Related Rates

Related rates problems all run on the same procedure: write the equation connecting the variables, differentiate it with respect to time, then substitute the values for the specific instant. The single rule that keeps you out of trouble is differentiating before you plug in numbers.

Related rates in calculus means finding how fast one quantity changes by using its relationship to another quantity whose rate you already know. If $y$ depends on $x$ and $x$ depends on $t$, then, assuming differentiability,

That chain rule is the engine. What makes these problems distinct is that they start from a geometric or physical situation, not a ready-made function.

When to use this method

Reach for related rates whenever two or more quantities change over time and a fixed equation connects them at every instant. If the radius of a circle changes, its area changes too; if the side of a cube changes, so does its volume. Common settings:

1. Geometry — circles, spheres, cones, ladders.
2. Physics — position, velocity, and other quantities changing together.
3. Engineering or chemistry — one measured quantity depending on another that changes over time.

The method is valid only while the relationship you wrote holds for the situation. If the model changes, the rate equation can change too.

The step-by-step procedure

1. Define the variables in the problem.
2. Write the equation that links them — start here, not with the rate you want.
3. Differentiate with respect to time $t$, applying the chain rule wherever a variable depends on time.
4. Substitute the instant: plug in the known values and rates for the specific moment.
5. Solve for the unknown rate and check units and sign.

The order of steps 3 and 4 is not optional. The variables are changing functions of time even when the equation does not show $t$ explicitly, which is why

Substitute too early and you can erase a changing variable before its derivative appears. In simple cases you might still land on the right answer by luck, but the method stops being reliable.

A full worked example

The radius of a circle is increasing at

How fast is the area increasing when $r = 5$ cm?

Step 2, write the relation:

Step 3, differentiate with respect to time:

Step 4, substitute the instant $r = 5$, $\frac{dr}{dt} = 3$:

Step 5, check units: radius is in centimeters, so area changes in

The original formula connected $A$ and $r$, not $A$ and $t$. Time entered only when we differentiated — the heart of related rates. This is also why the method leans on implicit differentiation: you are differentiating an equation with several linked variables, and each changing variable produces its own rate term.

Where each step tends to stall, and how to self-check

1. Step 4 too early: plugging in values before differentiating. If a variable like $r$ has already become a fixed number, its rate term vanishes — a sign you substituted out of order.
2. Step 3: forgetting that $r$ or $y$ depends on time, so its chain-rule factor goes missing.
3. Step 4: using the wrong instant. The problem asks for one specific moment, not a general average.
4. Step 5: ignoring units or signs. A shrinking quantity should produce a negative rate.
5. Step 2: writing a relation that does not match the geometry or physical setup.

A fast three-part check catches most of these: Did I write the relation before differentiating? Did every changing variable produce a rate term? Do the final units make sense?

Work the procedure yourself

Redo the circle example with $\frac{dr}{dt} = 1.5$ cm/s, evaluated at $r = 8$ cm. Then try a sphere-volume version and watch how changing $r^2$ to $r^3$ changes the final rate formula. For the next step, only take a similar problem into a solver after you have written the relationship and differentiated it yourself — that is the step the method depends on.