A ladder 10 metres long rests with one end against a vertical wall, the other on the floor. The lower end moves away from the wall at the rate of 2 metres/minute. The rate at which the upper end falls when its base is 6 metres away from the wall

Options

A. 3 metres/min
B. 2/3 metres/min
C. -3/2 metres/min
D. none of these

Explanation

To solve the problem step by step, we will use related rates in calculus.

### Step 1: Understand the problem and set up the variables
Let:
- \( x \) = distance of the lower end of the ladder from the wall (in meters)
- \( y \) = height of the upper end of the ladder from the ground (in meters)
- The length of the ladder is constant at 10 meters.

### Step 2: Establish the relationship using the Pythagorean theorem
Since the ladder, wall, and ground form a right triangle, we can use the Pythagorean theorem:
\[ x^2 + y^2 = 10^2 \]
This simplifies to:
\[ x^2 + y^2 = 100 \]

### Step 3: Differentiate with respect to time \( t \)
Differentiating both sides with respect to \( t \):
\[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(100) \]
Using the chain rule, this gives:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
Dividing through by 2:
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]

### Step 4: Substitute known values
We know from the problem that:
- \( \frac{dx}{dt} = 2 \) meters/minute (the lower end moves away from the wall)
- We need to find \( \frac{dy}{dt} \) when \( x = 6 \) meters.

### Step 5: Find \( y \) when \( x = 6 \)
Using the Pythagorean theorem:
\[ x^2 + y^2 = 100 \]
Substituting \( x = 6 \):
\[ 6^2 + y^2 = 100 \implies 36 + y^2 = 100 \implies y^2 = 64 \implies y = 8 \text{ meters} \]

### Step 6: Substitute \( x \) and \( y \) into the differentiated equation
Now substitute \( x = 6 \), \( y = 8 \), and \( \frac{dx}{dt} = 2 \) into the differentiated equation:
\[ 6(2) + 8 \frac{dy}{dt} = 0 \]
This simplifies to:
\[ 12 + 8 \frac{dy}{dt} = 0 \]
Solving for \( \frac{dy}{dt} \):
\[ 8 \frac{dy}{dt} = -12 \implies \frac{dy}{dt} = -\frac{12}{8} = -\frac{3}{2} \text{ meters/minute} \]

### Conclusion
The rate at which the upper end of the ladder falls when its base is 6 meters away from the wall is:
\[ \frac{dy}{dt} = -\frac{3}{2} \text{ meters/minute} \]