Why do I differentiate with respect to t in related rates?

All quantities change as functions of time t, so differentiating with respect to t gives the rates of change the problem asks for.

Why must I differentiate before substituting values?

If you substitute values before differentiating, constants disappear incorrectly. Always apply implicit differentiation first, then substitute known values.

What does a negative rate mean in related rates?

A negative rate means the quantity is decreasing — for example, a falling ladder height or dropping water level. Always check the sign makes physical sense.

How do I choose the right formula for a related rates problem?

Match the geometry: distances at right angles use the Pythagorean theorem, round objects use volume or area formulas, objects casting shadows use similar triangles.

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Related Rates Calculator

Choose a classic related rates scenario, enter your values, and get a complete 5-step solution — with an animated diagram showing exactly how the quantities change together, and a clear explanation of every differentiation step.

Background

Related rates problems ask: if two quantities are related by an equation, and one is changing at a known rate, how fast is the other changing? The key is implicit differentiation with respect to time t — every variable that changes with time gets a derivative, giving you an equation that relates the rates of change directly.

Set up your related rates problem

Step 1 — Choose a scenario

Each scenario has a specific geometric relationship. Pick the one that matches your problem.

Step 2 — Enter the values

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Result

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How to use this calculator

Choose the scenario that matches your problem — each one has a specific geometric setup and formula.
Enter the known values: the current measurements and the known rate of change.
Click Calculate to see the animated diagram, the 5-step solution, and the unknown rate of change.
Use the quick example chips to load classic textbook problems instantly.

The 5-step method for every related rates problem

Step 1 — Draw and label a diagram. Identify all quantities that change with time. Label them with variables.

Step 2 — Write the geometric equation that relates the variables (Pythagorean theorem, volume formula, similar triangles, etc.).

Step 3 — Differentiate both sides with respect to t using implicit differentiation. Every variable that changes with time gets a d/dt derivative.

Step 4 — Substitute the known values — the current measurements and the known rate — into the differentiated equation.

Step 5 — Solve for the unknown rate and state the answer with correct sign and units.

Example Problems & Step-by-Step Solutions

Example 1 — Sliding Ladder

A 10 ft ladder leans against a wall. The base slides away at 2 ft/s. How fast is the top sliding down when the base is 6 ft from the wall?

Step 1: x = base distance, y = height. x² + y² = 10² = 100.

Step 2: At x = 6: y = √(100−36) = 8 ft.

Step 3: Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0

Step 4: 2(6)(2) + 2(8)(dy/dt) = 0

Step 5: dy/dt = −24/16 = −1.5 ft/s (top slides down at 1.5 ft/s)

Example 2 — Inflating Balloon

Air is pumped into a spherical balloon at 100 cm³/s. How fast is the radius increasing when r = 5 cm?

Step 1: V = (4/3)πr³. Known: dV/dt = 100 cm³/s, r = 5 cm.

Step 2: Differentiate: dV/dt = 4πr²(dr/dt)

Step 3: Substitute: 100 = 4π(5²)(dr/dt)

Step 4: dr/dt = 100 / (100π) = 1/π ≈ 0.318 cm/s

Example 3 — Draining Conical Tank

A conical tank (radius = height always) drains at 2 m³/min. How fast does the water level drop when h = 4 m?

Step 1: Since r = h, V = (1/3)π(h)²(h) = (π/3)h³.

Step 2: Differentiate: dV/dt = πh²(dh/dt)

Step 3: −2 = π(16)(dh/dt)

Step 4: dh/dt = −2/(16π) = −1/(8π) ≈ −0.040 m/min

Example 4 — Shadow Length

A 6 ft person walks away from a 15 ft lamp at 4 ft/s. How fast does their shadow lengthen?

Step 1: Let x = distance from lamp, s = shadow length. By similar triangles: 15/(x+s) = 6/s.

Step 2: Cross multiply: 15s = 6(x+s) → 9s = 6x → s = (2/3)x.

Step 3: Differentiate: ds/dt = (2/3)(dx/dt) = (2/3)(4) = 8/3 ≈ 2.67 ft/s

Shadow lengthens at 8/3 ft/s, independent of position.

Frequently Asked Questions

Why do I differentiate with respect to t, not x?

Because all the quantities are functions of time t — the ladder height, balloon radius, water level — they all change as time passes. Differentiating with respect to t gives rates of change, which is exactly what the problem asks for.

Why must I find the missing variable before substituting?

You must differentiate first, then substitute. If you substitute values before differentiating, constants disappear incorrectly. The current values only go in after the chain rule has been applied to the equation.

What does a negative rate mean?

A negative rate means the quantity is decreasing. For the ladder, dy/dt < 0 means the height is falling. For the tank, dh/dt < 0 means the water level is dropping. Always check the sign makes physical sense.

What if the cone's radius and height aren't equal?

Use the given ratio r/h = constant from the problem, express r in terms of h, substitute into V = (1/3)πr²h before differentiating. This eliminates one variable so you only differentiate with respect to h and t.

How do I know which formula to use?

Match the geometry: distances in a right angle → Pythagorean theorem. A round object growing → volume or area formula. Objects at different heights casting shadows → similar triangles. The diagram always reveals the right relationship.

Can the rate of change itself be changing?

Yes, but in most introductory problems the given rate (like dx/dt) is constant. The unknown rate (like dy/dt) changes as the geometry changes — which is why the answer depends on the current position, not just the given rate.