Key Concepts and Formulas

• Pythagorean Theorem: In a right-angled triangle with sides $a$ and $b$, and hypotenuse $c$, we have $a^2 + b^2 = c^2$.
• Related Rates: If $x$ and $y$ are functions of time $t$, then differentiating an equation relating $x$ and $y$ with respect to $t$ gives a relationship between their rates of change, $\frac{dx}{dt}$ and $\frac{dy}{dt}$.
• Implicit Differentiation: Differentiating an equation implicitly with respect to a variable (in this case, time $t$) involves applying the chain rule to each term.

Step-by-Step Solution

Step 1: Define Variables and Diagram

Let $x$ be the horizontal distance of the bottom of the ladder from the wall, and $y$ be the vertical distance of the top of the ladder from the ground. The length of the ladder, $L$, is constant and equal to 2 meters = 200 cm. We have a right-angled triangle with sides $x$ and $y$ and hypotenuse $L$.

Step 2: State the Given Information

We are given that $\frac{dy}{dt} = -25$ cm/sec (negative since the top of the ladder is sliding down). We are also given that $y = 1$ m = 100 cm. We want to find $\frac{dx}{dt}$ when $y = 100$ cm.

Step 3: Apply the Pythagorean Theorem

We have the relationship $x^2 + y^2 = L^2$, where $L = 200$ cm. So, $x^2 + y^2 = 200^2 = 40000$.

Step 4: Differentiate with Respect to Time

Differentiate both sides of the equation $x^2 + y^2 = 40000$ with respect to time $t$:

$\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(40000)$

Using the chain rule, we get:

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

Step 5: Simplify the Equation

Divide both sides by 2:

$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$

Step 6: Find the Value of x when y = 100 cm

When $y = 100$ cm, we can find $x$ using the Pythagorean theorem:

$x^2 + (100)^2 = 200^2$ $x^2 + 10000 = 40000$ $x^2 = 30000$ $x = \sqrt{30000} = 100\sqrt{3} \text{ cm}$

Step 7: Substitute and Solve for dx/dt

Substitute $x = 100\sqrt{3}$, $y = 100$, and $\frac{dy}{dt} = -25$ into the equation $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$:

$(100\sqrt{3}) \frac{dx}{dt} + (100)(-25) = 0$ $100\sqrt{3} \frac{dx}{dt} = 2500$ $\frac{dx}{dt} = \frac{2500}{100\sqrt{3}} = \frac{25}{\sqrt{3}} \text{ cm/sec}$

Step 8: Rationalize the Denominator (Optional) $\frac{dx}{dt} = \frac{25}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{25\sqrt{3}}{3} \text{ cm/sec}$

Common Mistakes & Tips

• Units: Ensure all units are consistent (e.g., convert meters to centimeters).
• Sign Convention: Pay attention to the sign of the rates of change. Sliding down implies a negative rate of change for $y$.
• Implicit Differentiation: Remember to apply the chain rule correctly when differentiating with respect to time.

Summary

We used the Pythagorean theorem to relate the horizontal and vertical distances of the ladder, then differentiated with respect to time to relate their rates of change. Substituting the given values and solving for $\frac{dx}{dt}$, we found that the bottom of the ladder slides away from the wall at a rate of $\frac{25}{\sqrt{3}}$ cm/sec.

Final Answer

The final answer is $\boxed{\frac{25}{\sqrt{3}}}$, which corresponds to option (D).