A ladder 10 meters long rests with one end against a vertical wall, the other on the floor. The lower end move

3.2 · Q7

Q. A ladder 10 meters long rests with one end against a vertical wall, the other on the floor. The lower end moves away from the wall at the rate of 2 meters / minute. Find the rate at which the upper end falls when its base is 6 meters away from the wall.

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Concept understanding — Related Rates

Related Rates: When Two Things Change Together

Imagine you're blowing up a balloon. Your lungs push air in at a certain rate — say, 3 cubic centimetres per second. As the balloon grows, its radius increases. The question is: how fast is the radius increasing at the exact moment when the radius is 5 cm?

That's a related rates problem. Two quantities — the volume $V$ and the radius $r$ — are linked by a geometric formula ( $V = \frac{4}{3}\pi r^3$ ). You know how fast one is changing ( $dV/dt = 3$ ), and you want to find how fast the other is changing ( $dr/dt$ ) at a specific instant.

The core idea is simple: if two quantities are connected by an equation, their rates of change are also connected — by the derivative of that equation.

The Precise Statement

Let $x$ and $y$ be two quantities that both depend on time $t$ , and suppose they satisfy some equation $F(x, y) = 0$ (or $y = f(x)$ , etc.). Then:

Differentiate both sides of the equation with respect to time $t$ .
Use the chain rule wherever you see a variable that depends on $t$ .
The result is an equation linking $dx/dt$ and $dy/dt$ .

That's it. The "related rates" are $dx/dt$ and $dy/dt$ , and the chain rule is the tool that connects them.

\frac{d}{dt} \big[ \text{equation linking variables} \big] \quad\Longrightarrow\quad \text{equation linking rates}

The Chain Rule in Action

In the balloon example, the equation is $V = \frac{4}{3}\pi r^3$ . Differentiate both sides with respect to $t$ :

\frac{dV}{dt} = \frac{d}{dt}\left( \frac{4}{3}\pi r^3 \right)

The right side is a function of $r$ , and $r$ itself depends on $t$ . By the chain rule:

\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}

Now plug in what you know: $dV/dt = 3$ and $r = 5$ :

3 = 4\pi (5)^2 \frac{dr}{dt} \quad\Longrightarrow\quad 3 = 100\pi \frac{dr}{dt}

So:

\frac{dr}{dt} = \frac{3}{100\pi} \text{ cm/s}

That's the answer. The radius is growing at about 0.0095 cm/s when the balloon's radius is 5 cm.

Watch out

A common mistake is to plug in known values before differentiating. Don't. If you substitute $r = 5$ into $V = \frac{4}{3}\pi r^3$ first, you get a constant — and its derivative is zero. You lose the relationship between the rates. Always differentiate first, then substitute.

The General Recipe

For any related rates problem, follow these steps:

Identify all changing quantities and assign them variables. Note which rates you know and which you need.
Write an equation that relates the quantities at any time $t$ (not just at the instant of interest).
Differentiate implicitly with respect to $t$ . Use the chain rule for every variable that depends on $t$ .
Substitute the known values (including the specific instant's measurements) into the rate equation.
Solve for the unknown rate.

A Second Example: The Ladder Problem …

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With $x^2+y^2=100$ and $y=8$ when $x=6$ , differentiating gives $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=-1.5$ m/min, so the top slides down at $1.5$ m/min.

x^2+y^2=L^2

Set up the relation: $x^2+y^2=10^2=100.$
Height when $x=6$ :

y=\sqrt{100-6^2}=\sqrt{100-36}=\sqrt{64}=8\text{ m}.

Differentiate w.r.t. time $t$ :

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

Solve for $\dfrac{dy}{dt}$ : $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}.$ …

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Q.A ladder 10 meters long rests with one end against a vertical wall, the other on the floor. The lower end moves away from the wall at the rate of 2 meters / minute. Find the rate at which the upper end falls when its base is 6 meters away from the wall.

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