Math Problem Statement

A man of height 1.5 m walks away from a lamp post of height 4.5 m at a rate of 20 cm/s. How fast is the shadow lengthening when the man is 4.2 m from the lamp post?

Solution

This is a related rates problem involving similar triangles. Let us break it down step-by-step.

Given:

  1. Height of the lamp post = 4.5m4.5 \, \text{m}
  2. Height of the man = 1.5m1.5 \, \text{m}
  3. Speed of the man = 20cms=0.2m/s\frac{20 \, \text{cm}}{\text{s}} = 0.2 \, \text{m/s}
  4. Distance of the man from the lamp post at the moment in question = 4.2m4.2 \, \text{m}

We are tasked with finding the rate at which the shadow of the man is lengthening, i.e., dsdt\frac{ds}{dt}, where ss is the shadow's length.


Step 1: Relationship Between Variables

Using similar triangles: Height of lamp postDistance from lamp post to tip of shadow=Height of manLength of shadow\frac{\text{Height of lamp post}}{\text{Distance from lamp post to tip of shadow}} = \frac{\text{Height of man}}{\text{Length of shadow}}

Let:

  • xx: distance of the man from the lamp post,
  • ss: length of the man's shadow.

The total distance from the lamp post to the tip of the shadow is x+sx + s. Thus: 4.5x+s=1.5s\frac{4.5}{x + s} = \frac{1.5}{s}

Simplify: 4.5s=1.5(x+s)4.5s = 1.5(x + s) 4.5s=1.5x+1.5s4.5s = 1.5x + 1.5s 3s=1.5x3s = 1.5x s=0.5xs = 0.5x

So, the length of the shadow is always half the distance of the man from the lamp post.


Step 2: Differentiate with Respect to tt

Differentiate both sides of s=0.5xs = 0.5x with respect to tt: dsdt=0.5dxdt\frac{ds}{dt} = 0.5 \frac{dx}{dt}

Step 3: Substitution of Known Values

The man’s speed, dxdt=0.2m/s\frac{dx}{dt} = 0.2 \, \text{m/s}.

Substitute: dsdt=0.5×0.2\frac{ds}{dt} = 0.5 \times 0.2 dsdt=0.1m/s\frac{ds}{dt} = 0.1 \, \text{m/s}


Final Answer:

The shadow is lengthening at a rate of 0.1m/s0.1 \, \text{m/s}.


Would you like a deeper explanation, or do you have further questions? Here are some related questions to explore:

  1. How does the length of the shadow depend on the heights of the lamp post and man?
  2. What would happen to the shadow's rate if the man walked faster?
  3. How can this problem be modified for a moving lamp post?
  4. How does the concept of similar triangles simplify related rates problems?
  5. What if the heights of the lamp post and man were equal? What would happen to the shadow?

Tip: Always identify which variables are constant and which are changing when solving related rates problems!

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Math Problem Analysis

Mathematical Concepts

Similar Triangles
Related Rates
Differentiation

Formulas

s = 0.5x (derived using similar triangles)
ds/dt = 0.5 * dx/dt

Theorems

Properties of Similar Triangles

Suitable Grade Level

Grades 11-12