A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall a

Q: Wait, why is the rate of change of the height negative? Shouldn't it be positive?

That is a fantastic question! In physics, the sign of a rate of change tells us about the direction of motion. Since the ladder is sliding *down* the wall, the height $y$ is decreasing over time. If a quantity decreases as time passes, its derivative with respect to time must be negative. If we used a positive value, we would be modeling the ladder climbing up the wall, which is definitely not what is happening here!

Q: Can I use trigonometry instead of the Pythagorean theorem?

Absolutely! You could define the angle $\theta$ between the ladder and the ground. Then $y = L \sin(\theta)$ and $x = L \cos(\theta)$. Differentiating these with respect to time would also give you the correct answer. However, the Pythagorean theorem is often more direct because it avoids the need to calculate the angle $\theta$ at the specific instant. Both methods are elegant, but choose the one that feels most intuitive to you.

Q: Why does the derivative of the constant $L^2$ become zero?

Think about what the ladder represents. It is a rigid object, meaning its length does not change regardless of how it slides. In calculus, the derivative of any constant value is zero because a constant does not change as time passes. Since the length of the ladder is fixed at $2$ meters, its rate of change is zero, which simplifies our equation significantly!

Question

A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25 cm/sec., then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is :

Options:

(A) $25\sqrt{3}$

(B) $25$

(C) $25/\sqrt{3}$

(D) $25/3$

Sigma Percentile · Accepted Answer

### Analyzing the Geometric Anchor

To begin, we must translate the physical world into the language of mathematics. We have a vertical wall and horizontal ground, which naturally form a right-angled triangle.

The ladder, with a fixed length $L = 2$ m, acts as the hypotenuse. Let $y$ represent the height of the top of the ladder from the ground, and $x$ represent the distance of the bottom of the ladder from the wall.

Because the ladder is rigid, the relationship between these variables is locked by the Pythagorean theorem:
$$x^2 + y^2 = L^2$$

Substituting our known length, we get the master constraint equation:
$$x^2 + y^2 = 4$$

This equation is our anchor. It tells us that $x$ and $y$ are not independent; they are bound together in a geometric embrace. If one changes, the other must respond.

### The Snapshot of Motion

The problem asks us to find the rate of change at a very specific instant: when the top of the ladder is exactly $1$ m above the ground. Think of this as taking a high-speed photograph.

At this exact moment, $y = 1$. We must find the corresponding $x$ to complete our picture.

Plugging $y = 1$ into our constraint equation, we get:
$$x^2 + (1)^2 = 4$$

This simplifies to $x^2 = 3$, or $x = \sqrt{3}$ m. Now we have the full geometry of our snapshot: the ladder is at a position where the top is $1$ m high and the base is $\sqrt{3}$ m from the wall.

### The Dynamics of Change

Now, we introduce time. The ladder is sliding, meaning $x$ and $y$ are functions of time $t$. To find how fast the base is moving, we differentiate our constraint equation $x^2 + y^2 = 4$ with respect to $t$.

Using the chain rule, we get:
$$\frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(4)$$

This yields:
$$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$$

We can simplify this by dividing by $2$, leaving us with the 'Rate Equation':
$$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$$

This equation links the velocity of the top of the ladder ($\frac{dy}{dt}$) to the velocity of the bottom ($\frac{dx}{dt}$).

### The Final Calculation

We know the top of the ladder slides down at $25$ cm/sec. Because it is sliding **down**, the height $y$ is decreasing, so $\frac{dy}{dt} = -25$ cm/sec.

Now, we substitute our known values into the rate equation:
$$(\sqrt{3}) \frac{dx}{dt} + (1)(-25) = 0$$

Solving for $\frac{dx}{dt}$, we find:
$$\sqrt{3} \frac{dx}{dt} = 25$$

This leads us to the final result:
$$\frac{dx}{dt} = \frac{25}{\sqrt{3}} 	ext{ cm/sec}$$

The bottom of the ladder is sliding away from the wall at exactly **$\frac{25}{\sqrt{3}}$ cm/sec**. By simply understanding the geometric constraints and applying the power of calculus, we have predicted the motion of a physical object.

Select Answer:

Visualized Solution

Visualizing the Setup

Defining Variables $x$ and $y$

The Geometric Constraint

The Snapshot in Time: $y = 1$

Finding $x$ when $y = 1$

Understanding the Given Rate

Differentiating w.r.t. Time

Applying the Chain Rule

Substituting Known Values

The Raw Equation

Solving for $\frac{d x}{d t}$

Final Takeaway

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Geometric Anchor

The Snapshot of Motion

The Dynamics of Change

The Final Calculation

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