A ladder 10m long rest against a vertical wall with the lower end on the horizontal ground. The lower end of the ladder is pulled along the ground away from the wall at the rate of 3 cm/sec. The height of the upper end while it is descending at the rate of 4 cm/sec is

Options

A. `4sqrt3`m
B. 6 m
C. `5sqrt2` m
D. 8 m

Explanation

To solve the problem step by step, we will use the Pythagorean theorem and related rates.

### Step 1: Understand the problem
We have a ladder of length 10 m resting against a wall. The lower end of the ladder is being pulled away from the wall at a rate of 3 cm/sec, and the upper end is descending at a rate of 4 cm/sec. We need to find the height of the upper end of the ladder while it is descending.

### Step 2: Set up the variables
Let:
- \( x \) = the distance from the wall to the bottom of the ladder (horizontal distance)
- \( y \) = the height of the top of the ladder (vertical distance)
- The length of the ladder is constant at 10 m.

According to the Pythagorean theorem:
\[ x^2 + y^2 = 10^2 \]
or
\[ x^2 + y^2 = 100 \]

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

### Step 4: Substitute known rates
We know:
- \( \frac{dx}{dt} = 3 \) cm/sec (the rate at which the bottom of the ladder is moving away from the wall)
- \( \frac{dy}{dt} = -4 \) cm/sec (the rate at which the top of the ladder is descending, negative because \( y \) is decreasing)

Substituting these values into the differentiated equation:
\[ x(3) + y(-4) = 0 \]
This simplifies to:
\[ 3x - 4y = 0 \]
or
\[ 3x = 4y \quad \Rightarrow \quad x = \frac{4y}{3} \]

### Step 5: Substitute \( x \) back into the original equation
Now, substitute \( x = \frac{4y}{3} \) into the original Pythagorean equation:
\[ \left(\frac{4y}{3}\right)^2 + y^2 = 100 \]
This gives:
\[ \frac{16y^2}{9} + y^2 = 100 \]
To combine the terms, convert \( y^2 \) to a fraction:
\[ \frac{16y^2}{9} + \frac{9y^2}{9} = 100 \]
This simplifies to:
\[ \frac{25y^2}{9} = 100 \]

### Step 6: Solve for \( y \)
Multiply both sides by 9:
\[ 25y^2 = 900 \]
Now divide by 25:
\[ y^2 = 36 \]
Taking the square root:
\[ y = 6 \text{ m} \]

### Conclusion
The height of the upper end of the ladder while it is descending is 6 meters.

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