Related Rates Problems in Calculus: Step-by-Step Examples

A ladder is sliding down a wall. The bottom slides away from the wall at a nice, steady 2 feet per second.

Is the top of the ladder also sliding down at a steady pace?

Your gut says yes. Your gut is wrong. The top speeds up the closer the ladder gets to lying flat on the ground, even though the bottom never changes speed at all. By the end, the top is falling almost instantly, even faster than free fall.

That's related rates. Two things connected by a rule, where knowing the speed of one tells you the speed of the other, but the relationship between those speeds is rarely as simple as you'd guess.

What ties two changing things together?

Picture two people holding opposite ends of a rigid pole, one walking along the ground, one climbing a ladder. They're not directly connected by a rope pulling at a fixed rate. They're connected by geometry, by the fact that the pole has a fixed length.

Move one person, and the pole forces the other to move too. Not necessarily at the same speed. Not even in the same pattern. Just... consistently, according to whatever shape the constraint takes.

That's the whole idea behind related rates. We have two (or more) quantities that change over time, tied together by an equation that's always true. Not just at one moment, at every moment. And that equation is the "pole": rigid, fixed, unbreakable. If we know how fast one quantity moves, the equation tells us how fast the other one has to move to keep up.

But here's the thing, we almost never know either quantity as an explicit function of time. We don't have a formula for "height of the ladder top at time $t$ ." What we have is a relationship, and a rate. So how do we get from one to the other?

The ladder problem, step by step

Let's go back to that ladder. It's 10 feet long, leaning against a wall. Let $x$ be the distance from the wall to the base, and $y$ be the height of the top. Because the ladder has fixed length, Pythagoras gives us the rule that's always true:

x^2 + y^2 = 100

Now here's the trick. We don't solve for $y$ and then find its rate separately. We differentiate the whole equation with respect to time, treating $x$ and $y$ as functions of $t$ we haven't written down. Every term picks up a rate:

2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0

That's the chain rule doing its job. $x^2$ isn't just " $x$ squared", it's " $x(t)$ squared", so differentiating it gives $2x \cdot \frac{dx}{dt}$ , not just $2x$ .

Drag the base of the ladder. The base always slides away at a fixed 2 ft/s, but watch what happens to the top's speed as the base gets close to the wall.

dx/dt (base)2.0 ft/s

dy/dt (top)-1.50 ft/s

Height y8.00 ft

Drag the base around and watch the readout. When the base is close to the wall, $y$ is large and $x$ is small, so $\frac{dy}{dt}$ stays modest. But as $x$ grows and $y$ shrinks toward zero, that same formula, $\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}$ , blows up. A constant push at the bottom produces a wildly changing speed at the top.

Worked out at a specific moment: if the base is 6 ft from the wall (so the top is 8 ft up, since $6^2+8^2=100$ ) and the base moves out at 2 ft/s:

\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} = -\frac{6}{8}(2) = -1.5 \text{ ft/s}

The top is falling at 1.5 ft/s at that instant. Nowhere did we need a formula for $y(t)$ . We only needed the constraint, and one derivative.

The general recipe

Every related rates problem follows the same five moves, and once you see them as a checklist, the "trick" disappears:

Draw the picture. Label everything that changes with a variable.
Write the constraint. Find the one equation connecting your variables that's true at every instant, not just right now.
Differentiate with respect to time. Every variable becomes $\frac{d(\cdot)}{dt}$ via the chain rule. Constants disappear, as usual.
Plug in the instant you care about. Only after differentiating. Plugging in numbers too early is the single most common mistake, it locks the equation into one snapshot before you've extracted the general relationship between rates.
Solve for the rate you want.

Notice step 4's warning: plug in numbers after differentiating, never before. If you substitute $x=6$ into $x^2+y^2=100$ before differentiating, you get a single number, and derivatives of constants are zero. You'd lose the relationship entirely.

A case where area outpaces radius

Here's a different kind of constraint. A circular oil spill spreads across the surface of the water. Its radius grows steadily. Does its area grow steadily too?

A = \pi r^2 \quad \Rightarrow \quad \frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}

Same move as before: differentiate the whole equation, chain rule turns $r^2$ into $2r \cdot \frac{dr}{dt}$ .

Drag the edge of the circle to set its radius. The pump always grows the radius at 1 unit/s, but notice how fast the area piles on as the circle gets bigger.

dr/dt1.0 u/s

Area28.27 u²

dA/dt18.85 u²/s

Drag the edge outward. The radius grows at a fixed 1 unit per second the whole time, that part never changes. But $\frac{dA}{dt}$ depends on $r$ itself, so the bigger the spill already is, the faster its area piles on. Doubling the radius doubles the rate the area grows, even though the radius itself is growing no faster than before.

Worked example: the spill's radius grows at 3 m/min. How fast is the area increasing when $r=10$ m?

\frac{dA}{dt} = 2\pi(10)(3) = 60\pi \approx 188.5 \text{ m}^2/\text{min}

The one where the rate stays constant

Now for the interesting part: not every related rates problem gives you a changing rate. Sometimes the geometry conspires to keep things constant, and spotting why is the real skill.

A 6-foot-tall person walks away from a 15-foot lamppost. The lamp casts their shadow further and further across the ground. The triangle formed by the lamp, the person, and the shadow's tip is similar to the smaller triangle formed by the person and their own shadow, so:

\frac{15}{x+s} = \frac{6}{s} \quad \Rightarrow \quad s = \frac{6}{9}x = \frac{2}{3}x

where $x$ is the person's distance from the pole and $s$ is the shadow's length. Differentiate:

\frac{ds}{dt} = \frac{2}{3}\frac{dx}{dt}

Drag the person away from the lamppost. Watch the shadow, then check the readout below: does the tip's speed change as you drag?

dx/dt (walking)5.0 ft/s

ds/dt (shadow growth)3.33 ft/s

Tip speed (constant!)8.33 ft/s

Drag the person along the sidewalk and watch the tip-speed readout. Before you look at the number, guess: does the shadow's tip speed up as the person walks further from the light, the way the ladder's top did?

...it doesn't. The tip moves at a constant speed, no matter where the person stands. The reason is right there in the formula: $\frac{ds}{dt} = \frac{2}{3}\frac{dx}{dt}$ has no $x$ or $s$ left in it after differentiating, just constants. The similar-triangle constraint is linear in $x$ and $s$ , so its rate never depends on position. Compare that to the ladder's constraint, $x^2+y^2=100$ , which is quadratic, that's exactly why its rate depended on position and this one doesn't.

Worked example: walking at 5 ft/s, the shadow grows at $\frac{2}{3}(5) = 3.33$ ft/s, so the tip itself (person plus shadow) moves at $5 + 3.33 = 8.33$ ft/s, constant the whole walk.

The tank that fills unevenly

One more, and this one's the trickiest because the constraint is cubic. A conical tank, point down, is 12 feet tall with a 4-foot radius at the top rim. A pump fills it at a constant 5 cubic feet per minute. Does the water level rise at a constant rate too?

The tank's shape forces the radius at any water height $h$ to be $\frac{h}{3}$ (similar triangles again, cone-shaped this time), so the volume of water at height $h$ is:

V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{h}{3}\right)^2 h = \frac{\pi}{27}h^3

Differentiate with respect to time:

\frac{dV}{dt} = \frac{\pi}{9}h^2\,\frac{dh}{dt} \quad \Rightarrow \quad \frac{dh}{dt} = \frac{9}{\pi h^2}\frac{dV}{dt}

Drag the waterline, or hit play to fill the tank at a constant pump rate. Watch the level's speed as it passes the narrow bottom versus the wide top.

dV/dt (pump)5.0 ft³/min

dh/dt (level)1.59 ft/min

Volume3.14 ft³

Hit play and watch the waterline. Near the narrow point at the bottom, a small amount of water raises the level a lot, so $\frac{dh}{dt}$ is large. As the tank widens near the top, the same pump has to fill a much bigger cross-section to raise the level by the same amount, so $\frac{dh}{dt}$ shrinks. A constant pump produces a level that rises fast, then visibly slows down, exactly the opposite pattern from the ladder, where a constant push produced a rate that sped up.

Worked example: at $h=3$ ft, with the pump running at 5 ft³/min:

\frac{dh}{dt} = \frac{9(5)}{\pi(3)^2} = \frac{45}{9\pi} = \frac{5}{\pi} \approx 1.59 \text{ ft/min}

The short version

Related rates problems all reduce to the same three steps: find the equation that's always true, differentiate it with respect to time using the chain rule, then plug in the instant you care about. The chain rule is what turns " $x^2$ " into " $2x \cdot \frac{dx}{dt}$ ", carrying the unknown rate along for the ride even though you never wrote $x$ as a function of $t$ .

What changes from problem to problem is the shape of the constraint. A quadratic constraint, like the ladder's $x^2+y^2=100$ , gives you a rate that swings wildly depending on where you are. A linear constraint, like the shadow's similar triangles, gives you a rate that never changes at all. A cubic constraint, like the cone-shaped tank, gives you a rate that shrinks as the shape widens. The algebra is nearly identical every time. The behavior it produces depends entirely on the geometry you started with.

All visualizations are interactive React components running entirely in your browser, using exact closed-form formulas for each constraint, drag and play interactions included. No libraries beyond React.