A 7 -ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a of $$ 4 \mathrm{ft} / \mathrm{sec} $$, how fast is the bottom moving along the ground when the bottom of the ladder is 6 ft from the wall?

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We can solve this related rates problem by following these steps:

1. **Define the Variables:**
   - Let $$ x(t) $$ be the distance from the bottom of the ladder to the wall (along the ground).
   - Let $$ y(t) $$ be the height of the ladder on the wall.
   
2. **Relationship Between $$ x $$ and $$ y $$:**
   Since the ladder remains 7 feet long, by the Pythagorean theorem:
   $$
   x^2 + y^2 = 7^2 \implies x^2 + y^2 = 49
   $$
   
3. **Differentiate with Respect to Time $$ t $$:**
   $$
   \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(49)
   $$
   $$
   2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
   $$
   Simplifying:
   $$
   x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \implies \frac{dx}{dt} = -\frac{y}{x} \frac{dy}{dt}
   $$
   
4. **Given Values:**
   - $$ \frac{dy}{dt} = -4 $$ ft/s (the top of the ladder is sliding down).
   - $$ x = 6 $$ ft (distance from the wall).
   
5. **Find $$ y $$ When $$ x = 6 $$:**
   $$
   6^2 + y^2 = 49 \implies y^2 = 13 \implies y = \sqrt{13} \text{ ft}
   $$
   
6. **Calculate $$ \frac{dx}{dt} $$:**
   $$
   \frac{dx}{dt} = -\frac{\sqrt{13}}{6} \times (-4) = \frac{4\sqrt{13}}{6} = \frac{2\sqrt{13}}{3} \text{ ft/s}
   $$
   
**Conclusion:**

When the bottom of the ladder is 6 feet from the wall, it is moving away from the wall at a rate of $$\frac{2\sqrt{13}}{3}$$ feet per second.

**Final Answer:**

After simplifying, the bottom moves at (2 × √13)⁄3 ft per second.
When the bottom of the ladder is 6 feet from the wall, it is moving away from the wall at a rate of $$\frac{2\sqrt{13}}{3}$$ feet per second.

A 7 -ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a of \( 4 \mathrm{ft} / \mathrm{sec} \), how fast is the bottom moving along the ground when the bottom of the ladder is 6 ft from the wall?

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