Q.A man of height $2$ metres walks at a uniform speed of $5$ km/h away from a lamp post which is $6$ metres high. Find the rate at which the length of his shadow increases.

The key idea is Related Rates: use similar triangles to relate the variables, then differentiate with respect to time.

Step 1 – Set up the geometry.

Let the distance from the lamp post to the man be $l$ (in km), and the length of his shadow be $s$ (in km). The lamp post is $6$ m tall, the man is $2$ m tall. From the similar triangles $\triangle MSN \sim \triangle ASB$:

Step 2 – Solve for the relation.

Cross-multiplying: $2(l + s) = 6s \implies 2l + 2s = 6s \implies 2l = 4s \implies l = 2s$.

Step 3 – Differentiate with respect to time.

The man walks at $5$ km/h, so $\frac{dl}{dt} = 5$ km/h. Differentiating $l = 2s$:

This is a classic related rates problem. The core idea: two quantities (the man’s distance from the lamp post and the length of his shadow) change together because they are linked by geometry. When you know how fast one changes, you can find how fast the other changes — by differentiating the geometric relationship.

The geometry here is driven by light travelling in straight lines. The lamp at the top of the post casts a ray that just grazes the man’s head and hits the ground at the tip of his shadow. That ray, the lamp post, and the ground form a large right triangle. The man’s body and his shadow form a smaller, similar right triangle inside it. Similar triangles give a clean linear relation — no squares, no trig — which makes the differentiation trivial.

Let’s set it up step by step.

1. Draw and label the figure.

   Let $A$ be the foot of the lamp post, $B$ the lamp (so $AB = 6$ m). Let $M$ be the man’s feet, $N$ his head (so $MN = 2$ m). The point $S$ is the tip of his shadow on the ground.

   The distance from the lamp post to the man is $AM = l$. The shadow length is $MS = s$.

   The ray from $B$ through $N$ hits the ground at $S$, so $N$ lies on $BS$.

2. Write the similarity relation.

   Triangles $\triangle ASB$ and $\triangle MSN$ share the angle at $S$ and both have a right angle (at $A$ and $M$ respectively). So they are similar:

Here $AS = AM + MS = l + s$, and $MS = s$. Substituting:

which simplifies to

3. Solve for the relation between $l$ and $s$. Multiply through:

This is the key geometric link: the man’s distance from the post is always twice his shadow length.

4. Differentiate with respect to time. Both $l$ and $s$ change as the man walks. Differentiate $l = 2s$ implicitly:

The man walks away at a uniform speed of $5$ km/h, so $\frac{dl}{dt} = 5$ km/h (positive because $l$ increases).

5. Solve for $\frac{ds}{dt}$.

The units are consistent: km/h for speed, so the answer is in km/h. If the problem had asked in m/s, you’d convert: $2.5$ km/h $= \frac{2500}{3600} \approx 0.694$ m/s — but the given speed is in km/h, so the answer stays in km/h.

Method: Related Rates via Similar Triangles

The general technique for shadow/light-source problems, and any setup where similar right triangles link two changing lengths.

Steps

Step 1: Draw the two right triangles and identify what's similar

A light source at a fixed height, an object of fixed height, and the ground form a large triangle (source to the tip of the shadow) containing a smaller, similar triangle (the object to that same shadow tip). Similar triangles arise because both triangles share the angle at the shadow tip and both have a right angle at the ground.

Step 2: Write the similarity ratio and simplify to a linear relation

Set the ratio of corresponding sides of the two triangles equal, then simplify algebraically. For this family of problems it reduces to a simple proportional relation between the object's distance from the light source and the shadow's own length — always re-derive this ratio from the actual heights given, since it changes with the numbers.

Step 3: Differentiate the linear relation with respect to time

Because the relation between the two lengths is linear (not quadratic), differentiating it with respect to $t$ is immediate and introduces no product rule — each side's derivative is just a constant multiple of the other's rate.

Step 4: Substitute the known rate and solve

Substitute the given constant walking speed for the known rate, then solve for the unknown rate. Check the ratio makes physical sense: is the unknown quantity expected to grow faster or slower than the known one, given the relative heights involved?

Common Mistakes

Mistake 1: Assuming the shadow lengthens at the same speed as the man walks

Why it's wrong: it's tempting to think the shadow's tip moves at $5$ km/h simply because the man does, but the shadow length and the man's distance from the lamp post are two different quantities related by similar triangles, not identical to each other. Correct approach: derive the actual proportional relationship ($l=2s$ here) from the similar triangles first, and only then differentiate — the shadow turns out to grow at exactly half the man's walking speed in this setup.

Mistake 2: Confusing the "distance from the lamp post" with the "distance to the shadow tip"

Why it's wrong: the similarity ratio in this type of problem compares the whole distance from the light source to the shadow tip (post-to-man plus man-to-shadow-tip) against the man's height — using only the post-to-man distance in that ratio gives the wrong proportion entirely. Correct approach: carefully label the figure so that $AS = AM + MS$ is recognized as the full base of the large triangle, not just the segment from the post to the man.