A 10 -ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of $$ 2 \mathrm{ft} / \mathrm{sec} $$, how fast is the bottom moving along the ground when the bottom of the ladder is 5 ft from the wall?

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We start with the relation from the Pythagorean theorem for the ladder:

$$
x^2 + y^2 = 100
$$

where $$ x $$ is the distance along the ground and $$ y $$ is the height on the wall.

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

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

Simplify by dividing the entire equation by 2:

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

Solve for $$ \frac{dx}{dt} $$:

$$
\frac{dx}{dt} = -\frac{y}{x} \frac{dy}{dt}
$$

At the moment when the bottom of the ladder is 5 ft from the wall, we have $$ x = 5 $$. Use the Pythagorean theorem to find $$ y $$:

$$
5^2 + y^2 = 100 \quad \Rightarrow \quad 25 + y^2 = 100 \quad \Rightarrow \quad y^2 = 75 \quad \Rightarrow \quad y = 5\sqrt{3}
$$

Given that the top of the ladder is sliding down the wall at a rate of $$ \frac{dy}{dt} = -2 $$ ft/sec (negative since $$ y $$ is decreasing), substitute into the formula:

$$
\frac{dx}{dt} = -\frac{5\sqrt{3}}{5} (-2) = \sqrt{3} \cdot 2 = 2\sqrt{3} \text{ ft/sec}
$$

Thus, the bottom of the ladder is moving along the ground at a rate of $$ 2\sqrt{3} $$ ft/sec.
The bottom of the ladder is moving along the ground at a rate of $$ 2\sqrt{3} $$ ft/sec.

A 10 -ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of \( 2 \mathrm{ft} / \mathrm{sec} \), how fast is the bottom moving along the ground when the bottom of the ladder is 5 ft from the wall?

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