Expressing changing quantities as derivatives with respect to time, finding a geometric equation that relates them, differentiating it implicitly, and only then substituting the instant's values — worked through the sliding ladder, the inflating balloon, the filling cone and two objects approaching at right angles.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 4 — Applications of Derivatives
Related Rates
Objectives
Five outcomes. The calculus is one line of Section 3.8; the difficulty is entirely in the setup, and four of these five are about that.
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 296-307 — the section these objectives are drawn from
Warm-up
Section 3.6 differentiated a balloon's volume with respect to time by the chain rule, and Section 3.8 differentiated equations without solving them.
Discussion prompt
A sphere's volume is four thirds pi r cubed. If the radius depends on time, what does differentiating the volume with respect to time give, and where does the extra factor come from?
Hint: The volume depends on the radius, and the radius on time.
Answer:
\[ V = \tfrac{4}{3}\pi r^{3} \;\Longrightarrow\; \frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt} \]
The factor dr by dt comes from the chain rule, exactly as a y term produced dy by dx in Section 3.8. Here the independent variable is time, and every quantity that changes with time contributes its own rate.
That is the entire technique of this section. What is new is not the calculus but the discipline of setting the problem up — deciding what to name, what relation to write, and crucially when to substitute numbers.
Concept
Quantities linked by an equation have linked rates. Differentiating that equation with respect to time produces a relation among the rates, and the wanted rate is then found by substituting the values that hold at the instant in question.
related rates — Two or more quantities changing with time and connected by an equation. Differentiating the equation with respect to time relates their rates, so a known rate determines an unknown one.
\[ F(x, y) = 0 \;\Longrightarrow\; \frac{\partial F}{\partial x}\frac{dx}{dt} + \frac{\partial F}{\partial y}\frac{dy}{dt} = 0 \]
The order in step five is not stylistic. A quantity replaced by its instantaneous value becomes a constant, and constants have zero derivative — so substituting early erases exactly the rate being solved for.
Figure (svg): The five steps of a related rates problem, with the order that matters
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 296-298
Section
Section 1
Concept
Draw the situation, give every changing quantity a letter, and write down which rates are known and which is wanted. Numbers that hold only at the instant in question are set aside until the very end.
the setup — Naming each quantity, recording each rate as a derivative with respect to time with its correct sign, and distinguishing constants of the problem from quantities that change.
\[ \text{known: } \frac{dx}{dt}; \quad \text{wanted: } \frac{dy}{dt} \]
The distinction that matters most is between a quantity that is constant throughout — the ladder's length, the cone's proportions — and one that merely happens to have a particular value at the instant asked about.
Figure (svg): The five steps of a related rates problem, with the order that matters
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 296-300 — the problem-solving strategy
Picture it
The standard procedure.
Figure (svg): The five steps of a related rates problem, with the order that matters
Steps one to three are where the thinking is. Step four is a single application of Section 3.8, and step five is placed after it for a reason the next idea makes concrete.
Worked example
Example 4.1. The archetype of the whole section.
\[ \text{A 10 ft ladder slides down a wall. Its foot moves out at 2 ft/s. How fast is the top falling when the foot is 6 ft out?} \]
Name the quantities
Why: Distance out and height up, both changing.
\[ x\text{ and } y;\text{ the ladder is } a\text{ constant } 10 \]
Record the rates
Why: One known, one wanted.
\[ \,dx / \,dt = 2,\text{ want } \,dy / \,dt \]
Write the relation
Why: Pythagoras, true at every instant.
\[ x ^{2} + y ^{2} = 100 \]
Differentiate with respect to t
Why: Every changing letter gets a rate.
\[ 2 x(\,dx / \,dt) + 2 y(\,dy / \,dt) = 0 \]
Now substitute the instant's values
Why: At x = 6, Pythagoras gives y = 8.
\[ 12(2) + 16(\,dy / \,dt) = 0 \]
Solve
Why: Divide.
\[ \,dy / \,dt = -1.5 \text{ft} / s \]
Figure (svg): A ladder sliding down a wall, with the two changing lengths and the fixed hypotenuse marked
\[ \frac{dy}{dt} = -1.5 \text{ ft/s} \]
Verify: check the sign and what happens near the ground
Why: The answer is negative, meaning y is decreasing — the top is sliding DOWN, as it must be while the foot moves out. The magnitude is also plausible: the top falls more slowly than the foot moves out while the ladder is still fairly upright. Note what happens as x approaches 10: y approaches 0 and the equation forces dy by dt to become unbounded, predicting the top falls arbitrarily fast at the end. That is a genuine feature of the idealised model rather than an error.
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 298-299
Sorting
Ask whether it is the same at every instant.
Sort into buckets
Sort each quantity in the ladder problem.
The foot's speed is constant here because the problem says so; in other problems it would vary and would have to stay a letter too. Reading the wording carefully to see which rates are constant is part of the setup.
Worked example
Checkpoint 4.1. Which quantities are genuinely fixed?
\[ \text{In the ladder problem, which quantities are constants and which are not?} \]
Examine the ladder's length
Why: It does not change as the ladder slides.
\[ 10\text{ is } a\text{ genuine constant} \]
Examine the distance out
Why: It changes throughout.
Examine the height
Why: It changes throughout.
Examine the value 6
Why: It holds only at one instant.
Conclude
Why: Only the ladder's length may be substituted early.
Figure (svg): The solution to Worked example naming a constant correctly shown as a ladder of expressions, one row per legal move
\[ 10 \text{ is constant}; \; x, y \text{ vary}; \; 6 \text{ is an instant's value} \]
Verify: test the distinction by asking what changes if the instant changes
Why: At a different instant x would be 7 rather than 6, but the ladder would still be 10 feet long. That is the test: a genuine constant is the same at every instant, while an instantaneous value belongs to one moment only. Substituting 100 for the ladder's length squared is fine and was done from the start; substituting 6 for x before differentiating would have been fatal, as the next idea shows.
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 299-300
Trap
\[ x^{2}+y^{2}=100 \text{ with } x = 6, \; y = 8 \]
Substitute both before differentiating
Why: The student uses the instant's numbers immediately.
\[ 36 + 64 = 100 \;\Longrightarrow\; \frac{d}{dt}: \; 0 = 0 \]
The statement is true and completely empty. Both rates have vanished along with the variables that carried them.
\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0, \;\text{ THEN } x = 6, y = 8 \]
Keep every changing quantity as a letter until after differentiating
Why: A number has no rate; only a variable does.
The ladder's length may be substituted at once, because it is the same at every instant. The value 6 may not, because it describes one moment. Distinguishing a genuine constant from an instantaneous value is the single most important habit in this section.
Fill the middle
The ladder at the instant the foot is 6 feet out.
Fill in the blanks
6^8 + y^___ = 100 \;\Longrightarrow\; y = ___
Why: The height is 8 feet at that instant. This value is computed from the relation and substituted only after the differentiation is complete.
Ranking
A related rates problem from start to finish.
Put in order
Why: The order of d and e is the whole point. Reversing them turns a solvable problem into the empty statement zero equals zero, because a substituted number has no rate to contribute.
Prediction
Commit before reasoning.
Predict first
What happens if you substitute x = 6 before differentiating?
Correct: x becomes a constant, so dx by dt disappears and the equation says nothing.
\[ 36 + 64 = 100 \;\xrightarrow{\;d/dt\;}\; 0 = 0 \]
Why: Differentiation extracts rates from variables, and a number is not a variable. Substituting first replaces the whole relation with an arithmetic identity, whose derivative is zero equals zero — true and useless. The rates must be extracted while the quantities are still letters, and only then are the instant's values relevant. This single ordering is what most related rates errors come down to.
Section
Section 2
Concept
Differentiating the relation with respect to time is implicit differentiation with t as the independent variable. Each quantity that changes with time produces a chain-rule factor giving its own rate.
differentiating with respect to time — Applying the chain rule to each term of a relation, treating every quantity as a function of t. A term in x contributes a factor dx by dt, exactly as a y term contributed dy by dx in Section 3.8.
\[ \frac{d}{dt}\left[x^{2}\right] = 2x\frac{dx}{dt} \]
The Pythagorean relation appears so often that its differentiated form is worth recognising on sight: x times its rate plus y times its rate equals s times its rate, after the twos have cancelled.
Figure (svg): What goes wrong when a changing quantity is replaced by its instantaneous value too early
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 298-302 — differentiating the relation
Picture it
Substituting first, and differentiating first.
Figure (svg): What goes wrong when a changing quantity is replaced by its instantaneous value too early
The left column produces a true statement containing no rates at all. The right column produces an equation in the two rates, which is exactly what the problem asked for.
Worked example
Example 4.2. The factor turns out to be the surface area.
\[ \text{Air enters a spherical balloon at } 100 \text{ cm}^{3}\text{/s. How fast is the radius growing when } r = 5 \text{ cm?} \]
Name and record
Why: Volume and radius, both changing.
Write the relation
Why: The sphere's volume.
\[ V = (\frac{4}{3}) \pi r ^{3} \]
Differentiate with respect to t
Why: Chain rule on the cube.
Substitute the instant's values
Why: Now, not before.
\[ 100 = 4 \pi(25) (\,dr / \,dt) \]
Solve
Why: Divide.
\[ \,dr / \,dt = 1 / \pi \text{cm} / s \]
Figure (svg): A sphere with its radius growing, and the surface area factor that links the two rates
\[ \frac{dr}{dt} = \frac{1}{\pi} \approx 0.318 \text{ cm/s} \]
Verify: check the units and what happens for a larger balloon
Why: Cubic centimetres per second divided by square centimetres gives centimetres per second, which is a rate of length change — correct. Note the factor 4 pi r squared is the sphere's SURFACE AREA, so at a fixed inflow the radius grows more slowly the bigger the balloon is: at r equal to 10 the same 100 cubic centimetres per second gives only a quarter the radial growth. That is physically right, since the incoming air spreads over four times the area.
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 300-301
Fill the middle
The Pythagorean relation for two objects moving at right angles.
Fill in the blanks
x^\frac{ds}{dt}+y^___=s^___ \;\Longrightarrow\; x\frac______ + y\frac______ = s\,___
Why: The separation s is itself a function of time, so it contributes its own rate. The factors of 2 cancel throughout, leaving this tidy form worth recognising on sight.
Worked example
Checkpoint 4.2. The differentiated Pythagorean relation.
\[ \text{A car heads east at } 60 \text{ mph and another north at } 80 \text{ mph from the same crossroads. How fast are they separating after } 1 \text{ hour?} \]
Name and record
Why: Two distances and the separation.
\[ \,dx / \,dt = 60, \,dy / \,dt = 80,\text{ want } \,ds / \,dt \]
Write the relation
Why: Pythagoras.
\[ x ^{2} + y ^{2} = s ^{2} \]
Differentiate with respect to t
Why: Every letter changes.
\[ 2 x x' + 2 y y' = 2 s s' \]
Cancel the twos and substitute after one hour
Why: x = 60, y = 80, so s = 100.
\[ 60(60) + 80(80) = 100 s' \]
Solve
Why: The numerator is 10000.
\[ \,ds / \,dt = 100 \text{mph} \]
Figure (svg): Two objects moving at right angles, with the distance between them changing
\[ \frac{ds}{dt} = 100 \text{ mph} \]
Verify: notice why the answer is constant in time
Why: The answer is 100 mph, which is exactly the hypotenuse of the 60-80 velocity triangle. That is no coincidence: since both cars move at constant speeds from the same point, the separation grows linearly and its rate never changes. Substituting the values after two hours gives the same 100 mph, which is a useful check. When either speed varies, the separation rate genuinely does depend on the instant.
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 301-302
Error analysis
A student differentiates the Pythagorean relation for two moving objects.
Annotate
On: \( x^{2}+y^{2}=s^{2} \;\Longrightarrow\; 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2s \)
Every quantity that changes with time contributes a rate, including the one being solved for. A units check catches this instantly: the left side is a length times a speed and the right is a bare length.
Matching
Every changing letter gets a rate.
Match the pairs
Why: The first has zero on the right because 100 is a genuine constant; the fourth has a rate there because s changes. Comparing those two rows is the sharpest test of whether the constant-versus-variable distinction has landed.
Two truths and a lie
All three are about the differentiation step.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it is fatal rather than merely inefficient. Substituting an instant's value turns a variable into a constant, whose derivative is zero, so the rate you were solving for disappears entirely and the equation becomes zero equals zero.
Prediction
Commit before reasoning.
Predict first
Differentiating the sphere's volume with respect to time gives dV/dt equals what times dr/dt?
Correct: Four pi r squared — the surface area.
\[ \frac{d}{dr}\left[\tfrac43\pi r^{3}\right] = 4\pi r^{2} = \text{surface area} \]
Why: The power rule brings the 3 down and reduces the exponent, and the four thirds cancels against it to leave 4 pi r squared. That this equals the sphere's surface area is not a coincidence: growing the radius by a small amount adds a thin shell over the whole surface, whose volume is the area times the thickness. The same relationship holds for a circle, where differentiating the area gives the circumference.
Section
Section 3
Concept
When a relation contains two changing quantities but only one rate is known, geometry often links them. Substituting that link before differentiating leaves a single variable and a solvable equation.
reduction by geometry — Using a fixed proportion, usually from similar triangles, to express one changing quantity in terms of another. This substitution is legitimate before differentiating because the proportion holds at every instant.
\[ \frac{r}{h} = \frac{5}{10} \;\Longrightarrow\; r = \frac{h}{2} \;\text{ at every instant} \]
This is the one substitution that IS allowed before differentiating, and the reason is exactly the constant-versus-instantaneous distinction: the proportion is true at every moment, not just the one asked about.
Figure (svg): A conical tank filling, with the similar-triangles relation that eliminates one variable
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 302-305 — problems with several variables
Picture it
A conical tank, with the water's radius tied to its depth.
Figure (svg): A conical tank filling, with the similar-triangles relation that eliminates one variable
The tank's proportions are fixed, so the water's surface radius is always half its depth. Substituting that turns a volume formula in two variables into one in a single variable.
Worked example
Example 4.4. Eliminate first, then differentiate.
\[ \text{Water fills a cone of radius 5 and depth 10 at } 20 \text{ ft}^{3}\text{/min. How fast is the depth rising at } h = 4? \]
Write the volume relation
Why: It contains two variables.
\[ V = (\frac{1}{3}) \pi r ^{2} h \]
Use similar triangles to link them
Why: The proportion holds at every instant.
\[ \frac{r}{h} = \frac{5}{10},\text{ so } r = \frac{h}{2} \]
Substitute to eliminate r
Why: Now one variable only.
\[ V = (\frac{1}{3}) \pi(\frac{h}{2}) ^{2} h = (\frac{\pi}{12}) h ^{3} \]
Differentiate with respect to t
Why: Chain rule on the cube.
Substitute the instant's values and solve
Why: At h equal to 4.
\[ 20 = (\frac{\pi}{4}) (16) (d h / \,dt) \]
Figure (svg): A conical tank filling, with the similar-triangles relation that eliminates one variable
\[ \frac{dh}{dt} = \frac{5}{\pi} \approx 1.59 \text{ ft/min} \]
Verify: check the units and what happens as the tank fills
Why: Cubic feet per minute over square feet gives feet per minute, correct for a depth rate. And the factor h squared means the depth rises more slowly as the tank fills: at h equal to 8 the same inflow gives only a quarter the rise, because the surface is four times as wide. That is physically right and is the same area effect as the balloon. Note the similar-triangles substitution was made BEFORE differentiating, which is legitimate because the proportion holds at every instant.
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 303-304
Fill the middle
The cone's volume, after the similar-triangles substitution.
Fill in the blanks
V = \tfrac13\pi\left(\tfrac12___\right)^___h = \frac______}h^___
Why: One third times one quarter gives one twelfth. With a single variable the differentiation produces a single unknown rate, which the given inflow then determines.
Worked example
Checkpoint 4.4. Two substitutions, two different rules.
\[ \text{Explain why } r = h/2 \text{ may be substituted early but } h = 4 \text{ may not.} \]
Examine the proportion
Why: It comes from the tank's fixed shape.
Conclude for it
Why: Substituting loses nothing.
Examine the depth value
Why: It describes one moment only.
Conclude for it
Why: Substituting would freeze a variable.
State the general test
Why: Ask whether it holds at every instant.
Figure (svg): The solution to Worked example why the substitution order differs here shown as a ladder of expressions, one row per legal move
\[ r = \tfrac{h}{2} \text{ always}; \quad h = 4 \text{ only now} \]
Verify: test the rule on the ladder problem
Why: There, the relation x squared plus y squared equals 100 holds at every instant and was used from the start; the values 6 and 8 hold at one instant and waited. The same test decides both problems, and it is the only rule needed: ask whether the statement would still be true a second later. If yes, substitute freely; if no, keep the letters.
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 304-305
Trap
\[ V = \tfrac13\pi r^{2}h \;\Longrightarrow\; \frac{dV}{dt} = \tfrac13\pi\left(2rh\frac{dr}{dt} + r^{2}\frac{dh}{dt}\right) \]
Differentiate without eliminating r
Why: The student keeps both variables.
\[ \text{one equation, TWO unknown rates} \]
The problem gives no information about dr by dt, so this equation cannot be solved as it stands.
\[ r = \tfrac{h}{2} \;\Longrightarrow\; V = \tfrac{\pi}{12}h^{3} \;\Longrightarrow\; \frac{dV}{dt} = \tfrac{\pi}{4}h^{2}\frac{dh}{dt} \]
Use the geometry to eliminate a variable BEFORE differentiating
Why: The proportion holds at every instant, so it may be substituted freely.
Counting unknowns before differentiating is worth the few seconds. One equation can determine one unknown rate, so if two remain, a relation is missing — and for a cone, a similar-triangles relation is almost always what supplies it.
Sorting
Ask whether it holds at every instant.
Sort into buckets
Sort each substitution.
One question decides every case: would this still be true a second later? The cone's proportion would; its current depth would not. That single test replaces any list of rules.
Ranking
A related rates problem with two geometric variables.
Put in order
Why: Steps b and c both happen before the differentiation, which is what distinguishes them from step e. The legitimacy is the same in each case: a statement true at every instant may be used at any point, and one true at a single instant may not.
Prediction
Commit before reasoning.
Predict first
After differentiating you have one equation with two unknown rates. What does that indicate?
Correct: A geometric relation linking the variables has been missed.
\[ \text{2 unknown rates, 1 equation} \;\Longrightarrow\; \text{find another relation} \]
Why: One equation determines one unknown, so two unknown rates means information is missing — and in these problems it is almost always a fixed proportion from the shape, such as similar triangles in a cone or a fixed angle in a triangle. Counting unknowns before differentiating catches it early. Differentiating again would produce second derivatives and make matters worse, and the problem is certainly solvable once the relation is found.
Section
Section 4
Concept
A rate's sign records whether the quantity is growing or shrinking. The problem's wording determines it, and it must be entered correctly — the algebra cannot infer that a tank is draining rather than filling.
sign conventions for rates — A positive rate means the quantity is increasing; negative means decreasing. A quantity described as falling, draining or closing enters the equation with a negative rate even when the wording gives a positive number.
\[ \text{'draining at } 3\text{ L/min'} \;\Longrightarrow\; \frac{dV}{dt} = -3 \]
The answer's sign is then a genuine check. A ladder's top must slide down, a draining tank's level must fall, and an answer with the wrong sign signals an error in the setup rather than the arithmetic.
Figure (svg): The sign convention for rates, with increasing and decreasing quantities distinguished
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 299-306 — interpreting the sign and units
Picture it
How the wording translates.
Figure (svg): The sign convention for rates, with increasing and decreasing quantities distinguished
Three of the four are negative, which is worth noticing: a great many of these problems involve something falling, draining or closing, and the minus sign is easy to leave out.
Worked example
Example 4.5. The sign enters at the setup.
\[ \text{A cylindrical tank of radius } 3 \text{ ft drains at } 12 \text{ ft}^{3}\text{/min. How fast is the level falling?} \]
Record the rate with its sign
Why: Draining means the volume decreases.
Write the relation
Why: A cylinder of fixed radius.
\[ V = 9 \pi h \]
Differentiate with respect to t
Why: The radius is constant here.
Substitute and solve
Why: The negative rate carries through.
\[ -12 = 9 \pi(d h / \,dt) \]
State with units
Why: Feet per minute.
\[ d h / \,dt = -\frac{4}{3 \pi} \]
Figure (svg): The solution to Worked example a draining tank shown as a ladder of expressions, one row per legal move
\[ \frac{dh}{dt} = -\frac{4}{3\pi} \approx -0.42 \text{ ft/min} \]
Verify: check the sign and note what is unusual here
Why: The negative answer says the level is falling, which is what draining means — and had the minus been omitted at the setup, the answer would have claimed a draining tank was filling up. Note also that the rate does not depend on h at all: a cylinder has the same cross-section at every height, so the level falls at a constant rate. That is unlike the cone, where the rate depended on h squared, and the difference is entirely due to the shape.
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 305-306
Sorting
Ask whether the quantity is growing.
Sort into buckets
Sort each described rate.
Three of the five are negative, which is typical of this section. The wording almost always gives a positive magnitude, and translating it into the correct signed rate is a separate step that must be done deliberately.
Worked example
Checkpoint 4.5. The units confirm the quantity.
\[ \text{Verify the units in } \frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt} \text{ for the balloon.} \]
State the units of each factor
Why: Area and a length rate.
\[ \text{cm} ^{2} \times \text{cm} / s \]
Multiply
Why: The centimetres accumulate.
\[ \text{cm} ^{3} / s \]
Compare with the left side
Why: A volume rate.
\[ \text{cm} ^{3} / s,\text{ matching} \]
Note what a mismatch would mean
Why: A wrong power of r, usually.
Figure (svg): The solution to Worked example units as a check shown as a ladder of expressions, one row per legal move
\[ \text{cm}^{2}\cdot\frac{\text{cm}}{\text{s}} = \frac{\text{cm}^{3}}{\text{s}} \]
Verify: test the check by breaking the formula deliberately
Why: Had the factor been 4 pi r instead of 4 pi r squared — a plausible slip — the right side would come out as square centimetres per second, which is an AREA rate rather than a volume rate. The mismatch would be visible before any number was substituted. Units are the cheapest available check on a related rates relation, and they catch exactly the errors that arithmetic checking does not.
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 306-306
Error analysis
A student sets up a draining tank problem.
Annotate
On: \( \text{'drains at } 12\text{ ft}^{3}\text{/min'} \;\Longrightarrow\; \frac{dV}{dt} = 12 \)
The wording supplies the sign and the algebra cannot recover it. Reading each rate aloud as 'increasing' or 'decreasing' before writing it down is what makes this automatic.
Fill the middle
A tank losing twelve cubic feet of water each minute.
Fill in the blanks
\text-12 12 \;\Longrightarrow\; \frac______ = ___
Why: Draining means decreasing, so the rate is negative twelve. The wording gave a positive magnitude, and translating it is a deliberate step the algebra cannot do for you.
Matching
Output units over time.
Match the pairs
Why: Every rate carries the quantity's units divided by time, exactly as Section 3.4 established. Checking that both sides of a differentiated relation carry the same units is the fastest way to catch a wrong power in the geometry.
Prediction
Commit before reasoning.
Predict first
You compute that a draining tank's water level is rising. What has gone wrong?
Correct: A sign was entered wrongly at the setup.
\[ \frac{dV}{dt} = +12 \;\Longrightarrow\; \frac{dh}{dt} > 0: \text{ a filling tank, not a draining one} \]
Why: The algebra faithfully propagates whatever signs it is given, so a physically impossible answer points back to the input rather than the computation. In a draining problem the culprit is nearly always the given volume rate entered as positive. Checking the answer's sign against the physical situation is a complete test of the setup, and it costs nothing — which makes it worth doing on every one of these problems.
Section
Section 5
Concept
A related rates answer usually depends on the instant's values, and that dependence is informative. It says how the rate would change at a different moment, which is often the more useful conclusion.
structural reading — Interpreting the algebraic form of a related rates result rather than only its numerical value, to see how the answer would differ at other instants or for other geometries.
\[ \frac{dh}{dt} = \frac{4}{\pi h^{2}}\frac{dV}{dt} \;\Longrightarrow\; \text{slower as } h \text{ grows} \]
The cone and the balloon both produce rates inversely proportional to an area, and the cylinder produces one independent of height. Comparing them shows the geometry driving the physics.
Figure (svg): A sphere with its radius growing, and the surface area factor that links the two rates
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 300-307 — interpreting the results
Picture it
The balloon, with its shell.
Figure (svg): A sphere with its radius growing, and the surface area factor that links the two rates
In every one of these problems the linking factor is the area over which the change is spread. That is why a bigger balloon's radius grows more slowly and a cylinder's level falls at a constant rate.
Worked example
Example 4.6. The same inflow, three geometries.
\[ \text{For a cylinder, a cone and a sphere, how does the level or radius rate depend on the current size?} \]
Cylinder of radius R
Why: Volume is a constant times h.
Interpret
Why: No h appears.
Cone with r proportional to h
Why: Volume is a constant times h cubed.
\[ d h / \,dt\text{ proportional to } 1 / h ^{2} \]
Interpret
Why: The rate falls as it fills.
Sphere
Why: Volume is a constant times r cubed.
\[ \,dr / \,dt\text{ proportional to } 1 / r ^{2} \]
Figure (svg): The solution to Worked example comparing three shapes shown as a ladder of expressions, one row per legal move
\[ \text{cylinder: constant}; \quad \text{cone and sphere: } \propto \frac{1}{\text{area}} \]
Verify: check the common explanation
Why: In each case the rate is the inflow divided by the area over which it spreads. A cylinder's cross-section never changes, so the level rises steadily; a cone's widens as it fills, so the rise slows; a sphere's surface grows, so the radius grows more slowly. One sentence explains all three, and it also predicts the answer's form before any calculus is done — which is a useful way to check a result rather than merely obtain one.
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 306-307
Matching
The linking factor is always an area.
Match the pairs
Why: The first three are all the same statement — the rate is the inflow divided by a spreading area — with the area constant, growing and growing respectively. The fourth is different: a vanishing denominator rather than a growing one, and it signals the model breaking down rather than a physical fact.
Worked example
Checkpoint 4.6. What the ladder does at the end.
\[ \text{In the ladder problem, what happens to } \frac{dy}{dt} \text{ as the foot approaches } 10 \text{ ft?} \]
Write the general answer
Why: Solve for the wanted rate.
\[ \,dy / \,dt = -(\frac{x}{y}) (\,dx / \,dt) \]
Examine as x approaches 10
Why: Then y approaches 0.
Conclude
Why: The rate grows without bound.
Assess the model
Why: A real ladder would leave the wall.
Figure (svg): The solution to Worked example an answer that becomes unbounded shown as a ladder of expressions, one row per legal move
\[ \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} \;\to\; -\infty \text{ as } y \to 0 \]
Verify: decide whether this is a flaw
Why: The mathematics is correct: if the foot really moved at a constant 2 feet per second all the way out, the top would have to fall arbitrarily fast at the end. What fails is the assumption, since no real ladder stays in contact with the wall under those conditions. Reading the structure of the answer revealed the model's limit, which a single numerical answer at x equal to 6 would have hidden entirely.
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 307-307
Trap
\[ \frac{dh}{dt} = \frac{5}{\pi} \text{ ft/min at } h = 4 \]
Report the number and stop
Why: The student answers the literal question only.
True, but it conceals that this rate applies to one instant and would be four times smaller at twice the depth.
\[ \frac{dh}{dt} = \frac{4}{\pi h^{2}}\frac{dV}{dt} \;\Longrightarrow\; \text{at } h = 4: \; \frac{5}{\pi} \]
Keep the general form, then evaluate
Why: The formula answers every instant at once.
Deriving the general expression costs nothing extra — the substitution is the last step either way — and it turns a single number into an understanding of how the filling behaves throughout. It also makes checks available, such as noticing that the rate must fall as the tank fills.
Fill the middle
The cone's depth rate, solved in general.
Fill in the blanks
\frac2___ = \frac______h^___\frac______ \;\Longrightarrow\; \frac______ = \frac______}}}\frac______
Why: The depth rate is inversely proportional to h squared, so it falls sharply as the tank fills. That structural fact is invisible in a single numerical answer.
Two truths and a lie
All three are about interpreting the answer.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and the physical picture refutes it. As a cone fills, its surface widens, so the same inflow spreads over a larger area and the level rises more SLOWLY. The algebra agrees: the rate is inversely proportional to the depth squared, so at twice the depth it is a quarter as fast.
Prediction
Commit before reasoning.
Predict first
In every one of these filling problems, the volume rate equals an area times a length rate. Why?
Correct: Because adding volume means spreading a thin layer over a surface.
\[ \Delta V \approx A\,\Delta h \;\Longrightarrow\; \frac{dV}{dt} = A\frac{dh}{dt} \]
Why: A small increase in depth or radius adds a layer whose volume is the surface area times the layer's thickness, so dividing by the elapsed time gives the area times the length rate. That is why differentiating a sphere's volume gives its surface area, and a circle's area gives its circumference. The chain rule is the mechanism that produces the factor, but the geometric reason is what makes the result predictable before any differentiation.
Comparison
Fill the blanks. The relation changes; the procedure does not.
Comparison matrix
| Situation | The relation | What links the rates |
|---|---|---|
| Sliding ladder | x^2 + y^2 = 100 | x x' + y y' = 0 |
| Inflating balloon | V = (4/3) pi r^3 | V' = 4 pi r^2 r' |
| Filling cone | V = (pi/12) h^3, after eliminating r | V' = (pi/4) h^2 h' |
| Two cars at right angles | x^2 + y^2 = s^2 | x x' + y y' = s s' |
In every row the linking factor is an area or a length from the geometry, and in every row the substitution of the instant's numbers happens only after that link is established.
Pattern
Given a problem about linked changing quantities.
If two unknown rates remain after step four, a relation was missed in step three — for a cone or a triangle, it is almost always a similar-triangles proportion.
Stewart, Calculus: Early Transcendentals 8e, §3.9 Related Rates §3.9, pp. 245-250
Check
The ladder. Differentiate before substituting.
Check your understanding
A 10 ft ladder's foot moves out at 2 ft/s. How fast is the top falling when the foot is 6 ft out?
Answer: A
Why: With x = 6 and y = 8, the relation 12(2) + 16 y' = 0 gives y' = -1.5.
Check
The balloon. The factor is the surface area.
Check your understanding
Air enters a sphere at 100 cm^3/s. How fast is the radius growing at r = 5?
Answer: A
Why: 100 = 4 pi (25) dr/dt gives dr/dt = 1/pi.
Check
Order of operations. The one rule that matters.
Check your understanding
Why must you differentiate before substituting the instant's values?
Answer: A
Why: Substituting first turns the relation into an arithmetic identity whose derivative is 0 = 0.
Real world
An oil spill spreads as a circular slick of uniform thickness 2 millimetres. Oil escapes from the wreck at 0.5 cubic metres per minute, and a response team needs to know how fast the slick's edge is advancing when its radius is 40 metres.
Discussion prompt
Set the problem up and solve it, then say what the answer's structure tells the team about the next few hours.
Hint: Volume is thickness times area, and the thickness is constant.
Answer:
\[ V = 0.002\,\pi r^{2} \;\Longrightarrow\; \frac{dV}{dt} = 0.004\pi r\frac{dr}{dt} \]
Substituting the inflow and the instant's radius:
\[ 0.5 = 0.004\pi(40)\frac{dr}{dt} \;\Longrightarrow\; \frac{dr}{dt} = \frac{0.5}{0.16\pi} \approx 0.995 \text{ m/min} \]
So the edge is advancing at just under a metre per minute at that moment.
The structure is what the team actually needs. Solving in general gives dr by dt inversely proportional to r, so the advance slows as the slick grows — at 80 metres it will be half as fast, at 160 metres a quarter. The same oil is spreading around an ever longer perimeter.
That has a direct operational consequence: containment booms placed early must be deployed against a fast-moving edge, while later the edge is nearly stationary but the perimeter to be covered is far longer. Reporting only the single figure of 0.995 metres per minute would have hidden the trade-off entirely, which is why the general form is worth deriving even when one instant is asked about.
Commit first
Answer, then rate your confidence honestly.
Predict first
In a related rates problem, when may a number be substituted?
Correct: Only after differentiating, unless the value holds at every instant.
\[ \text{true always: substitute freely}; \quad \text{true now: substitute last} \]
Why: A genuine constant — the ladder's length, the cone's proportions, the slick's thickness — may be substituted at once, because it is the same at every moment and nothing is lost. A value describing one instant must wait, because substituting it turns a variable into a constant and destroys the rate being solved for. Keeping everything as letters is unnecessarily awkward, and substituting everything at the start makes the problem unsolvable.
Explain it
They substituted the instant's values first, got zero equals zero, and think the problem is broken.
Discussion prompt
In four sentences or fewer, explain what happened.
Hint: Ask them what the derivative of a number is.
Answer:
Ask them what the derivative of 36 plus 64 equals 100 could possibly be. It is zero equals zero, because every symbol in it is a fixed number and numbers do not change.
By substituting 6 for x they told the equation that x never changes — but the whole problem is about how fast x is changing. The rates have to be extracted while the quantities are still letters, and the instant's values are only relevant afterwards, when you evaluate the rate relation at that particular moment.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For substitution, ask whether the statement would still be true a second later. For the relation, look for Pythagoras, a volume formula or a similar-triangles proportion — those three cover nearly everything. For elimination, count unknown rates before differentiating and hunt for a relation if there are two. For signs, read each rate aloud as increasing or decreasing before writing it. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the five steps of the procedure with the last two boxed together and a note on why their order matters. Below, work the ladder problem completely: draw the wall and two ladder positions, label x, y and the fixed 10, differentiate the relation, then substitute and solve, marking clearly which numbers entered before the differentiation and which after. Beside it, show what substituting first would have produced. In the middle of the page, draw the cone with its similar triangles, derive r equals h over 2, eliminate r, and solve for the depth rate at h equal to 4 — then write the general form and one sentence on how the rate changes as the tank fills. At the bottom, write the four sign situations from this section with their signs, and beside them the differentiated Pythagorean relation in its tidy form. In a margin, write the one question that decides whether a value may be substituted early.
If your cone answer does not contain h squared in a denominator somewhere, check the elimination step — the whole physical story of a cone filling more slowly as it goes is carried by that h squared.
Recap
Five things, and four of them are about the setup rather than the calculus.
| If you see | Then |
|---|---|
| A value true at every instant | Substitute it freely, even early |
| A value true at one instant | Keep the letter until after differentiating |
| Two unknown rates after differentiating | A geometric relation was missed |
| A cone or a triangle | Look for similar triangles |
| 'Draining', 'falling', 'closing' | The rate is negative |
| A volume rate and a length rate | The linking factor is an area |
| A physically impossible answer | A sign was entered wrongly at the setup |
Section 4.2 stays with the tangent line but asks a different question: not how fast something changes, but how well the tangent stands in for the curve nearby — which turns the derivative into a tool for approximation and for estimating error.
OpenStax Calculus Volume 1, §4.1 Related Rates §4.1, pp. 296-307 — everything on these slides traces back here
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