What Are Related Rates?
For example, if air is being pumped into a spherical balloon, the volume and the radius are both changing — and they are related by the formula V = (4/3)πr³. If you know how fast the volume is increasing (dV/dt), you can find how fast the radius is increasing (dr/dt) at any given moment.
The word "related" refers to the fact that the rates dV/dt and dr/dt are connected through the geometric relationship between V and r. The key tool is implicit differentiation with respect to time. You take a known equation relating the quantities, differentiate every term with respect to t (using the chain rule), and then solve for the unknown rate.
Where they show up
Related rates problems appear throughout Calculus I and are a staple of AP Calculus AB and BC exams. They also have real-world applications in physics (velocity and acceleration), engineering (fluid flow), and biology (growth rates).
The 5-Step Method
Every related rates problem can be solved with the same systematic approach.
- Step 1 — Draw a diagram.
Sketch the situation and label all quantities that change with time using variables (not fixed numbers). This is the most important step because it prevents the most common mistakes.
- Step 2 — Identify what you know and what you want.
Write down every given rate (
dx/dt,dV/dt, etc.) and clearly state which rate you are solving for. Note the specific instant at which you need the answer. - Step 3 — Write an equation relating the variables.
Find a geometric, physical, or algebraic equation that connects the changing quantities. This equation should not contain any rates yet — it relates the quantities themselves (lengths, areas, volumes, angles, etc.). If the equation has too many variables, use a constraint to eliminate one before differentiating.
- Step 4 — Differentiate both sides with respect to time
t.Apply the chain rule to every term. For example,
d/dt[r²] = 2r · (dr/dt). This step converts the static equation into a rate equation. - Step 5 — Substitute known values and solve.
Plug in all known quantities and rates at the specific instant, then solve for the unknown rate. Never substitute specific numbers before differentiating — you need the general rate equation first.
Following these five steps methodically will get you through any related rates problem, no matter how complicated it looks.
Ladder Sliding Down a Wall
A 10-foot ladder leans against a vertical wall. The base of the ladder slides away from the wall at 2 ft/s. How fast is the top of the ladder sliding down the wall when the base is 6 feet from the wall?
Step 1: Draw a right triangle. The ladder is the hypotenuse (10 ft, constant). Let x = distance from the base of the ladder to the wall, and y = height of the top of the ladder on the wall.
Step 2: Given dx/dt = 2 ft/s. Find dy/dt when x = 6.
Step 3: By the Pythagorean theorem, x² + y² = 10² = 100.
Step 4: Differentiate with respect to t: 2x(dx/dt) + 2y(dy/dt) = 0.
Step 5: When x = 6, find y: y² = 100 − 36 = 64, so y = 8. Substitute: 2(6)(2) + 2(8)(dy/dt) = 0. Simplify: 24 + 16(dy/dt) = 0, so dy/dt = −24/16 = −3/2 ft/s.
The negative sign means the top is sliding down, which makes physical sense.
The top of the ladder slides down at 1.5 ft/sNotice that we found y at the specific instant after differentiating, not before. This is crucial — substituting y = 8 into the original equation before differentiating would have eliminated y entirely and made it impossible to find dy/dt.
Inflating Balloon Problem
Air is pumped into a spherical balloon at a rate of 100 cm³/s. How fast is the radius increasing when the radius is 5 cm?
Step 1: Draw a sphere with radius r.
Step 2: Given dV/dt = 100 cm³/s. Find dr/dt when r = 5 cm.
Step 3: The volume of a sphere is V = (4/3)πr³.
Step 4: Differentiate with respect to t: dV/dt = 4πr²(dr/dt).
Step 5: Substitute dV/dt = 100 and r = 5: 100 = 4π(25)(dr/dt) = 100π(dr/dt). Solve: dr/dt = 100/(100π) = 1/π ≈ 0.318 cm/s.
The radius is increasing at 1/π cm/s (about 0.318 cm/s) when r = 5 cmAn interesting observation: as the balloon gets larger, the same volume of air causes a smaller increase in radius. This makes sense geometrically — the surface area grows as r², so the same volume increment is spread over a larger surface. If you re-solve at r = 10, you get dr/dt = 100/(400π) = 1/(4π), which is four times slower. Related rates problems often reveal these non-obvious physical insights.
Cone Filling with Water
Water pours into an inverted cone (point down) at 3 m³/min. The cone has a height of 10 m and a top radius of 5 m. How fast is the water level rising when the water is 4 m deep?
Step 1: Draw the inverted cone. Let r = radius of the water surface and h = depth of the water.
Step 2: Given dV/dt = 3 m³/min. Find dh/dt when h = 4 m.
Step 3: The volume of a cone is V = (1/3)πr²h. This equation has three variables (V, r, h), but we only want dV/dt and dh/dt. We need to eliminate r. By similar triangles, the cone's proportions give r/h = 5/10 = 1/2, so r = h/2. Substitute: V = (1/3)π(h/2)²h = (1/3)π(h²/4)h = πh³/12.
Step 4: Differentiate: dV/dt = (3πh²/12)(dh/dt) = (πh²/4)(dh/dt).
Step 5: Substitute dV/dt = 3 and h = 4: 3 = (π(16)/4)(dh/dt) = 4π(dh/dt). Solve: dh/dt = 3/(4π) ≈ 0.239 m/min.
The water level rises at 3/(4π) m/min when the depth is 4 mThe key insight in this problem was using similar triangles to eliminate the radius variable. This technique appears in nearly every cone-related rates problem. Without it, you would have three variables and two unknown rates, making the problem unsolvable with one equation.
Tips and Common Errors
This is the most common and most costly mistake. If you plug in x = 6 before differentiating, you turn x into a constant, and its derivative becomes zero. You lose the dx/dt term and get a wrong answer. Always differentiate first, then substitute.
When differentiating r³ with respect to t, the answer is 3r²(dr/dt), not 3r². Every variable that changes with time needs a d/dt factor.
Make sure your equation actually relates the quantities whose rates you know and want. If the problem involves a circle, use circle formulas. If it involves a right triangle, use the Pythagorean theorem. A wrong starting equation guarantees a wrong answer.
A negative rate means the quantity is decreasing. If a ladder's top slides down, dy/dt < 0. Don't drop the negative sign — it carries physical meaning.
Pro tips
If dV/dt is in cm³/s and r is in cm, then dr/dt must come out in cm/s. If the units don't match, you have an error.
After finding your answer, do a quick sanity check. Is the rate positive or negative? Is the magnitude reasonable? Does it match the physics of the situation?
For instant feedback on related rates problems, scan your work with Solver AI to see every step of the implicit differentiation and solution.