A ladder $10\text{ m}$ long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of $1\text{ m/s}$, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is $6\text{ m}$ from the wall?
1. Let \( t \) be the time in seconds. Define \( x(t) \) as the distance from the bottom of the ladder to the wall, and \( y(t) \) as the height of the top of the ladder from the ground.
2. According to the Pythagorean theorem, the relationship between \( x(t) \) and \( y(t) \) is given by:
   \[
   x(t)^2 + y(t)^2 = 10^2 = 100
   \]
3. Differentiate both sides of the equation with respect to \( t \):
   \[
   \frac{d}{dt} \left( x(t)^2 + y(t)^2 \right) = \frac{d}{dt} (100)
   \]
4. Applying the chain rule to the left-hand side:
   \[
   2x(t) \frac{dx(t)}{dt} + 2y(t) \frac{dy(t)}{dt} = 0
   \]
5. Simplify the equation:
   \[
   x(t) \frac{dx(t)}{dt} + y(t) \frac{dy(t)}{dt} = 0
   \]
6. From the problem statement, we know that \( x(t) = 6 \) meters and \( \frac{dx(t)}{dt} = 1 \) m/s. We need to find \( \frac{dy(t)}{dt} \) when \( x(t) = 6 \).
7. Using the Pythagorean theorem to find \( y(t) \) when \( x(t) = 6 \):
   \[
   y(t) = \sqrt{100 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 \text{ meters}
   \]
8. Substitute \( x(t) = 6 \), \( y(t) = 8 \), and \( \frac{dx(t)}{dt} = 1 \) into the differentiated equation:
   \[
   6 \cdot 1 + 8 \cdot \frac{dy(t)}{dt} = 0
   \]
9. Solve for \( \frac{dy(t)}{dt} \):
   \[
   6 + 8 \frac{dy(t)}{dt} = 0
   \]
   \[
   8 \frac{dy(t)}{dt} = -6
   \]
   \[
   \frac{dy(t)}{dt} = -\frac{6}{8} = -\frac{3}{4} \text{ m/s}
   \]
10. The negative sign indicates that the top of the ladder is sliding down the wall. The rate at which the top of the ladder is sliding down the wall is \( \frac{3}{4} \text{ m/s} \).

The final answer is \( \boxed{\frac{3}{4} \text{ m/s}} \).