frequent error
1	draw and label	"Let $r$ be the radius at time $t$ ."	a symbol used but never defined
2	write the relation	"Since the cones are similar, $r = \frac{h}{2}$ ."	two variables left where the context allows one
3	differentiate with respect to $t$	"Differentiating gives $\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$ ."	a constant given a rate
4	substitute the instant	"At $h = 5$ and $\frac{dV}{dt} = 3$ ."	substitution before differentiating
5	solve and label	"Therefore $\frac{dh}{dt} \approx 0.153$ m/min."	a bare number

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