Related Rates: Distance, Shadows & Velocity Decomposition

In this lesson, you will tackle real related rates problems from start to finish – two cars at a crossroad, a boat being pulled toward a cliff, and even the mystery of how fast your shadow moves when you walk away from a streetlight. Along the way, you will discover a cool shortcut from physics called “velocity decomposition” that sometimes lets you skip the algebra entirely. By the end, you will have a reliable step-by-step method for any related rates problem.

TipWhy Related Rates Matter

Related rates problems ask: “If one quantity is changing, how fast is a connected quantity changing?” This shows up constantly in the real world:

• Navigation: Air traffic controllers track how fast the distance between two planes is changing to prevent collisions
• Construction: Engineers calculate how fast a shadow moves to plan lighting for buildings and stadiums
• Rescue operations: Coast guard teams figure out how quickly to reel in a towline to bring a boat to shore at the right speed
• Medicine: Doctors model how fast a drug concentration in the blood is changing based on IV drip rates
• Sports: Coaches analyze how quickly the gap between two runners is closing during a relay race

Today we’ll solve several classic related rates problems using both calculus (implicit differentiation) and physics (velocity decomposition).

Topics Covered

• Setting up related rates problems: identifying variables, given rates, and sought rates
• Using the Pythagorean theorem \(s^2 = x^2 + y^2\) to relate distances
• Differentiating implicit equations: \(s\,ds = x\,dx + y\,dy\)
• Dividing by \(dt\) to convert differentials into rates
• Two cars at a crossroad: finding \(\dfrac{ds}{dt}\) from \(\dfrac{dx}{dt}\) and \(\dfrac{dy}{dt}\)
• Velocity decomposition: projecting motion onto the radial direction
• Boat-and-cliff problem: reeling in a rope connected to a boat
• Shadow problems using similar triangles to relate variables
• Distinguishing variables from constants in problem setup

Lecture Video

Key Frames from the Lecture

What You Need to Know First

NoteWhat is implicit differentiation?

When you have an equation like \(x^2 + y^2 = s^2\) and both \(x\), \(y\), and \(s\) depend on time \(t\), you can’t just “take the derivative with respect to \(x\).” Instead, you differentiate each term with respect to its own variable:

\[2x\,dx + 2y\,dy = 2s\,ds\]

Then divide both sides by \(dt\) to get rates:

\[2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2s\frac{ds}{dt}\]

This is implicit differentiation – treating all changing quantities as variables simultaneously.

NoteWhat is the Pythagorean theorem?

For a right triangle with legs \(a\) and \(b\) and hypotenuse \(c\):

\[a^2 + b^2 = c^2\]

This is the go-to equation for distances in perpendicular directions. For related rates, the key is recognizing which sides are variables and which are constants.

NoteWhat are similar triangles?

Two triangles are similar if they have the same angles. When triangles are similar, their corresponding sides are proportional:

\[\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\]

Shadow problems almost always involve similar triangles formed by a light source, an object, and the shadow.

NoteWhat is vector decomposition?

Any velocity vector can be broken into two perpendicular components. If you want to know how motion affects a particular direction (say, the distance between two objects), you project the velocity onto that direction:

\[v_{\text{radial}} = v \cos\theta\]

where \(\theta\) is the angle between the velocity and the direction you care about. The perpendicular component \(v \sin\theta\) does not affect the distance – it only causes rotation.

The Related Rates Procedure

Every related rates problem follows the same three-step recipe:

Step 1 – Set up variables and rates. Clearly identify:

• What are the variables (quantities that change)?
• What rates are given (\(\frac{dx}{dt}\), \(\frac{dy}{dt}\), etc.)?
• What rate are we trying to find?
• Watch the sign: if a quantity is decreasing, its rate is negative.

Step 2 – Relate the variables. Find a geometric equation connecting the variables (Pythagorean theorem, similar triangles, etc.). Do not plug in numbers yet – the equation must hold at every instant, not just at the given moment.

Step 3 – Differentiate, then plug in. Differentiate the equation (take differentials of each variable), divide by \(dt\), and only then substitute the given numerical values.

Problem 1: Two Cars at a Crossroad

Setup

Two cars are at a crossroad. Car A drives north at \(60\) mph, and Car B drives west (toward the intersection) at \(40\) mph. At the moment of interest, Car A is \(500\) m north and Car B is \(375\) m east. How fast is the distance between them changing?

The green dashed line is the distance \(s\) between the two cars. The triangle has legs \(x = 375\) and \(y = 500\), forming a 3-4-5 ratio.

Solution by Calculus (Related Rates)

Variables and rates:

• \(y\) = position of Car A (north), \(\frac{dy}{dt} = 60\) mph (increasing)
• \(x\) = position of Car B (east), \(\frac{dx}{dt} = -40\) mph (decreasing – moving toward the origin)
• \(s\) = distance between the cars; find \(\frac{ds}{dt}\)

Relate the variables using the Pythagorean theorem:

\[s^2 = x^2 + y^2\]

Differentiate (take differentials, then divide by \(dt\)):

\[2s\,ds = 2x\,dx + 2y\,dy \quad \Longrightarrow \quad s\frac{ds}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}\]

Plug in. Notice \(x : y = 375 : 500 = 3 : 4\), so \(s\) corresponds to \(5\) in the ratio. We only need the ratios:

\[\frac{ds}{dt} = \frac{x}{s}\cdot\frac{dx}{dt} + \frac{y}{s}\cdot\frac{dy}{dt} = \frac{3}{5}(-40) + \frac{4}{5}(60) = -24 + 48 = 24 \text{ mph}\]

ImportantKey Idea: Pythagorean Related Rates

When two objects move along perpendicular directions, the Pythagorean theorem links their positions, and differentiating it links their speeds to the rate the distance between them changes.

\[\boxed{\frac{ds}{dt} = 24 \text{ mph}}\]

The distance is increasing at \(24\) mph.

Solution by Velocity Decomposition (Physics Way)

Instead of algebra, decompose each car’s velocity into the radial direction (along the line connecting A and B) and the tangential direction (perpendicular to it). Only the radial component affects the distance between the cars.

Car A moves at \(60\) mph upward. The triangle \(OAB\) has ratio \(3:4:5\), so the projection of Car A’s velocity onto the radial direction uses \(\cos\theta = \frac{4}{5}\):

\[v_{r,A} = 60 \times \frac{4}{5} = 48 \text{ mph (increasing distance)}\]

Car B moves at \(40\) mph to the left. Projecting onto the radial direction with \(\sin\theta = \frac{3}{5}\):

\[v_{r,B} = 40 \times \frac{3}{5} = 24 \text{ mph (decreasing distance)}\]

The net rate of change:

\[\frac{ds}{dt} = 48 - 24 = 24 \text{ mph}\]

Same answer – but we could almost eyeball it once we spotted the \(3\)-\(4\)-\(5\) ratio.

Problem 2: Boat and Cliff (Rope Problem)

Setup

A person stands atop a \(50\) m cliff and pulls in a rope attached to a boat at \(2\) m/s. At the moment of interest, the boat is \(120\) m from the base of the cliff. How fast is the boat approaching the cliff?

The dashed red line is the rope of length \(r = 130\). The cliff height \(h = 50\) is constant – only \(x\) and \(r\) are variables.

Solution by Related Rates

Variables and rates:

• \(r\) = length of the rope (hypotenuse), \(\frac{dr}{dt} = -2\) m/s (rope is getting shorter)
• \(x\) = horizontal distance from boat to cliff base; find \(\frac{dx}{dt}\)
• \(h = 50\) m is a constant (the cliff height does not change, so \(dh = 0\))

Relate the variables:

\[r^2 = x^2 + h^2 = x^2 + 50^2\]

Differentiate (since \(h\) is constant, its differential is zero):

\[2r\,dr = 2x\,dx \quad \Longrightarrow \quad \frac{dx}{dt} = \frac{r}{x}\cdot\frac{dr}{dt}\]

Find the ratio. At the moment: \(x = 120\), \(h = 50\). Notice \(120 : 50 = 12 : 5\), so the sides are in ratio \(5 : 12 : 13\), giving \(r/x = 13/12\).

\[\frac{dx}{dt} = \frac{13}{12} \times (-2) = -\frac{13}{6} \text{ m/s}\]

ImportantKey Idea: When One Side Is Constant, the Problem Simplifies

In the boat-and-cliff problem, the cliff height never changes, so its differential is zero. This means only two variables remain, and you get a clean ratio between the rope-pull speed and the boat speed.

\[\boxed{\frac{dx}{dt} = -\frac{13}{6} \approx -2.17 \text{ m/s}}\]

The boat approaches the cliff at \(\frac{13}{6}\) m/s – faster than the rope is pulled, because the rope is at an angle.

Solution by Velocity Decomposition

The rope shrinks at \(2\) m/s – that is the radial component of the boat’s velocity. The boat moves horizontally, so its full velocity \(v\) projects onto the rope direction as \(v \cdot \frac{12}{13}\). Setting this equal to \(2\):

\[v = 2 \times \frac{13}{12} = \frac{13}{6} \text{ m/s}\]

Eyeball it once you spot the \(5\)-\(12\)-\(13\) triangle.

Problem 3: Shadow of a Walking Person

Setup

A lamppost is \(5\) m tall. A \(2\) m tall person walks away from the lamp at \(3\) m/s. When the person is \(3\) m from the lamp, how fast is the tip of their shadow moving?

Drag the slider \(t\) to move the person. Watch how the shadow tip (green dot) always moves proportionally – the ratio \(s/x = 5/3\) never changes.

Solution

Variables and constants:

• \(x\) = distance from person to lamppost (variable), with \(\frac{dx}{dt} = 3\) m/s
• \(s\) = distance from shadow tip to lamppost (variable); find \(\frac{ds}{dt}\)
• Heights \(H = 5\) m (lamp) and \(h = 2\) m (person) are constants

Relate using similar triangles. The big triangle (lamp to shadow tip, height \(5\), base \(s\)) is similar to the small triangle (person to shadow tip, height \(2\), base \(s - x\)):

\[\frac{x}{s - x} = \frac{H - h}{h} = \frac{5 - 2}{2} = \frac{3}{2}\]

Cross multiply:

\[2x = 3(s - x) = 3s - 3x \quad \Longrightarrow \quad 5x = 3s\]

Differentiate:

\[5\,dx = 3\,ds \quad \Longrightarrow \quad \frac{ds}{dt} = \frac{5}{3}\cdot\frac{dx}{dt} = \frac{5}{3} \times 3 = 5 \text{ m/s}\]

ImportantKey Idea: Similar Triangles Give Constant-Ratio Rates

When similar triangles create a linear relationship between two variables (like \(5x = 3s\)), their rates of change are locked in the same ratio. That is why the shadow tip speed is constant no matter where the person stands.

\[\boxed{\frac{ds}{dt} = 5 \text{ m/s}}\]

A striking result: the shadow tip moves at a constant \(5\) m/s regardless of where the person is. Because the relationship between \(s\) and \(x\) is linear, their rates are in a fixed ratio.

Problem 4: Shadow on a Wall

Setup

A lamp is \(3\) m high, and a \(2.5\) m tall person stands between the lamp and a wall \(4\) m away. The person walks toward the wall at \(3\) m/s and is currently \(1\) m from the lamp. How fast is the shadow moving on the wall?

Solution

Variables and constants:

• \(x\) = distance from person to lamp (variable), \(\frac{dx}{dt} = 3\) m/s
• \(y\) = height of shadow on the wall (variable); find \(\frac{dy}{dt}\)
• Lamp height \(= 3\) m, person height \(= 2.5\) m, wall distance \(= 4\) m are all constants

Relate using similar triangles. The light at height \(3\) casts a ray over the person’s head (height \(2.5\)) onto the wall. By similar triangles:

\[\frac{x}{4 - x} = \frac{3 - 2.5}{y - 3} \quad \Longrightarrow \quad x(y - 3) = 0.5(4 - x) \quad \Longrightarrow \quad xy = 2 + 2.5x\]

Differentiate using the product rule on \(xy\), then divide by \(dt\):

\[y\frac{dx}{dt} + x\frac{dy}{dt} = 2.5\frac{dx}{dt}\]

Plug in. At \(x = 1\): \(y = 2 + 2.5 = 4.5\) m.

\[4.5(3) + 1 \cdot \frac{dy}{dt} = 2.5(3) \quad \Longrightarrow \quad \frac{dy}{dt} = 7.5 - 13.5 = -6\]

\[\boxed{\frac{dy}{dt} = -6 \text{ m/s}}\]

The negative sign means the shadow is moving down the wall at \(6\) m/s. Unlike Problem 3, the \(xy\) product makes this relationship nonlinear – the shadow speed depends on position.

Common Mistakes to Avoid

1. Plugging in numbers too early. If you substitute \(x = 3\) before differentiating, you get \(\frac{d}{dt}(3) = 0\) – nonsense. The equation must hold at every instant.

2. Confusing variables and constants. The cliff height \(h = 50\) never changes, so \(dh = 0\). The person’s position \(x\) does change, so \(dx \neq 0\). Label everything at the start.

3. Forgetting the sign. If a quantity is decreasing, its rate is negative. Car B approaching means \(\frac{dx}{dt} < 0\); rope pulled in means \(\frac{dr}{dt} < 0\).

4. Taking the derivative instead of the differential. Give each variable its own differential (\(2r\,dr = 2x\,dx\)), then divide by \(dt\). Do not decide “with respect to what” prematurely.

Cheat Sheet

What you need	Formula
Pythagorean relation	\(s^2 = x^2 + y^2\)
Differentiated Pythagorean	\(s\dfrac{ds}{dt} = x\dfrac{dx}{dt} + y\dfrac{dy}{dt}\)
Similar triangles	\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\) (corresponding sides proportional)
Radial projection	\(v_{\text{radial}} = v\cos\theta\) (component along the distance direction)
Related rates procedure	(1) Set up variables and rates, (2) Relate variables, (3) Differentiate and plug in
Product rule differential	\(d(xy) = y\,dx + x\,dy\)
Key principle	Only the radial component of velocity changes the distance between two objects

The Related Rates Workflow

\[\text{Identify variables} \;\to\; \text{Geometric equation} \;\xrightarrow{\text{differentiate}}\; \text{Differential equation} \;\xrightarrow{\div\, dt}\; \text{Rate equation} \;\xrightarrow{\text{plug in}}\; \text{Answer}\]