Topic 4.4 Notes – Introduction to Related Rates

What a Related Rates Problem Is

A related rates problem involves:

• Two or more variables that are changing over time
• An equation connecting those variables
• A known rate like $\frac{dx}{dt}$
• An unknown rate like $\frac{dy}{dt}$

Every variable depends on the same independent variable, almost always time $t$.

If one quantity depends on another, and that depends on time, the chain rule explains what happens:

$$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$$

That layered dependence is the whole idea. When you differentiate with respect to $t$, every changing variable picks up a $\frac{d(\cdot)}{dt}$.

Quick reminder from Unit 2: this is just implicit differentiation, but now you’re interpreting everything as a rate.

The Core Strategy

Strong students follow the same structure every time.

1) Define Variables Clearly

Assign variables to all changing quantities.

Identify:

• Known rates (example: $\frac{dr}{dt} = 2$ cm/sec)
• Values at the specific instant
• The unknown rate you’re solving for

Include units. Units often save points on FRQs.

2) Write an Equation Connecting the Variables

Use geometry or a formula from the problem.

Common relationships:

• Circle: $A = \pi r^2$
• Sphere: $V = \frac{4}{3}\pi r^3$
• Right triangle: $x^2 + y^2 = L^2$
• Rectangle: $A = lw$

If extra variables appear, use another equation to eliminate them before differentiating.

For example, a sliding ladder forms a right triangle.

Sliding ladder right triangle model

The 10 ft ladder is the hypotenuse, so $x^2 + y^2 = 100$.

3) Differentiate With Respect to Time

Differentiate both sides with respect to $t$.

Every variable gets a chain rule factor.

Example:

$$x^2 + y^2 = 100$$

Differentiate:

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

Notice how $x^2$ became $2x\frac{dx}{dt}$.
That chain rule factor is what students forget most often.

4) Substitute After Differentiating

Only now plug in:

• Known values of variables
• Known rates

If you need a missing value (like $y$), go back to the original equation and solve for it first.

Then solve algebraically for the unknown rate.

Always interpret the sign:

• Positive → increasing
• Negative → decreasing

And attach units like ft/sec or cm²/sec.

Differentiation Rules That Show Up

The chain rule is always present.

Example:

$$\frac{d}{dt}(r^2) = 2r\frac{dr}{dt}$$

Sometimes you also need:

Product Rule

If area depends on two changing sides:

$$A = xy$$

Then:

$$\frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}$$

This shows up a lot with rectangles or expanding surfaces.

Quotient Rule

If a relationship involves division:

$$Q = \frac{x}{y}$$

Then:

$$\frac{dQ}{dt} = \frac{y\frac{dx}{dt} - x\frac{dy}{dt}}{y^2}$$

If two variables are multiplying or dividing, expect extra differentiation rules beyond the chain rule.

Common Mistakes That Cost Points

• Plugging in numbers before differentiating
  You’ll lose the variable and its rate.
• Forgetting chain rule factors
  $\frac{d}{dt}(y^2) = 2y\frac{dy}{dt}$, not just $2y$.
• Ignoring direction
  If something is shrinking, its rate is negative.
• Dropping units
  AP graders expect them in context problems.

On multiple choice, missing a negative sign is the fastest way to pick the trap answer. On FRQs, missing the chain rule usually loses most of the points for the part.