The Shadow & The Ladder: Classic JEE Related Rates Problems

Why These Problems Are JEE Classics

The Shadow Problem and Ladder Problem have appeared in various forms in JEE Main and Advanced for decades. They perfectly test your understanding of:

Geometric visualization and right triangle trigonometry
Implicit differentiation with respect to time
Chain rule applications in physical contexts
Rate relationships between changing quantities

🎯 The Core Concept

Related rates problems involve finding the rate at which one quantity changes with respect to time, given the rate at which another related quantity changes. The key is establishing the geometric relationship between variables before differentiating.

JEE Classic Medium

The Classic Ladder Problem

Problem Statement:

A ladder 5 meters long is leaning against a wall. The bottom of the ladder is pulled away from the wall at a rate of 2 m/s. How fast is the top of the ladder sliding down the wall when the bottom is 3 meters from the wall?

📐 Visualize: Right triangle with ladder as hypotenuse

Wall (y-axis) | Ladder (5m) | Ground (x-axis)

dx/dt = 2 m/s, find dy/dt when x = 3m

Step-by-Step Solution:

Step 1: Establish the relationship

Using Pythagoras theorem: $x^2 + y^2 = 5^2$

Where x = distance from wall, y = height on wall

Step 2: Differentiate with respect to time

$\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(25)$

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

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

Step 3: Find y when x = 3

$3^2 + y^2 = 25$

$y^2 = 16 \Rightarrow y = 4$ (taking positive since height)

Step 4: Substitute known values

$3(2) + 4\frac{dy}{dt} = 0$

$6 + 4\frac{dy}{dt} = 0$

$4\frac{dy}{dt} = -6$

$\frac{dy}{dt} = -\frac{3}{2}$ m/s

🎯 Interpretation:

The negative sign indicates the top is sliding down at 1.5 m/s when the bottom is 3m from the wall.

💡 Key Insights

The relationship between x and y is always maintained by the fixed ladder length
Negative rate means the quantity is decreasing
At x = 0, dy/dt = 0 (ladder vertical, top not moving)
As x → 5, dy/dt → -∞ (ladder nearly horizontal, top falls rapidly)

JEE Advanced 2019 Hard

The Classic Shadow Problem

Problem Statement:

A man 1.8 meters tall walks away from a lamp post 6 meters high at a speed of 1.2 m/s. How fast is the tip of his shadow moving? How fast is the length of his shadow increasing?

💡 Visualize: Similar triangles formed by lamp, man, and shadow

Lamp (6m) | Man (1.8m) | Shadow tip

dx/dt = 1.2 m/s, find ds/dt and d(shadow length)/dt

Step-by-Step Solution:

Step 1: Set up variables and similar triangles

Let x = distance from lamp to man

Let s = distance from lamp to shadow tip

Let L = shadow length = s - x

Using similar triangles: $\frac{6}{s} = \frac{1.8}{s - x}$

Step 2: Solve the proportion for s

$6(s - x) = 1.8s$

$6s - 6x = 1.8s$

$4.2s = 6x$

$s = \frac{10}{7}x$

Step 3: Find rate of shadow tip movement

Differentiate: $\frac{ds}{dt} = \frac{10}{7}\frac{dx}{dt}$

$\frac{ds}{dt} = \frac{10}{7}(1.2) = \frac{12}{7} \approx 1.714$ m/s

Step 4: Find rate of shadow length increase

Shadow length L = s - x = $\frac{10}{7}x - x = \frac{3}{7}x$

$\frac{dL}{dt} = \frac{3}{7}\frac{dx}{dt} = \frac{3}{7}(1.2) = \frac{3.6}{7} \approx 0.514$ m/s

🎯 Interpretation:

The shadow tip moves faster than the man (1.714 m/s vs 1.2 m/s), while the shadow length increases at 0.514 m/s.

💡 Key Insights

The shadow tip always moves faster than the person
The ratio depends only on the heights ratio, not the walking speed
Similar triangles give the proportional relationship between variables
For very tall lamp posts, shadow tip speed → person's speed

🚀 Universal Related Rates Framework

Step-by-Step Approach

Draw and label a clear diagram
Define variables and given rates
Find geometric relationship between variables
Differentiate with respect to time
Substitute known values
Solve for the unknown rate
Interpret the sign and magnitude

Common Geometric Relationships

Pythagoras theorem: $a^2 + b^2 = c^2$
Similar triangles: $\frac{a}{b} = \frac{c}{d}$
Circle formulas: $A = πr^2$, $C = 2πr$
Volume formulas: Sphere, cone, cylinder
Trigonometric ratios: sin, cos, tan in right triangles

Advanced Variation: The Moving Shadow

JEE Advanced Level:

A man 1.8m tall walks at 1.5 m/s toward a building. A lamp on the ground 10m from the building casts his shadow on the building. How fast is the shadow height changing when the man is 5m from the building?

Conceptual Approach:

Step 1: Complex geometry setup

This involves two sets of similar triangles - one for the ground shadow and one for the building shadow

Step 2: Multiple related rates

The man's position affects both where his shadow hits the building and the height of that shadow

Step 3: Chain rule application

Requires differentiating a composite relationship involving both horizontal and vertical components

💡 Advanced Insight:

JEE Advanced often combines multiple geometric concepts in single related rates problems, testing your ability to handle complexity systematically.

⚠️ Common Related Rates Mistakes

Conceptual Errors

Forgetting the chain rule when differentiating
Mixing up similar triangles in shadow problems
Ignoring negative signs that indicate decreasing quantities
Substituting too early before differentiating

Procedural Errors

Not drawing a clear diagram with all variables labeled
Using wrong geometric relationship for the situation
Algebra mistakes when solving proportions
Units inconsistency in final answer

✅ Prevention Strategy

Always follow the 7-step framework and double-check your geometric relationship before differentiating. Verify that your answer makes sense physically (e.g., shadow tip should move faster than the person).

📝 Practice Your Skills

Try these JEE-style related rates problems:

1. A spherical balloon is inflated at 10 cm³/s. How fast is the radius increasing when radius is 5cm?

Hint: Use $V = \frac{4}{3}πr^3$ and chain rule

2. A 13m ladder slides down with top moving at 0.5 m/s. How fast is bottom moving when top is 5m high?

Hint: This is the ladder problem in reverse

3. Water pours into conical tank (vertex down) at 2 m³/min. How fast is water level rising when depth is 3m? (Radius = 2m, height = 6m)

Hint: Use similar triangles for radius-height relationship

Key Takeaways for JEE Success

🎯 Must-Know Patterns

Ladder problems → Pythagoras theorem
Shadow problems → Similar triangles
Volume problems → Geometric formulas
All require → Chain rule application

🚀 Exam Strategy

Always start with a clear diagram
Write the geometric relationship explicitly
Differentiate before substituting values
Interpret the sign of your answer

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