Related rates

Understanding Related Rates Problems

In related rates problems, you are given a situation where several quantities change with respect to time, and you are asked to find how fast one quantity is changing at a specific moment, given information about how other related quantities are changing at that same moment.

The key ideas are:

• All rates are derivatives with respect to time $t$.
• The quantities are related by an equation (often geometric, but not always).
• You differentiate the relation with respect to $t$, using the chain rule where needed.
• You then substitute the known values at the instant of interest and solve for the unknown rate.

You should already be familiar with derivatives, differentiation rules, and basic applications (from the parent chapter). Here we focus on the structure and method unique to related rates.

General Strategy for Related Rates

Most related rates problems can be approached with a standard sequence of steps.

1. Read and understand the problem
• Identify the quantities that depend on time: lengths, areas, volumes, angles, positions, etc.
• Clarify what is given (known rates like $\dfrac{dx}{dt}$, values of variables at a moment) and what you are asked to find (an unknown rate, like $\dfrac{dy}{dt}$ at a specific time or condition).
2. Introduce variables and draw a diagram when possible
• Assign symbols to the time-dependent quantities: $x(t)$, $y(t)$, $r(t)$, $V(t)$, etc.
• Draw a clear, labeled picture if the situation is geometric (triangles, cones, circles, ladders, tanks, etc.).
• Note that the variables usually represent instantaneous measurements that can change with time.
3. Find an equation relating the variables
• Use geometry (similar triangles, Pythagorean theorem, area/volume formulas), trigonometry, or algebraic relations to connect the variables.
• At this stage, write an equation involving the variables themselves, not their derivatives.
• Typically, do not plug in the specific moment’s values yet. Keep variables symbolic for differentiation.
4. Differentiate implicitly with respect to time $t$
• Apply $\dfrac{d}{dt}$ to both sides of the relation.
• Use the chain rule: every time you differentiate a quantity that depends on $t$, you multiply by its derivative with respect to $t$.
• For example:
• If $y = x^2$, then differentiating with respect to $t$ gives
  $$\frac{dy}{dt} = 2x \frac{dx}{dt}.$$
• If $A = \pi r^2$, then
  $$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}.$$
• If $s^3 = V$, then
  $s^2 \frac{ds}{dt} = \frac{dV}{dt}.$$
5. Substitute all known values at the specific instant
• Now plug in all the information given in the problem: values of variables at that moment and any known rates.
• If a needed quantity is not given directly, you may have to find it by using the original relation first (before inserting into your differentiated equation).
6. Solve for the unknown rate
• Solve the resulting equation for the desired derivative (rate).
• Include units and sign (positive for increasing, negative for decreasing) in your final answer.
7. Interpret the result
• Make sure the sign and magnitude of your answer make sense in the context of the word problem.
• Answer in words, not just in symbols or numbers.

Common Patterns and Examples of Relations

Related rates problems typically fall into recognizable patterns. The specific calculus technique (implicit differentiation with respect to $t$) is the same; only the underlying relations differ.

1. Changing lengths: Pythagorean-type relations

Many problems involve right triangles whose sides change over time.

Typical relation:
$$x^2 + y^2 = z^2.$$

When you differentiate with respect to $t$:
$$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt}.$$

Often, you can simplify by dividing by $2$:
$$x \frac{dx}{dt} + y \frac{dy}{dt} = z \frac{dz}{dt}.$$

Example structure: A moving point

A point moves so that its $x$-coordinate and $y$-coordinate change. You might be asked how fast its distance from the origin is changing.

• Variables: $x(t)$, $y(t)$, and distance $r(t)$ from origin.
• Relation: $x^2 + y^2 = r^2$.
• Rates: $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ given; find $\dfrac{dr}{dt}$ at a moment.

The steps are exactly as in the general strategy, using the Pythagorean relation.

2. Similar triangles and geometric proportions

Sometimes lengths are related by a proportion.

If two triangles are similar, for example:
$$\frac{x}{a} = \frac{y}{b}.$$

You can rewrite this relationship in a convenient algebraic form before differentiating, such as:
$$bx = ay.$$

Then differentiate:
$$b \frac{dx}{dt} = a \frac{dy}{dt}.$$

These occur often in problems involving:

• Shadows growing or shrinking as a person walks toward or away from a light.
• Ladders sliding against walls when heights and distances form similar triangles.

3. Areas and volumes changing over time

Geometry provides formulas for area and volume; in related rates problems those areas/volumes change as linear dimensions change.

Circles

For a circle with radius $r(t)$ and area $A(t)$:
$$A = \pi r^2 \quad \Rightarrow \quad \frac{dA}{dt} = 2\pi r \frac{dr}{dt}.$$

For circumference $C(t)$:
$$C = 2\pi r \quad \Rightarrow \quad \frac{dC}{dt} = 2\pi \frac{dr}{dt}.$$

Spheres

For a sphere with radius $r(t)$ and volume $V(t)$:
$$V = \frac{4}{3}\pi r^3 \quad \Rightarrow \quad \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}.$$

For surface area $S(t)$:
$$S = 4\pi r^2 \quad \Rightarrow \quad \frac{dS}{dt} = 8\pi r \frac{dr}{dt}.$$

Cones

For a right circular cone (height $h$, radius $r$, volume $V$):
$$V = \frac{1}{3}\pi r^2 h.$$

If both $r$ and $h$ can change:
$$\frac{dV}{dt} = \frac{1}{3}\pi \left( 2r h \frac{dr}{dt} + r^2 \frac{dh}{dt} \right).$$

Often, one dimension (say $h$) is related to $r$ through similar triangles, such as $r = k h$, where $k$ is a constant. Then you can write $V$ in terms of a single variable before differentiating, which usually simplifies the calculation.

4. Product and quotient relations

Sometimes the relation is algebraic but not directly geometric.

If $Q = xy$ and both $x$ and $y$ depend on $t$:
$$\frac{dQ}{dt} = x \frac{dy}{dt} + y \frac{dx}{dt}.$$

If $R = \dfrac{x}{y}$ with $x(t)$, $y(t)$:
$$\frac{dR}{dt} = \frac{y \frac{dx}{dt} - x \frac{dy}{dt}}{y^2}.$$

These appear, for instance, in economic, physical, or population models when one quantity is the product or ratio of others.

5. Angle and trigonometric relations

In some problems, angles change while lengths also change.

Typical right triangle:

• Opposite side $y(t)$, adjacent side $x(t)$, angle $\theta(t)$.

Relations might include:
$$\tan \theta = \frac{y}{x}, \quad \sin \theta = \frac{y}{r}, \quad \cos \theta = \frac{x}{r}.$$

For example, from $\tan \theta = \dfrac{y}{x}$, differentiate with respect to $t$:
$$\sec^2 \theta \,\frac{d\theta}{dt}
= \frac{x \frac{dy}{dt} - y \frac{dx}{dt}}{x^2}.$$

Here, chain rule and derivatives of trigonometric functions are used together with related rates ideas.

Important Practical Details

When to plug in numbers

A common question is: should you plug in the specific numerical values before differentiating or after?

• Often, it is safest and clearest to:
• Keep all variables symbolic while you differentiate.
• Then substitute all known values after the differentiation step.

This avoids accidentally treating changing quantities as constants.

However, if the relation contains a quantity that is constant for all time (e.g., a fixed ladder length, a fixed radius, a fixed height of a light post), you can substitute that constant early or leave it in; it will behave correctly either way in differentiation.

Dealing with missing information

Sometimes a value needed for substitution (like a length at the specific moment) is not given directly. In that case:

1. Use the original geometrical or algebraic relation (not yet differentiated) to find the missing value at the given instant.
2. Then plug that into the differentiated equation.

Example pattern:

• Given: $\dfrac{dx}{dt}$ and $x$ at the instant.
• Need: $y$ at that instant.
• Use $x^2 + y^2 = 25$ (or another relation) to find $y$.
• Then substitute into the rate equation to solve for $\dfrac{dy}{dt}$.

Signs and direction

The sign of a rate tells you about direction:

• Positive rate: quantity is increasing with time.
• Negative rate: quantity is decreasing with time.

In word problems:

• Distances that are “shortening” or “decreasing” will typically have negative derivatives.
• Distances that are “increasing,” volumes “filling,” heights “rising,” etc. usually have positive derivatives.

Be careful to interpret the language of the problem when deciding whether to input a rate as positive or negative.

Units

Each derivative has combined units:

• If $x$ is measured in meters and $t$ in seconds, then $\dfrac{dx}{dt}$ has units $\text{m/s}$.
• If $V$ is volume (e.g., $\text{m}^3$), then $\dfrac{dV}{dt}$ might be in $\text{m}^3/\text{s}$.

Always express the final answer with correct units and in terms of the situation (e.g., “meters per second,” “square centimeters per minute,” “cubic meters per hour”).

Worked Example (Structure Only)

Consider a typical “filling tank” structure to see how all the ideas fit together.

1. Variables and relation
• Let $h(t)$ be the height of the water, $V(t)$ its volume.
• Suppose (from geometry) $V$ is a known function of $h$, such as
  $$V = k h^3,$$
  where $k$ is a constant based on the tank’s shape.
2. Differentiate with respect to $t$

$$\frac{dV}{dt} = 3k h^2 \frac{dh}{dt}.$$

3. Substitute known values
• If $\dfrac{dV}{dt}$ and $h$ at a certain instant are known, plug them into the equation along with $k$.
4. Solve for $\dfrac{dh}{dt}$

$$\frac{dh}{dt} = \frac{1}{3k h^2} \frac{dV}{dt}.$$

5. Interpret
• Attach units and interpret in words: “At that instant, the water level is rising at $\dots$ meters per minute.”

The same pattern appears in many kinds of related rates problems, just with different relations between the variables.

Common Pitfalls and How to Avoid Them

1. Forgetting that variables depend on time
• Every relevant quantity that can change should be treated as a function of $t$.
• Do not differentiate $x^2$ as if $x$ were a constant; always use $\dfrac{d}{dt}(x^2) = 2x \dfrac{dx}{dt}$ if $x$ changes with $t$.
2. Plugging in specific variable values too early
• If you substitute a particular $x$ or $y$ into your equation before differentiating, you may accidentally treat it as a constant.
• Safer: keep variables, differentiate, then plug in.
3. Losing the units or ignoring sign
• Rates must have units combining the units of the quantity and time.
• Reflect whether the rate is naturally positive or negative based on the story.
4. Using incorrect formulas or relations
• Carefully identify the correct geometric or algebraic relation among the changing quantities.
• Double-check formulas (areas, volumes, trigonometric relations) before differentiating.

Summary

Related rates problems use derivatives to connect how different quantities change with time:

• Model the situation with variables that depend on $t$.
• Find an equation relating those variables.
• Differentiate implicitly with respect to $t$, using the chain rule.
• Substitute all known values at the specific instant.
• Solve for the unknown rate, being mindful of sign, units, and context.

The core calculus idea is simple: if quantities are related by an equation, then their rates of change are related by the derivative of that equation with respect to time.