This deck handles two or more quantities that change in time and are linked by an equation. It gives the master procedure and then works the seven classic problems: the ladder, the ripple, the balloon, the cone tank, the separating vehicles, the streetlight shadow, and the angle of elevation. It targets the four errors that cost the most points - substituting the instant's numbers before differentiating, dropping the chain-rule rate factor, using a relation that holds only at one instant, and getting the sign or the units of a rate wrong.
Subject: Calculus I · 129 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 11
When one thing changes, what else has to change - and how fast?
Objectives
By the end of this deck you can:
Everything in this deck is one idea wearing seven costumes: the chain rule, applied to time.
Warm-up
Discussion prompt
Before we open Related Rates: without looking back, what was the main idea of Implicit and Logarithmic Differentiation, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck shows how to differentiate a curve you cannot solve for y, and how to tame ugly products, quotients, and variable exponents by taking a logarithm first. It targets the four classic errors: differentiating a y term as if it were a constant, skipping the product rule on an x-times-y term, failing to collect every dy/dx term before dividing, and forcing the power rule onto a variable base with a variable exponent.
Section
Part 1
Concept
Blow air into a spherical balloon. The volume grows - but so does the radius, and so does the surface area. They are not free to move independently: fix the volume and the radius is forced.
A related-rates problem hands you the speed of one of those quantities and asks for the speed of another.
related rates — Two or more quantities that both change over time and are tied together by an equation. Knowing how fast one of them changes forces how fast the others change.
Counterexample
Discussion prompt
Blow air into a spherical balloon. The volume grows - but so does the radius, and so does the surface area. They are not free to move independently: fix the volume and the radius is forced.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
A related-rates problem hands you the speed of one of those quantities and asks for the speed of another.
Intuition
Picture two meshed gears. Turn the small one and the big one has to turn too - you do not get to pick its speed. The teeth force a ratio between them.
In a related-rates problem, the linking equation is the teeth. It decides how a speed on one side shows up as a speed on the other.
So the whole job is: find the teeth, then read off the ratio at the instant you care about.
Analogy
Discussion prompt
Explain Two gears, one turn by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Picture two meshed gears. Turn the small one and the big one has to turn too - you do not get to pick its speed. The teeth force a ratio between them.
Concept
You have known this formula since geometry class.
\[ A = \pi r^2 \]
In geometry class the radius is a fixed number. In a related-rates problem the radius is a function of time, which makes the area a function of time as well.
\[ A(t) = \pi\left[r(t)\right]^2 \]
Saying that out loud once, before you touch anything, is what makes the next step legal.
Explain it
Discussion prompt
Explain Time is the hidden variable to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
You have known this formula since geometry class.
Picture it
Animation
Shows: Time is the hidden variable — a rendered Manim animation.
Rendered with Manim.
Takeaway: Every quantity is secretly a function of time. That is why chain rules appear.
Intuition
A related-rates problem hands you a single photograph: the numbers at one instant.
But rates live in the movie, not the photograph. You have to differentiate the relationship that holds through the whole movie, and only then look at the photograph.
That ordering - movie first, photograph second - is the entire game. Almost every lost point in this chapter comes from doing it backwards.
Picture it
Animation
Shows: A photograph versus a movie — a rendered Manim animation.
Rendered with Manim.
Takeaway: Substituting one frame's numbers early freezes the movie into a photograph.
Concept
You apply the time-derivative operator to both sides of the linking equation, treating every letter in it as a function of time.
\[ \frac{d}{dt}\Big[\ \cdot\ \Big] \]
\[ \frac{d}{dt}\left[r^2\right] = 2r\,\frac{dr}{dt} \]
The extra factor on the right is the chain rule at work: the outer power rule, times the derivative of the inside - and the inside is the radius, which depends on time.
That trailing factor is the single most-forgotten piece of notation in first-semester calculus. Guard it.
Concept
Every related-rates problem hands you exactly three things. Find and label all three before you compute anything.
If you cannot name all three in words, you are not ready to differentiate. Most of the difficulty in this topic is finding the third one.
Ranking
Put in order
Put the moves of Warm-up: differentiate three relations with respect to time into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The radius is a function of time, so the power rule on the right produces a rate factor for the radius.
Worked example
Do not solve anything yet. Just apply the time derivative and keep every rate factor.
Differentiate the circle-area relation
Why: The radius is a function of time, so the power rule on the right produces a rate factor for the radius.
\[ A = \pi r^2 \quad\Longrightarrow\quad \frac{dA}{dt} = 2\pi r\,\frac{dr}{dt} \]
Differentiate the sphere-volume relation
Why: Same move, one power higher: the three from the exponent multiplies the constant four thirds down to four, and the radius keeps its rate factor.
\[ V = \frac{4}{3}\pi r^3 \quad\Longrightarrow\quad \frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt} \]
Differentiate a Pythagorean relation with a fixed hypotenuse
Why: Both legs vary, so both earn a rate factor. The right side is a genuine constant, and the derivative of a constant is zero.
\[ x^2 + y^2 = 100 \quad\Longrightarrow\quad 2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0 \]
Check the units on the first result
Why: The left side is a rate of area: square feet per second. The right side is a length times a length per second, which is also square feet per second. The units balance, so no rate factor was lost.
\[ \frac{\text{ft}^2}{\text{s}} \;=\; \text{ft}\cdot\frac{\text{ft}}{\text{s}} \]
Picture it
Animation
Shows: Each line of the worked example "Warm-up: differentiate three relations with respect to time", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The left side is a rate of area: square feet per second. The right side is a length times a length per second, which is also square feet per second. The units balance, so no rate factor was lost.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A circular ripple's radius grows at 3 inches per second. How fast is the area growing when the radius is 4 inches?
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It answers 'how much area per extra inch of radius', not 'how much area per second'.
Same ripple. Keep the chain-rule factor that the time derivative produces.
Why: It answers 'how much area per extra inch of radius', not 'how much area per second'. Look at the units: inches, not square inches per second. The answer is off by exactly the missing factor of 3.
Trap
A circular ripple's radius grows at 3 inches per second. How fast is the area growing when the radius is 4 inches?
\[ \frac{dA}{dt} \stackrel{?}{=} 2\pi r = 2\pi(4) = 8\pi \approx 25.13 \]
This differentiated with respect to the radius, not time
Why: It answers 'how much area per extra inch of radius', not 'how much area per second'. Look at the units: inches, not square inches per second. The answer is off by exactly the missing factor of 3.
Same ripple. Keep the chain-rule factor that the time derivative produces.
\[ \frac{dA}{dt} = 2\pi r\,\frac{dr}{dt} = 2\pi(4)(3) = 24\pi \]
Report with units
Why: The area is growing at twenty-four pi, about 75.40 square inches per second. Square inches per second is a rate of area, which is what was asked for.
Notation
Annotate
From Trap: dropping the rate factor — read this one piece at a time. What is each part doing?
On: \( \frac{dA}{dt} \stackrel{?}{=} 2\pi r = 2\pi(4) = 8\pi \approx 25.13 \)
Concept
A rate is a derivative, and a derivative is a ratio of units. Write the units down as you name each rate.
| Quantity | Its rate | Typical units |
|---|---|---|
| Radius | dr/dt | ft per s |
| Area | dA/dt | sq ft per s |
| Volume | dV/dt | cu ft per min |
| Height | dh/dt | ft per min |
| Angle | d(theta)/dt | rad per s |
Units are a free error-checker. If the units of your answer are wrong, either a rate factor is missing or an extra one crept in.
Pattern
Step through it
Step through Every rate carries units one row at a time. What is driving the change, and what would the row after the last one be?
Intuition
The problem statement tells you exactly which derivative is which. Learn the phrasebook and the setup writes itself.
| The problem says | It means |
|---|---|
| is increasing at 3 ft per second | that rate is positive 3 |
| is decreasing at 3 ft per second | that rate is negative 3 |
| is pumped in at 2 cubic ft per min | the volume rate is positive 2 |
| is leaking out at 2 cubic ft per min | the volume rate is negative 2 |
| how fast is ... changing | solve for that derivative |
| at the instant when ... | these numbers go in AFTER differentiating |
Circle the phrase 'at the instant when'. It marks the numbers you are not allowed to use early.
Comparison
Comparison matrix
From Translating the words into derivatives: refill the It means column from what you know. The rest of the table is as it appeared.
| The problem says | It means |
|---|---|
| is increasing at 3 ft per second | that rate is positive 3 |
| is decreasing at 3 ft per second | that rate is negative 3 |
| is pumped in at 2 cubic ft per min | the volume rate is positive 2 |
| is leaking out at 2 cubic ft per min | the volume rate is negative 2 |
| how fast is ... changing | solve for that derivative |
| at the instant when ... | these numbers go in AFTER differentiating |
Concept
A rate is positive when the quantity is growing and negative when it is shrinking. Nothing else is encoded in the sign.
\[ \frac{dh}{dt} = -0.4 \ \text{ft/min} \]
That reads: the height is falling at four tenths of a foot per minute. The magnitude is the speed, the sign is the direction.
When you write the final sentence, translate the sign into a word - rising, falling, growing, shrinking, approaching, separating. A grader wants the word.
Picture it
Animation
Shows: The sign of a rate is a direction — a rendered Manim animation.
Rendered with Manim.
Takeaway: Read the sign back into the story before you write the final sentence.
Concept
Some quantities in the picture genuinely never change: a ladder's length, a tank's shape, the fixed distance from an observer to a launch pad.
Those are constants and their derivatives are zero. Anything that could be different one second later is a variable and must carry a rate.
\[ \frac{d}{dt}\left[100\right] = 0 \qquad \text{but} \qquad \frac{d}{dt}\left[x^2\right] = 2x\,\frac{dx}{dt} \]
The most common setup mistake in the whole chapter is demoting a variable to a constant just because you happen to know its value at one instant.
Section
Part 2
Constraint
Discussion prompt
Run The six-step related-rates procedure with this step confiscated:
Eliminate extra variables. If a variable has no known rate and is tied to another by a ratio that holds at every instant, substitute it out now.
Is it still possible? If it is, say what takes its place and what it costs you. If it is not, say exactly what that step was providing that nothing else does.
Hint: A step you can drop for free was never load-bearing. If you cannot drop it, name the thing that goes wrong the moment it is gone.
Answer:
Pattern
Step five before step six, always. Steps one and four are where nearly all the thinking lives; step five is mechanical.
Edge cases
Discussion prompt
The six-step related-rates procedure works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
Step five before step six, always. Steps one and four are where nearly all the thinking lives; step five is mechanical.
Picture it
Animation
Shows: The related-rates procedure with the substitute-last step highlighted.
Step four is where problems are won or lost.
Takeaway: Name the quantities, relate them, differentiate with respect to time, and substitute the instant's values LAST. Substituting early turns a variable into a constant and destroys the problem.
Concept
This is the rule the whole deck rests on, so here is the reason in one line.
Substituting a number for a variable tells the derivative that the variable never changes. Its rate collapses to zero, and the information you needed is destroyed.
\[ r = 5 \ \text{(too early)} \;\Longrightarrow\; A = 25\pi \;\Longrightarrow\; \frac{dA}{dt} = 0 \]
The instant's numbers describe one frame of the film. Differentiation needs the film.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A circular ripple's radius grows at 2 feet per second. Find how fast the area is growing at the instant the radius is 5 feet.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The area is now the fixed number twenty-five pi, and the derivative of a constant is zero.
Same ripple. Leave the radius as a letter and differentiate the relation first.
Why: The area is now the fixed number twenty-five pi, and the derivative of a constant is zero.
Trap
A circular ripple's radius grows at 2 feet per second. Find how fast the area is growing at the instant the radius is 5 feet.
\[ A = \pi r^2 = \pi(5)^2 = 25\pi \]
Plugged in the radius before differentiating
Why: The area is now the fixed number twenty-five pi, and the derivative of a constant is zero.
\[ \frac{dA}{dt} = \frac{d}{dt}\left[25\pi\right] = 0 \]
The answer claims a spreading ripple has a frozen area. An answer of zero is almost always this mistake.
Same ripple. Leave the radius as a letter and differentiate the relation first.
\[ A = \pi r^2 \quad\Longrightarrow\quad \frac{dA}{dt} = 2\pi r\,\frac{dr}{dt} \]
Now substitute the instant's values
Why: The differentiated relation is true at every instant, so it is safe to freeze the radius at 5 feet only at this point.
\[ \frac{dA}{dt} = 2\pi(5)(2) = 20\pi \approx 62.83 \ \text{ft}^2/\text{s} \]
Say it in words
Why: The area of the ripple is increasing at about 62.83 square feet per second at that instant.
Break the constraint
Discussion prompt
The rule this trap just fixed:
The differentiated relation is true at every instant, so it is safe to freeze the radius at 5 feet only at this point.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
The area is now the fixed number twenty-five pi, and the derivative of a constant is zero.
Step zero
Discussion prompt
The spreading slick, start to finish — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Name the variables and their rates
Answer:
Worked example
An oil slick spreads in a circle. Its radius is increasing at 3 meters per minute. How fast is the area of the slick increasing at the instant the radius is 25 meters?
Name the variables and their rates
Why: Both the radius and the area change with time. The radius rate is given; the area rate is what the question wants.
\[ \frac{dr}{dt} = 3 \ \text{m/min}, \qquad \frac{dA}{dt} = \ ? \]
Write the linking equation
Why: The slick is a circle, so its area is completely determined by its radius. This is the equation that holds at every instant.
\[ A = \pi r^2 \]
Differentiate both sides with respect to time
Why: Power rule on the outside times the derivative of the inside; the inside is the radius, a function of time, so it contributes its rate.
\[ \frac{dA}{dt} = 2\pi r\,\frac{dr}{dt} \]
Substitute the instant's values
Why: Only now is it safe to freeze the radius at 25 meters, because the relation above already accounts for the fact that it moves.
\[ \frac{dA}{dt} = 2\pi(25)(3) = 150\pi \]
State the answer with units and a direction
Why: The area of the slick is increasing at one hundred fifty pi square meters per minute.
\[ \frac{dA}{dt} = 150\pi \approx 471.24 \ \text{m}^2/\text{min} \]
Check the size of the answer against the picture
Why: The edge of the slick is about 157 meters around, and it pushes outward 3 meters each minute, so the new ring of oil is roughly 157 times 3, about 471 square meters per minute. That is exactly the answer, so the magnitude and units are right.
\[ 2\pi(25)\cdot 3 = 150\pi \approx 471.24 \]
Picture it
Animation
Shows: Each line of the worked example "The spreading slick, start to finish", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The edge of the slick is about 157 meters around, and it pushes outward 3 meters each minute, so the new ring of oil is roughly 157 times 3, about 471 square meters per minute. That is exactly the answer, so the magnitude and units are right.
Elimination
Eliminate the wrong options
A pebble makes a circular ripple whose radius grows at 4 meters per minute. How fast is the enclosed area growing at the instant the radius is 10 meters?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Differentiate the area relation with respect to time to get dA/dt equal to 2 pi r times dr/dt, then substitute the radius 10 and the rate 4: two times pi times 10 times 4 is 80 pi, about 251.33 square meters per minute.
Check
Set it up on paper before you pick. Name the rates, differentiate, then substitute.
Check your understanding
A pebble makes a circular ripple whose radius grows at 4 meters per minute. How fast is the enclosed area growing at the instant the radius is 10 meters?
Answer: A
Why: Differentiate the area relation with respect to time to get dA/dt equal to 2 pi r times dr/dt, then substitute the radius 10 and the rate 4: two times pi times 10 times 4 is 80 pi, about 251.33 square meters per minute.
Concept
A related-rates answer is a sentence, not a bare number. It has three parts: the quantity, the direction, and the rate with units.
The area of the slick is increasing at about 471 square meters per minute.
Engineers and graders read the same sentence. A number with no units and no direction is not an answer to the question that was asked.
Section
Part 3
Picture it
Figure (svg): A right triangle with horizontal leg x, vertical leg y, and hypotenuse z, with the right angle marked
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Whenever two sides of a right triangle are changing, the Pythagorean theorem is your linking equation.
Concept
Whenever two sides of a right triangle are changing, the Pythagorean theorem is your linking equation.
Figure (svg): A right triangle with horizontal leg x, vertical leg y, and hypotenuse z, with the right angle marked
\[ x^2 + y^2 = z^2 \]
Differentiating with respect to time gives a relation among all three rates. The twos cancel, which makes the working form very clean.
\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \quad\Longrightarrow\quad x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt} \]
If the hypotenuse is a fixed length - a ladder, a rope, a rod - then its rate is zero and the right side vanishes.
Picture it
Animation
Shows: The Pythagorean link — a rendered Manim animation.
Rendered with Manim.
Takeaway: Distance problems almost always hand you this equation.
Picture it
Figure (svg): A ladder leaning against a vertical wall, with x the distance from the wall to the ladder's foot and y the height of its top
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
A 10-foot ladder leans against a vertical wall. The bottom of the ladder is pulled away from the wall at 1 foot per second. How fast is the top sliding down the wall at the instant the bottom is 6 feet from the wall?
Worked example
A 10-foot ladder leans against a vertical wall. The bottom of the ladder is pulled away from the wall at 1 foot per second. How fast is the top sliding down the wall at the instant the bottom is 6 feet from the wall?
Figure (svg): A ladder leaning against a vertical wall, with x the distance from the wall to the ladder's foot and y the height of its top
Name the variables and rates
Why: Let the distance from the wall to the foot of the ladder be one variable and the height of the top be another. Both change; the ladder's length does not.
\[ \frac{dx}{dt} = 1 \ \text{ft/s}, \qquad \frac{dy}{dt} = \ ? \]
Write the linking equation
Why: The wall, the ground, and the ladder form a right triangle whose hypotenuse is the ladder, and the ladder cannot stretch.
\[ x^2 + y^2 = 10^2 = 100 \]
Find the missing side at the instant
Why: The question gives only the base, so use the Pythagorean theorem to get the height that goes with it. Store this number for step six, not for step five.
\[ y = \sqrt{100 - 6^2} = \sqrt{64} = 8 \ \text{ft} \]
Differentiate both sides with respect to time
Why: Both legs are functions of time so both earn rate factors. The one hundred is a true constant because the ladder's length never changes, so the right side is zero.
\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \]
Solve for the wanted rate before plugging in
Why: Isolating the unknown rate first keeps the algebra clean and makes the minus sign impossible to lose.
\[ \frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt} \]
Substitute the instant's values
Why: With the base at 6 feet, the height at 8 feet, and the base rate at 1 foot per second, the top is moving at negative three quarters of a foot per second - that is, sliding down at three quarters of a foot per second.
\[ \frac{dy}{dt} = -\frac{6}{8}(1) = -\frac{3}{4} \ \text{ft/s} \]
Verify by substituting back into the differentiated equation
Why: The two terms must cancel exactly, and they do. The negative sign also matches the picture: as the foot slides out, the top must come down.
\[ 2(6)(1) + 2(8)\left(-\tfrac{3}{4}\right) = 12 - 12 = 0 \ \checkmark \]
Reverse engineer
Discussion prompt
Work backwards. The example finished here:
Verify by substituting back into the differentiated equation
What was it asked to do, and what must it have been given? Reconstruct the problem from its answer.
Hint: Every quantity in the result had to enter somewhere. Account for each one.
Answer:
A 10-foot ladder leans against a vertical wall. The bottom of the ladder is pulled away from the wall at 1 foot per second. How fast is the top sliding down the wall at the instant the bottom is 6 feet from the wall?
Intuition
The bottom of the ladder moves at a steady 1 foot per second. The top does not move at a steady anything.
| base (ft) | height (ft) | rate of the top (ft/s) |
|---|---|---|
| 1.0 | 9.95 | -0.10 |
| 6.0 | 8.00 | -0.75 |
| 8.0 | 6.00 | -1.33 |
| 9.6 | 2.80 | -3.43 |
As the ladder flattens, the height shrinks toward zero, and the height sits in the denominator of the rate formula. Small denominator, huge rate.
So there is no such thing as 'the speed of the top'. A related-rates answer always belongs to one instant, and the problem always tells you which one.
Pattern
Step through it
Step through Why the top races down at the end one row at a time. What is driving the change, and what would the row after the last one be?
Anomaly
Predict first
A student writes this, and it looks reasonable:
Same ladder: 10 feet long, the foot pulled away at 1 foot per second, currently 6 feet from the wall.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This says the top of the ladder climbs the wall at three quarters of a foot per second while the foot slides away - the ladder would have to be growing.
Differentiate the true equation and leave both variable terms on the same side, where the constant sends the other side to zero.
Why: This says the top of the ladder climbs the wall at three quarters of a foot per second while the foot slides away - the ladder would have to be growing.
Trap
Same ladder: 10 feet long, the foot pulled away at 1 foot per second, currently 6 feet from the wall.
\[ 2x\frac{dx}{dt} \stackrel{?}{=} 2y\frac{dy}{dt} \]
Moved the second term across as if the two rates had the same sign
Why: This says the top of the ladder climbs the wall at three quarters of a foot per second while the foot slides away - the ladder would have to be growing.
\[ \frac{dy}{dt} = \frac{x}{y}\cdot\frac{dx}{dt} = \frac{6}{8}(1) = +\frac{3}{4} \ \text{ft/s} \]
Differentiate the true equation and leave both variable terms on the same side, where the constant sends the other side to zero.
\[ x^2 + y^2 = 100 \;\Longrightarrow\; 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \]
Solve and keep the sign the algebra gives you
Why: Because the two terms must add to zero, one of the rates has to be negative. As the base grows, the height must shrink: the top slides down at three quarters of a foot per second.
\[ \frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt} = -\frac{6}{8}(1) = -\frac{3}{4} \ \text{ft/s} \]
Translation
\( 2x\frac{dx}{dt} \stackrel{?}{=} 2y\frac{dy}{dt} \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Prediction
Predict first
A 13-foot ladder leans against a wall. Its foot is pulled away from the wall at 2 feet per second. How fast is the top moving at the instant the foot is 5 feet from the wall?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: Down at 5/6 of a foot per second
Why: When the foot is 5 feet out the height is the square root of 169 minus 25, which is 12. Differentiating gives the top's rate as the negative of the base over the height times the base rate: negative 5 over 12 times 2, which is negative 5 sixths of a foot per second, so the top is sliding down.
Check
Find the missing leg first, then differentiate, then substitute.
Check your understanding
A 13-foot ladder leans against a wall. Its foot is pulled away from the wall at 2 feet per second. How fast is the top moving at the instant the foot is 5 feet from the wall?
Answer: A
Why: When the foot is 5 feet out the height is the square root of 169 minus 25, which is 12. Differentiating gives the top's rate as the negative of the base over the height times the base rate: negative 5 over 12 times 2, which is negative 5 sixths of a foot per second, so the top is sliding down.
Concept
When two objects travel along perpendicular paths, the distance between them is the hypotenuse - and now all three sides are variables.
\[ z^2 = x^2 + y^2 \quad\Longrightarrow\quad z\frac{dz}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt} \]
Pick what each letter measures once and never change it. The usual choice is: each letter is that object's distance to the crossing point.
With that choice, an object driving toward the crossing point has a shrinking distance, so its rate is negative even though it is moving forward.
Estimation
Predict first
Car A is 0.3 miles west of an intersection driving east toward it at 50 miles per hour. Car B is 0.4 miles south of the same intersection driving north toward it at 60 miles per hour. How fast is the distance between the cars changing at that instant?
Commit before you compute: what does Two cars closing on an intersection come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the sign and the size
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Negative is right: both cars are approaching, so the gap must shrink.
Worked example
Car A is 0.3 miles west of an intersection driving east toward it at 50 miles per hour. Car B is 0.4 miles south of the same intersection driving north toward it at 60 miles per hour. How fast is the distance between the cars changing at that instant?
Let each letter be a distance to the intersection
Why: Both cars are approaching, so both of those distances are shrinking, which makes both given rates negative.
\[ \frac{dx}{dt} = -50 \ \text{mi/h}, \qquad \frac{dy}{dt} = -60 \ \text{mi/h} \]
Write the linking equation and find the current separation
Why: The two roads are perpendicular, so the cars and the intersection form a right triangle at every instant. At this instant the separation is the square root of 0.09 plus 0.16.
\[ z^2 = x^2 + y^2, \qquad z = \sqrt{0.3^2 + 0.4^2} = 0.5 \ \text{mi} \]
Differentiate both sides with respect to time
Why: All three lengths change, so all three earn rate factors. Dividing out the common factor of two leaves the working form.
\[ z\frac{dz}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt} \]
Substitute the instant's values
Why: Each car's contribution is its distance times its rate; both are negative because both distances are shrinking.
\[ 0.5\,\frac{dz}{dt} = (0.3)(-50) + (0.4)(-60) = -15 - 24 = -39 \]
Solve and state the answer
Why: The distance between the cars is decreasing at 78 miles per hour at that instant - they are closing on each other.
\[ \frac{dz}{dt} = \frac{-39}{0.5} = -78 \ \text{mi/h} \]
Verify the sign and the size
Why: Negative is right: both cars are approaching, so the gap must shrink. And 78 sits between the faster single speed, 60, and the sum of the speeds, 110, which is exactly where a closing rate has to land when the cars are not driving straight at each other.
Picture it
Animation
Shows: Each line of the worked example "Two cars closing on an intersection", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Negative is right: both cars are approaching, so the gap must shrink. And 78 sits between the faster single speed, 60, and the sum of the speeds, 110, which is exactly where a closing rate has to land when the cars are not driving straight at each other.
Intuition
Each car eats into the gap, but only the part of its motion aimed along the line joining them counts.
Rewriting the differentiated relation shows the formula doing that projection for you.
\[ \frac{dz}{dt} = \frac{x}{z}\cdot\frac{dx}{dt} + \frac{y}{z}\cdot\frac{dy}{dt} \]
Each weight is a fraction between zero and one, so no single car ever contributes its full speed unless it is heading straight at the other.
In the last example those weights were six tenths and eight tenths: thirty plus forty-eight gives seventy-eight.
Check
Both cars start at the same point at the same time, so first find where each one is after two hours.
Check your understanding
Two cars leave an intersection at the same moment: one drives east at 30 miles per hour, the other drives north at 40 miles per hour. How fast is the distance between them increasing two hours later?
Answer: A
Why: After two hours the cars are 60 and 80 miles out, so the separation is 100 miles. The differentiated Pythagorean relation gives 100 times the separation rate equal to 60 times 30 plus 80 times 40, which is 5000, so the rate is 50 miles per hour.
Section
Part 4
Concept
For a sphere, a single number controls the whole object: the radius.
\[ V = \frac{4}{3}\pi r^3, \qquad S = 4\pi r^2 \]
\[ \frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt} \]
Read that out loud: the volume rate equals the surface area times the radius rate. Adding volume is laying a thin shell of new material over the entire surface.
Picture it
Animation
Shows: The sphere: one variable decides everything — a rendered Manim animation.
Rendered with Manim.
Takeaway: Surface area appears on its own — that is why the radius slows as the balloon grows.
Step zero
Discussion prompt
The inflating balloon — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Name the rates
Answer:
Worked example
Air is pumped into a spherical balloon at 100 cubic centimeters per second. How fast is the radius increasing at the instant the radius is 25 centimeters?
Name the rates
Why: The pump gives the volume rate; the question wants the radius rate. Air is going in, so the volume rate is positive.
\[ \frac{dV}{dt} = 100 \ \text{cm}^3/\text{s}, \qquad \frac{dr}{dt} = \ ? \]
Write the linking equation
Why: The balloon stays spherical, so its volume is determined by its radius at every instant.
\[ V = \frac{4}{3}\pi r^3 \]
Differentiate both sides with respect to time
Why: The three from the exponent cancels the three in the denominator, and the radius contributes its rate factor.
\[ \frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt} \]
Substitute and solve
Why: Now freeze the radius at 25 centimeters. The surface area there is four pi times 625, or 2500 pi square centimeters.
\[ 100 = 4\pi(25)^2\,\frac{dr}{dt} = 2500\pi\,\frac{dr}{dt} \]
State the answer
Why: The radius is increasing at one over twenty-five pi centimeters per second, about 0.0127 centimeters per second - slow, because that air has a lot of surface to cover.
\[ \frac{dr}{dt} = \frac{100}{2500\pi} = \frac{1}{25\pi} \approx 0.0127 \ \text{cm/s} \]
Verify by putting the answer back into the differentiated relation
Why: Multiplying the surface area by the radius rate must return the pump rate, and it does: the pi terms cancel and 2500 over 25 is 100.
\[ 2500\pi\cdot\frac{1}{25\pi} = 100 \ \text{cm}^3/\text{s} \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "The inflating balloon", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Multiplying the surface area by the radius rate must return the pump rate, and it does: the pi terms cancel and 2500 over 25 is 100.
Intuition
Pump at a steady rate and the radius shoots up at first, then crawls. Nothing about the pump changed - the geometry did.
| radius (cm) | surface area (sq cm) | radius rate (cm/s) |
|---|---|---|
| 5 | 314.2 | 0.3183 |
| 10 | 1256.6 | 0.0796 |
| 25 | 7854.0 | 0.0127 |
| 50 | 31415.9 | 0.0032 |
The same cubic centimeter of air has to spread over a larger and larger skin, so it lifts the surface by less and less.
Double the radius and the surface area quadruples, so the radius rate drops to a quarter. The table shows exactly that.
Pattern
Step through it
Step through Why the balloon's radius slows down one row at a time. What is driving the change, and what would the row after the last one be?
Hypothesis
Predict first
The melting snowball is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Turn the words into signed rates
Why: Decreasing means negative. The question asks about the diameter, not the radius, so keep track of which one you are solving for.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
A snowball melts so that its surface area decreases at 1 square centimeter per minute. How fast is the diameter decreasing at the instant the diameter is 10 centimeters?
Turn the words into signed rates
Why: Decreasing means negative. The question asks about the diameter, not the radius, so keep track of which one you are solving for.
\[ \frac{dS}{dt} = -1 \ \text{cm}^2/\text{min}, \qquad \frac{dD}{dt} = \ ? \]
Write the surface-area relation and differentiate it
Why: Surface area is determined by the radius, so start there and convert to the diameter at the end.
\[ S = 4\pi r^2 \quad\Longrightarrow\quad \frac{dS}{dt} = 8\pi r\,\frac{dr}{dt} \]
Substitute the instant's radius
Why: A diameter of 10 centimeters means a radius of 5 centimeters, so the coefficient is eight pi times 5, or forty pi.
\[ -1 = 8\pi(5)\,\frac{dr}{dt} = 40\pi\,\frac{dr}{dt} \]
Solve for the radius rate
Why: Negative, as it must be: a melting snowball shrinks.
\[ \frac{dr}{dt} = -\frac{1}{40\pi} \ \text{cm/min} \]
Convert to the diameter rate
Why: The diameter is twice the radius at every instant, so its rate is twice the radius rate. This is the step the question was really testing.
\[ D = 2r \;\Longrightarrow\; \frac{dD}{dt} = 2\frac{dr}{dt} = -\frac{1}{20\pi} \approx -0.0159 \ \text{cm/min} \]
Verify against the given rate
Why: Put the radius rate back into the differentiated surface relation: eight pi times 5 times negative one over forty pi is negative 1 square centimeter per minute, which is exactly what the problem said. The diameter is shrinking at about 0.0159 centimeters per minute.
\[ 8\pi(5)\left(-\frac{1}{40\pi}\right) = -1 \ \text{cm}^2/\text{min} \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "The melting snowball", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Put the radius rate back into the differentiated surface relation: eight pi times 5 times negative one over forty pi is negative 1 square centimeter per minute, which is exactly what the problem said. The diameter is shrinking at about 0.0159 centimeters per minute.
Check
Watch the sign of the given rate, and watch which power of the radius shows up after differentiating.
Check your understanding
Air escapes from a spherical balloon at 50 cubic centimeters per minute. How fast is the radius changing at the instant the radius is 10 centimeters?
Answer: A
Why: Differentiating the sphere-volume relation gives the volume rate as four pi r squared times the radius rate. With the volume rate at negative 50 and the radius at 10, the coefficient is 400 pi, so the radius rate is negative 50 over 400 pi, which is negative 1 over 8 pi, about negative 0.0398 centimeters per minute.
Concept
Water sitting in an upright cone-shaped tank has both a depth and a surface radius, and both of them are changing as the tank fills.
\[ V = \frac{1}{3}\pi r^2 h \]
You are given one rate and asked for one rate, but this equation contains two unknown rates. Differentiating right now would leave you with one equation and two unknowns.
You need one more fact about the geometry before you differentiate.
Picture it
Animation
Shows: A cone volume rewritten in one variable before differentiating.
One variable is one chain rule.
Takeaway: Use the shape's fixed proportions to eliminate a variable before differentiating. Two variables means two chain rules and usually an error.
Picture it
Figure (svg): An inverted cone-shaped tank of top radius R and height H, partly filled with water of surface radius r and depth h
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
However full the tank is, the water itself forms a cone that is similar to the tank - same angle at the point, same proportions.
Intuition
Figure (svg): An inverted cone-shaped tank of top radius R and height H, partly filled with water of surface radius r and depth h
However full the tank is, the water itself forms a cone that is similar to the tank - same angle at the point, same proportions.
Similar cones have the same radius-to-height ratio, and the tank's own numbers hand it to you.
\[ \frac{r}{h} = \frac{R}{H} \]
This ratio is true at every instant, not just the one in the question. That is exactly what makes it legal to use before differentiating.
Picture it
Animation
Shows: The ladder relation differentiated with respect to time.
The relation holds at every instant, so it can be differentiated.
Takeaway: The equation relating the quantities is true at every instant, which is exactly what licenses differentiating both sides with respect to time.
Concept
Use the fixed ratio to rewrite the volume in terms of the one variable you actually care about.
\[ r = \frac{R}{H}\,h \quad\Longrightarrow\quad V = \frac{1}{3}\pi\left(\frac{R}{H}h\right)^2 h = \frac{\pi R^2}{3H^2}\,h^3 \]
Now the relation has one variable, and differentiating produces exactly one rate factor - the one you were asked about.
The alternative, differentiating with both variables and then hunting down the second rate, does work. It is just twice the algebra and twice the chances to slip.
The test for legality is simple: does this relation hold at every instant, or only at the one in the question? Only the first kind may be used early.
Ranking
Put in order
Put the moves of Filling a cone-shaped tank into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The inflow gives the volume rate; the question asks for the rate of the depth.
Worked example
A tank shaped like a cone with its point down is 16 feet tall and 4 feet in radius at the top. Water flows in at 2 cubic feet per minute. How fast is the water level rising at the instant the water is 8 feet deep?
Name the rates
Why: The inflow gives the volume rate; the question asks for the rate of the depth.
\[ \frac{dV}{dt} = 2 \ \text{ft}^3/\text{min}, \qquad \frac{dh}{dt} = \ ? \]
Use similar cones to eliminate the surface radius
Why: The tank's radius-to-height ratio is 4 to 16, so the water's radius is always one quarter of its depth. This holds while the tank fills, not just at one moment.
\[ \frac{r}{h} = \frac{4}{16} = \frac{1}{4} \quad\Longrightarrow\quad r = \frac{h}{4} \]
Rewrite the volume with depth alone
Why: Substituting removes the second variable, so the differentiated equation will have only the rate you want.
\[ V = \frac{1}{3}\pi\left(\frac{h}{4}\right)^2 h = \frac{\pi h^3}{48} \]
Differentiate both sides with respect to time
Why: Power rule on the cube, times the depth's rate factor. The three over forty-eight reduces to one sixteenth.
\[ \frac{dV}{dt} = \frac{3\pi h^2}{48}\,\frac{dh}{dt} = \frac{\pi h^2}{16}\,\frac{dh}{dt} \]
Substitute the instant's depth and solve
Why: At a depth of 8 feet the coefficient is pi times 64 over 16, which is four pi.
\[ 2 = \frac{\pi(8)^2}{16}\,\frac{dh}{dt} = 4\pi\,\frac{dh}{dt} \quad\Longrightarrow\quad \frac{dh}{dt} = \frac{1}{2\pi} \]
State the answer with units
Why: The water level is rising at one over two pi feet per minute, about 0.159 feet per minute, at that instant.
\[ \frac{dh}{dt} = \frac{1}{2\pi} \approx 0.159 \ \text{ft/min} \]
Verify with the two-variable form of the relation
Why: At a depth of 8 the radius is 2 and its rate is one quarter of the depth rate, or one over eight pi. Feeding those into the product-rule version of the cone volume returns 2 cubic feet per minute, the given inflow.
\[ \frac{1}{3}\pi\!\left[2rh\frac{dr}{dt} + r^2\frac{dh}{dt}\right] = \frac{\pi}{3}\!\left[\frac{32}{8\pi} + \frac{4}{2\pi}\right] = \frac{\pi}{3}\cdot\frac{6}{\pi} = 2 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Filling a cone-shaped tank", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At a depth of 8 the radius is 2 and its rate is one quarter of the depth rate, or one over eight pi. Feeding those into the product-rule version of the cone volume returns 2 cubic feet per minute, the given inflow.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Same tank: 16 feet tall, 4 feet in radius at the top, filling at 2 cubic feet per minute, water currently 8 feet deep. At that depth the surface radius is 2 feet, so put the 2 in now.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The radius equals 2 feet at this one moment; a second later it is not 2.
Use the similar-cones ratio instead. That relation is true the whole time the tank is filling, so it survives differentiation.
Why: The radius equals 2 feet at this one moment; a second later it is not 2. Freezing it kills the radius rate and inflates the answer by a factor of three.
Trap
Same tank: 16 feet tall, 4 feet in radius at the top, filling at 2 cubic feet per minute, water currently 8 feet deep. At that depth the surface radius is 2 feet, so put the 2 in now.
\[ V = \frac{1}{3}\pi(2)^2 h = \frac{4\pi}{3}h \]
Differentiated a relation that is only true at one instant
Why: The radius equals 2 feet at this one moment; a second later it is not 2. Freezing it kills the radius rate and inflates the answer by a factor of three.
\[ 2 = \frac{4\pi}{3}\,\frac{dh}{dt} \;\Longrightarrow\; \frac{dh}{dt} = \frac{3}{2\pi} \approx 0.477 \ \text{ft/min} \]
Use the similar-cones ratio instead. That relation is true the whole time the tank is filling, so it survives differentiation.
\[ r = \frac{h}{4} \;\Longrightarrow\; V = \frac{\pi h^3}{48} \]
Differentiate, then substitute the depth
Why: Now the radius is still allowed to grow, and the answer comes out three times smaller: about 0.159 feet per minute.
\[ 2 = \frac{\pi h^2}{16}\,\frac{dh}{dt} \;\Longrightarrow\; \frac{dh}{dt} = \frac{1}{2\pi} \approx 0.159 \ \text{ft/min} \]
Notation
Annotate
From Trap: freezing a variable with a one-instant fact — read this one piece at a time. What is each part doing?
On: \( 2 = \frac{\pi h^2}{16}\,\frac{dh}{dt} \;\Longrightarrow\; \frac{dh}{dt} = \frac{1}{2\pi} \approx 0.159 \ \text{ft/min} \)
Commit first
Predict first
A tank shaped like a cone with its point down is 8 meters tall and 2 meters in radius at the top. Water flows in at 3 cubic meters per minute. How fast is the water level rising when the water is 4 meters deep?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: 3 over pi meters per minute
Why: The ratio of radius to depth is 2 to 8, so the radius is one quarter of the depth and the volume is pi times depth cubed over 48. Differentiating gives the volume rate as pi times depth squared over 16 times the depth rate; at a depth of 4 that coefficient is pi, so the depth rate is 3 over pi, about 0.955 meters per minute.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
Get the radius-to-depth ratio from the tank's own dimensions first, and be careful which height goes into the final substitution.
Check your understanding
A tank shaped like a cone with its point down is 8 meters tall and 2 meters in radius at the top. Water flows in at 3 cubic meters per minute. How fast is the water level rising when the water is 4 meters deep?
Answer: A
Why: The ratio of radius to depth is 2 to 8, so the radius is one quarter of the depth and the volume is pi times depth cubed over 48. Differentiating gives the volume rate as pi times depth squared over 16 times the depth rate; at a depth of 4 that coefficient is pi, so the depth rate is 3 over pi, about 0.955 meters per minute.
Section
Part 5
Picture it
Figure (svg): A streetlight of height 15 on the left, a shorter person standing at distance x, and the light ray from the lamp over the person's head down to the shadow tip
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
A streetlight, a person walking under it, and the person's shadow make two nested right triangles with exactly the same angles.
Concept
A streetlight, a person walking under it, and the person's shadow make two nested right triangles with exactly the same angles.
Figure (svg): A streetlight of height 15 on the left, a shorter person standing at distance x, and the light ray from the lamp over the person's head down to the shadow tip
Similar triangles means matching sides are in the same ratio. Set the two ratios equal and you have an equation in the two changing lengths.
\[ \frac{\text{lamp height}}{\text{lamp to shadow tip}} = \frac{\text{person height}}{\text{shadow length}} \]
The two heights are constants here. Only the two horizontal lengths vary, so only they earn rate factors.
Picture it
Animation
Shows: Similar triangles give the linking equation — a rendered Manim animation.
Rendered with Manim.
Takeaway: The shape's fixed proportions are the constraint you were missing.
Intuition
'How fast is the shadow lengthening?' and 'how fast is the tip of the shadow moving?' are different questions with different answers.
The shadow's length is measured from the person's feet, so it depends only on the shadow.
The tip's position is measured from the lamppost, so it is the person's distance plus the shadow's length. The tip moves faster, because the person is walking too.
Decide which one the problem is asking for before you differentiate anything. Circle the word 'tip' or the word 'length' in the problem.
Step zero
Discussion prompt
The streetlight shadow — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Name the two changing lengths
Answer:
Worked example
A streetlight is 15 feet tall. A person 6 feet tall walks away from its base at 5 feet per second. How fast is the shadow lengthening, and how fast is the tip of the shadow moving?
Name the two changing lengths
Why: Let the person's distance from the lamppost be one variable and the shadow's length be another. Both heights are fixed, so they are constants.
\[ \frac{dx}{dt} = 5 \ \text{ft/s}, \qquad \frac{ds}{dt} = \ ? \]
Set up similar triangles
Why: The big triangle runs from the top of the lamp to the shadow tip; the small one from the top of the person's head to the same tip. Same angles, so matching sides are proportional.
\[ \frac{15}{x+s} = \frac{6}{s} \]
Clear the fractions and simplify first
Why: Cross-multiplying and collecting the shadow terms gives a much simpler relation than the original proportion, and it holds at every instant.
\[ 15s = 6x + 6s \;\Longrightarrow\; 9s = 6x \;\Longrightarrow\; s = \tfrac{2}{3}x \]
Differentiate with respect to time
Why: Two thirds is a constant multiplier, so the shadow's rate is two thirds of the walking rate.
\[ \frac{ds}{dt} = \tfrac{2}{3}\cdot\frac{dx}{dt} = \tfrac{2}{3}(5) = \frac{10}{3} \approx 3.33 \ \text{ft/s} \]
Now answer the second question: the tip
Why: The tip's distance from the lamppost is the person's distance plus the shadow's length, so its rate is the sum of the two rates.
\[ p = x + s = \tfrac{5}{3}x \;\Longrightarrow\; \frac{dp}{dt} = \tfrac{5}{3}(5) = \frac{25}{3} \approx 8.33 \ \text{ft/s} \]
Verify the proportion at a sample instant
Why: When the person is 6 feet from the lamp the shadow is 4 feet and the tip is 10 feet out. The two ratios agree at 1.5, so the relation is right. The rates also agree: 5 plus ten thirds is twenty-five thirds.
\[ \frac{15}{10} = 1.5 = \frac{6}{4}, \qquad 5 + \frac{10}{3} = \frac{25}{3} \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "The streetlight shadow", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: When the person is 6 feet from the lamp the shadow is 4 feet and the tip is 10 feet out. The two ratios agree at 1.5, so the relation is right. The rates also agree: 5 plus ten thirds is twenty-five thirds.
Anomaly
Predict first
A student writes this, and it looks reasonable:
How fast is the tip of the shadow moving? The 15-foot lamp, the 6-foot walker, 5 feet per second.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The shadow's length is measured from the person's feet, and the person's feet are moving.
Measure the tip from the fixed lamppost, which is the only thing in the picture that is not moving.
Why: The shadow's length is measured from the person's feet, and the person's feet are moving. The tip's speed cannot be smaller than the walker's own 5 feet per second, so ten thirds is impossible on its face.
Trap
How fast is the tip of the shadow moving? The 15-foot lamp, the 6-foot walker, 5 feet per second.
\[ s = \tfrac{2}{3}x \;\Longrightarrow\; \frac{ds}{dt} = \frac{10}{3} \ \text{ft/s} \]
Reported the shadow's length rate as the tip's speed
Why: The shadow's length is measured from the person's feet, and the person's feet are moving. The tip's speed cannot be smaller than the walker's own 5 feet per second, so ten thirds is impossible on its face.
Measure the tip from the fixed lamppost, which is the only thing in the picture that is not moving.
\[ p = x + s = x + \tfrac{2}{3}x = \tfrac{5}{3}x \]
Differentiate the tip's position
Why: The tip advances at the walker's 5 feet per second plus the shadow's ten thirds, giving twenty-five thirds, about 8.33 feet per second - faster than the walker, as it must be.
\[ \frac{dp}{dt} = \tfrac{5}{3}(5) = \frac{25}{3} \approx 8.33 \ \text{ft/s} \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Check
Set up the proportion, simplify it, then decide whether the question wants the shadow or its tip.
Check your understanding
A lamppost is 20 feet tall. A person 5 feet tall walks away from its base at 4 feet per second. How fast is the tip of the person's shadow moving along the ground?
Answer: A
Why: Similar triangles give 20 over the quantity distance plus shadow equals 5 over shadow, so 15 times the shadow equals 5 times the distance and the shadow is one third of the distance. The tip is therefore four thirds of the distance, and its rate is four thirds of 4, which is 16 thirds, about 5.33 feet per second.
Concept
When the question asks how fast an angle is turning, the linking equation is a trig ratio instead of a geometry formula. Everything else is identical.
\[ \tan\theta = \frac{\text{opposite}}{\text{adjacent}} \]
\[ \frac{d}{dt}\left[\tan\theta\right] = \sec^2\theta\,\frac{d\theta}{dt} \]
The angle earns a rate factor exactly like every other variable. Choose the trig function whose two sides are the ones you know something about.
Picture it
Animation
Shows: Angles change in time too — a rendered Manim animation.
Rendered with Manim.
Takeaway: Radians always — the trig derivatives assume it.
Intuition
The derivative formulas for sine, cosine, and tangent are only true when the angle is measured in radians. In degrees they are wrong by a constant factor.
So an angular rate that falls out of a related-rates problem is automatically in radians per second. Convert to degrees per second only at the very end, and only if the question asks.
\[ 1 \ \text{rad} = \frac{180^{\circ}}{\pi} \approx 57.3^{\circ} \]
A useful sanity check: angular rates in these problems are usually small numbers, because one radian is a large turn.
Estimation
Predict first
A balloon rises straight up at 5 feet per second from a launch point 100 feet from an observer on the ground. How fast is the observer's angle of elevation increasing at the instant the balloon is 100 feet high?
Commit before you compute: what does Angle of elevation to a rising balloon come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with the geometry of the line of sight
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. An angular rate is the speed across the line of sight divided by the distance along it.
Worked example
A balloon rises straight up at 5 feet per second from a launch point 100 feet from an observer on the ground. How fast is the observer's angle of elevation increasing at the instant the balloon is 100 feet high?
Name the variable and the constant
Why: The balloon's height changes; the 100 feet of ground between the observer and the launch point never does. That constant is what makes the tangent the right choice.
\[ \frac{dh}{dt} = 5 \ \text{ft/s}, \qquad \frac{d\theta}{dt} = \ ? \]
Write the trig relation
Why: The height is the opposite side and the fixed 100 feet is the adjacent side, so the tangent of the elevation angle links them.
\[ \tan\theta = \frac{h}{100} \]
Differentiate both sides with respect to time
Why: The left side needs the chain rule for the tangent, the right side is one hundredth of the height rate.
\[ \sec^2\theta\,\frac{d\theta}{dt} = \frac{1}{100}\cdot\frac{dh}{dt} \]
Find the trig value at the instant
Why: When the height equals the ground distance, the tangent is 1 and the angle is forty-five degrees, so the secant squared is 2.
\[ \tan\theta = 1 \;\Longrightarrow\; \theta = \frac{\pi}{4} \;\Longrightarrow\; \sec^2\theta = 2 \]
Substitute and solve
Why: Two times the angular rate equals five hundredths, so the angular rate is one fortieth of a radian per second, about 1.43 degrees per second.
\[ 2\,\frac{d\theta}{dt} = \frac{5}{100} \;\Longrightarrow\; \frac{d\theta}{dt} = \frac{1}{40} = 0.025 \ \text{rad/s} \]
Verify with the geometry of the line of sight
Why: An angular rate is the speed across the line of sight divided by the distance along it. Here the balloon's velocity makes a forty-five degree angle with the line of sight, so the crossing speed is 5 times the cosine of forty-five degrees, and the line of sight is 100 times the square root of 2 long. The ratio is 0.025 radians per second, matching.
\[ \frac{5\cos\left(\pi/4\right)}{100\sqrt{2}} = \frac{3.5355}{141.42} = 0.025 \ \text{rad/s} \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Angle of elevation to a rising balloon", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: An angular rate is the speed across the line of sight divided by the distance along it. Here the balloon's velocity makes a forty-five degree angle with the line of sight, so the crossing speed is 5 times the cosine of forty-five degrees, and the line of sight is 100 times the square root of 2 long. The ratio is 0.025 radians per second, matching.
Section
Part 6
Concept
If the quantity you want is a product of two things that are both changing, differentiating in time needs the product rule.
\[ A = LW \quad\Longrightarrow\quad \frac{dA}{dt} = \frac{dL}{dt}\,W + L\,\frac{dW}{dt} \]
Each term is one factor holding still while the other moves. Add the two contributions and you have the total rate.
The tempting shortcut - multiplying the two rates together - is never right, and it does not even have the right units.
Step zero
Discussion prompt
A rectangle with both sides changing — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Turn the words into signed rates
Answer:
Worked example
A rectangle's length is increasing at 3 centimeters per second while its width is decreasing at 2 centimeters per second. How fast is the area changing at the instant the length is 20 centimeters and the width is 10 centimeters?
Turn the words into signed rates
Why: Increasing is positive, decreasing is negative. That single sign decides whether the final answer is growth or shrinkage.
\[ \frac{dL}{dt} = 3 \ \text{cm/s}, \qquad \frac{dW}{dt} = -2 \ \text{cm/s} \]
Differentiate the area relation with the product rule
Why: Both factors are functions of time, so neither one can be treated as a constant.
\[ A = LW \quad\Longrightarrow\quad \frac{dA}{dt} = \frac{dL}{dt}\,W + L\,\frac{dW}{dt} \]
Substitute the instant's values
Why: The lengthening adds 30 square centimeters per second while the narrowing removes 40, so the two effects fight and the narrowing wins.
\[ \frac{dA}{dt} = (3)(10) + (20)(-2) = 30 - 40 = -10 \ \text{cm}^2/\text{s} \]
State the answer
Why: The area is decreasing at 10 square centimeters per second at that instant, even though one side is getting longer.
\[ \frac{dA}{dt} = -10 \ \text{cm}^2/\text{s} \]
Verify with a small step forward in time
Why: After one hundredth of a second the sides are about 20.03 and 9.98, giving an area of 199.8994 instead of 200. That is a change of about negative 0.1006 over 0.01 seconds, or about negative 10.06 square centimeters per second - the derivative, to rounding.
\[ \frac{(20.03)(9.98) - 200}{0.01} = \frac{-0.1006}{0.01} \approx -10.06 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "A rectangle with both sides changing", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: After one hundredth of a second the sides are about 20.03 and 9.98, giving an area of 199.8994 instead of 200. That is a change of about negative 0.1006 over 0.01 seconds, or about negative 10.06 square centimeters per second - the derivative, to rounding.
Concept
The method never used the fact that the equations came from shapes. Any equation linking quantities that vary in time behaves the same way.
\[ R = pq \quad\Longrightarrow\quad \frac{dR}{dt} = \frac{dp}{dt}\,q + p\,\frac{dq}{dt} \]
Revenue is price times quantity sold. If the price is falling while sales are rising, that is the sliding ladder wearing a business suit.
Boyle's law for a compressed gas, resistances in parallel, and the concentration of a mixing tank are all related-rates problems too.
Explain it
Discussion prompt
Explain Nothing here is really about geometry to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
The method never used the fact that the equations came from shapes. Any equation linking quantities that vary in time behaves the same way.
Ranking
Put in order
Put the moves of A compressed gas: Boyle's law into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Boyle's law says the product is a fixed number for this sample.
Worked example
A sample of gas is compressed at constant temperature, so the product of its pressure and its volume stays constant. At the instant the pressure is 100 kilopascals and the volume is 600 cubic centimeters, the pressure is increasing at 20 kilopascals per minute. How fast is the volume changing?
Write the linking equation
Why: Boyle's law says the product is a fixed number for this sample. Both factors are variables; only their product is constant.
\[ PV = C \]
Differentiate both sides with respect to time
Why: Product rule on the left because both factors move; the derivative of the constant on the right is zero.
\[ \frac{dP}{dt}\,V + P\,\frac{dV}{dt} = 0 \]
Solve for the wanted rate before substituting
Why: Isolating first keeps the minus sign visible, and the minus sign is the physics: squeezing harder must shrink the volume.
\[ \frac{dV}{dt} = -\frac{V}{P}\cdot\frac{dP}{dt} \]
Substitute the instant's values
Why: Six hundred over one hundred is 6, and 6 times 20 is 120, with a minus sign in front.
\[ \frac{dV}{dt} = -\frac{600}{100}(20) = -120 \ \text{cm}^3/\text{min} \]
Verify that the two product-rule terms cancel
Why: They must sum to zero because the product is constant, and they do: 12000 minus 12000. The volume is decreasing at 120 cubic centimeters per minute.
\[ (20)(600) + (100)(-120) = 12000 - 12000 = 0 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "A compressed gas: Boyle's law", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: They must sum to zero because the product is constant, and they do: 12000 minus 12000. The volume is decreasing at 120 cubic centimeters per minute.
Concept
Before you circle an answer, run this list. It costs almost no time and it catches almost every lost point.
An answer of exactly zero, or an answer whose units are one factor of time off, is almost always one of the last two questions failing.
Analogy
Discussion prompt
Explain Sanity checks that take ten seconds by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Before you circle an answer, run this list. It costs almost no time and it catches almost every lost point.
Pattern
| Symptom in your answer | What went wrong | The fix |
|---|---|---|
| It came out zero | You substituted the instant's values before differentiating | Differentiate first, substitute last |
| Units are missing a per-second | A chain-rule rate factor was dropped | Every variable gets its own rate factor |
| The sign is backwards | A shrinking quantity was given a positive rate | Decreasing means the rate is negative |
| Off by a fixed multiple | A variable was frozen using a relation true at only one instant | Only relations true at every instant may be used early |
| Correct number, wrong question | Shadow length instead of shadow tip, radius instead of diameter | Reread the last sentence of the problem |
The middle column is the diagnosis. If you can name which of the five you made, you will not repeat it.
Trade off
Comparison matrix
From The five classic errors and how to spot them: every row here is a choice with a cost. Fill the The fix column, then say which row you would actually pick and what you give up for it.
| Symptom in your answer | What went wrong | The fix |
|---|---|---|
| It came out zero | You substituted the instant's values before differentiating | Differentiate first, substitute last |
| Units are missing a per-second | A chain-rule rate factor was dropped | Every variable gets its own rate factor |
| The sign is backwards | A shrinking quantity was given a positive rate | Decreasing means the rate is negative |
| Off by a fixed multiple | A variable was frozen using a relation true at only one instant | Only relations true at every instant may be used early |
| Correct number, wrong question | Shadow length instead of shadow tip, radius instead of diameter | Reread the last sentence of the problem |
Prediction
Predict first
Which step contains the FIRST error in this solution?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: Step 2
Why: Step 2 substitutes the instant's radius before differentiating, which turns the volume into the fixed number five hundred pi over three and destroys the radius rate. Everything after that is a correct consequence of a broken step. The right route is to differentiate first, getting the volume rate as four pi r squared times the radius rate, and only then put in the radius 5.
Check
Here is a student's solution to this problem: air is pumped into a spherical balloon at 100 cubic centimeters per second; find how fast the radius is increasing at the instant the radius is 5 centimeters.
\[ \textbf{Step 1:}\quad V = \tfrac{4}{3}\pi r^3 \]
\[ \textbf{Step 2:}\quad r = 5 \;\Longrightarrow\; V = \tfrac{4}{3}\pi(125) = \tfrac{500\pi}{3} \]
\[ \textbf{Step 3:}\quad \frac{dV}{dt} = \frac{d}{dt}\left[\tfrac{500\pi}{3}\right] = 0 \]
\[ \textbf{Step 4:}\quad \text{so } \frac{dr}{dt} = 0 \]
Check your understanding
Which step contains the FIRST error in this solution?
Answer: B
Why: Step 2 substitutes the instant's radius before differentiating, which turns the volume into the fixed number five hundred pi over three and destroys the radius rate. Everything after that is a correct consequence of a broken step. The right route is to differentiate first, getting the volume rate as four pi r squared times the radius rate, and only then put in the radius 5.
Elimination
Eliminate the wrong options
Which sentence correctly reports that result?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The negative sign says the height is decreasing and the magnitude 0.4 is the speed, so the level is falling at four tenths of a foot per minute. The units, feet per minute, confirm that the quantity being described is a height and not a volume.
Check
A tank is draining. At one instant a student correctly computes the rate of change of the water's height as negative four tenths of a foot per minute.
Check your understanding
Which sentence correctly reports that result?
Answer: A
Why: The negative sign says the height is decreasing and the magnitude 0.4 is the speed, so the level is falling at four tenths of a foot per minute. The units, feet per minute, confirm that the quantity being described is a height and not a volume.
Intuition
Write the relationship that is true at every instant, differentiate it with respect to time, and only then look at the single instant the problem asks about.
Every problem in this deck - ladder, ripple, balloon, snowball, cone, cars, shadow, angle, rectangle, gas - obeyed that one sentence and nothing else.
The variety is in finding the linking equation. The calculus never changes.
Counterexample
Discussion prompt
Write the relationship that is true at every instant, differentiate it with respect to time, and only then look at the single instant the problem asks about.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Every problem in this deck - ladder, ripple, balloon, snowball, cone, cars, shadow, angle, rectangle, gas - obeyed that one sentence and nothing else.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Rates That Travel Together · The Master Procedure · Pythagorean Problems · Volumes That Fill and Empty · Shadows, Triangles, and Angles · Products, Signs, and Sanity. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
| Situation | Linking equation | Differentiated form |
|---|---|---|
| Spreading circle | area equals pi times radius squared | dA/dt = 2 pi r dr/dt |
| Ladder or two perpendicular paths | x squared plus y squared equals z squared | x dx/dt + y dy/dt = z dz/dt |
| Sphere filling or melting | volume equals four thirds pi radius cubed | dV/dt = 4 pi r squared dr/dt |
| Cone tank | radius is a fixed fraction of depth | dV/dt = (pi R squared / H squared) h squared dh/dt |
| Shadow or similar triangles | matching sides in equal ratio | one rate is a constant multiple of the other |
| Angle of elevation | tangent of angle equals opposite over adjacent | sec squared times d(theta)/dt |
| Product of two changing amounts | quantity equals first times second | product rule in time |
Next up: linear approximation and differentials, which take the same derivative and use it to estimate values instead of rates.
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