Q.A ladder, $5$ m long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of $10$ cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when the lower end of the ladder is $2$ m from the wall is:
(A) $\dfrac{1}{10}$ radian/sec
(B) $\dfrac{1}{20}$ radian/sec
(C) $20$ radian/sec
(D) $10$ radian/sec

Concept: Related Rates — we relate the rate of change of the angle to the given rate of the top sliding down.

Let the ladder make angle $\theta$ with the floor. The top is at height $y$ from the floor, so

$\sin\theta = \frac{y}{5}$.

Given $\frac{dy}{dt} = -10\ \text{cm/s} = -0.1\ \text{m/s}$ (negative because $y$ decreases).

Differentiate with respect to $t$:

$\cos\theta \cdot \frac{d\theta}{dt} = \frac{1}{5} \cdot \frac{dy}{dt}$.

When the lower end is $2$ m from the wall, the base $x = 2$. Then $y = \sqrt{5^2 - 2^2} = \sqrt{21}$ m, so

$\cos\theta = \frac{x}{5} = \frac{2}{5}$.

Substitute: …

We have a 5 m ladder sliding down a wall. The top moves downward at 10 cm/s (which is 0.1 m/s — always convert to consistent units). The question asks: when the foot of the ladder is 2 m from the wall, how fast is the angle between the ladder and the floor changing? And note the phrasing: “the rate at which the angle … is decreasing” — so we expect a negative rate, and the answer choices are all positive magnitudes.

1. Set up the geometry and variables

Let the foot of the ladder be at distance $x$ from the wall, and let the top be at height $y$ above the floor. The ladder length is fixed at 5 m:

Let $\theta$ be the angle between the ladder and the floor. From the right triangle:

Either works. Since we are given $x = 2$ m and we know $dx/dt$ is related to $dy/dt$, it’s convenient to use $\cos\theta = x/5$.

2. Relate the rates

Differentiate $\cos\theta = \frac{x}{5}$ with respect to time $t$:

So

We need $d\theta/dt$ when $x = 2$. At that instant, find $y$ from $x^2 + y^2 = 25$:

Then $\sin\theta = y/5 = \sqrt{21}/5$.

3. Find $dx/dt$

The top slides down at 10 cm/s = 0.1 m/s. That means $dy/dt = -0.1$ (negative because $y$ decreases). We need $dx/dt$. Differentiate $x^2 + y^2 = 25$:

So

At $x = 2$, $y = \sqrt{21}$, and $dy/dt = -0.1$: …

Method: Related Rates Involving an Angle (Trigonometric Setups)

When the rate you're asked for is the rate of change of an angle (not a length), you need a trigonometric relation connecting the angle to the sides of a right triangle, plus the Pythagorean relation between those sides if only one side's rate is directly given.

Steps

Step 1: Set up a right-triangle relation between the angle and the sides.

Identify a fixed hypotenuse (here, the ladder's length) and let $x$, $y$ be the two legs (distance from wall, height on wall). Choose whichever trig ratio uses the side whose rate you're given or can find most directly, e.g.

Step 2: Differentiate this relation implicitly with respect to time.

Both $\theta$ and $x$ depend on $t$, so the chain rule applies to each:

Step 3: If the given rate belongs to the OTHER leg, use the Pythagorean relation to find the rate you actually need.

Since the hypotenuse (ladder length) is fixed, $x^2+y^2=(\text{hypotenuse})^2$. Differentiate this too:

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

Common Mistakes

Mistake 1: Not converting cm/s to m/s before substituting

Why it's wrong: The ladder's length and the base distance are given in metres, but the sliding rate is given as $10$ cm/s — mixing units (using $10$ instead of $0.1$ in the metre-based equation) gives an answer off by a factor of $100$. Correct approach: convert every given rate/length to the same unit system before writing or solving the related-rates equation.

Mistake 2: Mixing up which side pairs with which trig ratio

Why it's wrong: With $x$ = base distance from the wall and $y$ = height up the wall, $\cos\theta=x/5$ and $\sin\theta=y/5$ — swapping these (e.g. using $\sin\theta=x/5$) effectively swaps the adjacent and opposite sides and produces a differentiated equation that doesn't match the actual geometry. Correct approach: sketch the right triangle once, label which leg is adjacent to $\theta$ (the floor) and which is opposite, and stay consistent with that labelling throughout. …