Q.Find the value of the following: A ladder $5 \text{ m}$ long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $2 \text{cm/s}$. How fast is its height on the wall decreasing when the foot of the ladder is $4 \text{ m}$ away from the wall ?

Concept: Related Rates — we relate the rates of change of the ladder’s foot distance and its height using the Pythagorean constraint.

Let $x$ be the distance of the foot from the wall, and $y$ the height of the top on the wall. The ladder length is constant:

Differentiate both sides with respect to time $t$:

$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \quad\Rightarrow\quad \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}.$ …

Why related rates works here

We have two quantities that change with time: the distance of the ladder’s foot from the wall (call it $x$) and the height of the ladder’s top on the wall (call it $y$). The ladder’s length is fixed at 5 m, so $x$ and $y$ are always linked by $x^2 + y^2 = 5^2$.

When we are told how fast $x$ changes ($dx/dt = 2\ \text{cm/s}$), we can find how fast $y$ changes ($dy/dt$) by differentiating that relationship. The key idea: the geometry gives a constraint; time‑differentiation turns that constraint into a relation between rates.

Step‑by‑step solution

1. Set up variables and the constraint

Let $x$ be the distance (in metres) from the wall to the foot of the ladder.

Let $y$ be the height (in metres) of the top of the ladder on the wall.

The ladder is 5 m long, so by Pythagoras:

2. Identify given rates and the moment of interest

We are given $\frac{dx}{dt} = 2\ \text{cm/s}$.

Important: The ladder length is in metres, but the rate is in cm/s. Convert everything to consistent units. Let’s work in metres and seconds:

We need $\frac{dy}{dt}$ when $x = 4\ \text{m}$.

3. Differentiate the constraint with respect to time

Differentiate $x^2 + y^2 = 25$ implicitly (remember $x$ and $y$ are functions of $t$):

Divide through by 2:

4. Solve for the unknown rate

From the equation above:

The negative sign makes sense: as $x$ increases, $y$ decreases — the top slides down.

5. Find $y$ when $x = 4$

From $x^2 + y^2 = 25$:

(Only the positive root matters here — height is positive.)

6. Plug in the numbers …

Method: The Sliding-Ladder Related-Rates Setup

This method applies to any "sliding object of fixed length" problem — a ladder against a wall, a rope over a pulley — where a right-triangle constraint links two changing distances.

Steps

Step 1: Assign variables to the two changing distances and note the fixed length

Let $x$ be the distance from the wall to the foot of the ladder, and $y$ the height of the top on the wall. The ladder's own length is constant.

Step 2: Write the Pythagorean constraint

where $\ell$ is the (constant) ladder length.

Step 3: Differentiate both sides with respect to time $t$

The minus sign is structural: as the foot moves away from the wall ($x$ increasing), the top must slide down ($y$ decreasing).

Step 4: Work in ONE consistent unit system throughout …

Common Mistakes

Mistake 1: Mixing metres and centimetres without converting

Why it's wrong: the ladder length and foot distance are given in metres, but the given rate $\frac{dx}{dt}=2\ \text{cm/s}$ is in centimetres per second — plugging $2$ directly into an equation set up in metres (or vice versa) gives an answer wrong by a factor of $100$. Correct approach: convert every quantity to ONE consistent unit before substituting — working entirely in centimetres (ladder $=500\,\text{cm}$, $x=400\,\text{cm}$) is usually cleanest since the given rate is already in cm/s.

Mistake 2: Dropping the negative sign, or misreporting it as an increase …