Question

a 5-meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at 6 meters per minute. at a certain instant, the top of the ladder is 3 meters from the ground. what is the rate of change of the area formed by the ladder at that instant (in square meters per minute)?

Answer 1

The rate of change of the area formed by a 5-meter ladder sliding down a wall, when the top of the ladder is 3 meters from the ground, is 9 square meters per minute.

Step-by-step explanation:

To find the rate of change of the area of the right-angled triangle formed by the ladder, the wall, and the ground at the instant when the ladder's bottom is moving away from the wall at 6 meters per minute, we can use the concept of related rates. At the instant where the ladder is 3 meters high (and since the ladder is 5 meters long), we can calculate the base of the triangle using the Pythagorean theorem.

Let the base of the triangle be x and the height be y. At the specific instant we're considering, y = 3 m. Using Pythagorean theorem:

x² + y² = L²
x² + 3² = 5²
x² + 9 = 25
x² = 16
x = 4 m

The area A of the right-angled triangle is A = (1/2) * base * height. Therefore, A = (1/2) * x * y. Differentiating both sides with respect to time t, we get dA/dt = (1/2) * (x * dy/dt + y * dx/dt). We know that dy/dt = 0 m/min because the top of the ladder is not moving up or down at the instant in question, and dx/dt is the speed at which the bottom of the ladder moves away from the wall, which is 6 m/min.

Thus, dA/dt = (1/2) * (4 m * 0 m/min + 3 m * 6 m/min) = 9 m²/min. So, the area of the triangle is increasing at a rate of 9 square meters per minute at that instant.

Answer 2

The rate of change of the area formed by the ladder when the top is 3 meters above the ground and the bottom is moving away from the wall at 6 meters per minute is 9 square meters per minute.

Step-by-step explanation:

To determine the rate of change of the area formed by the ladder, we can use the concept of similar triangles and rates of change.

At the instant where the ladder's bottom is moving away from the wall at 6 meters per minute and the top of the ladder is 3 meters from the ground, we can represent the situation with a right-angled triangle.

The ladder itself is the hypotenuse, with a fixed length of 5 meters.

If we denote the distance of the bottom of the ladder from the wall as x, and the height of the ladder on the wall as h, the area A of the triangle at this instant is given by:

A = (1/2) * x * h

Since we know the top of the ladder is 3 meters from the ground, we use the Pythagorean theorem to find x:

5^2 = x^2 + 3^2

Solving for x, we get:

x = √(5^2 - 3^2) = 4 meters

Therefore, the area at that instant is:

A = (1/2) * 4 * 3 = 6 square meters

The rate of change of the area can be found by differentiating the area with respect to time.

Since x is changing at 6 meters per minute, we use the chain rule for products:

dA/dt = (1/2) * (dx/dt * h + x * dh/dt)

The top of the ladder is sliding down the wall, and thus h is decreasing. However, at the instant we're considering, we only care about the rate at which the distance x is increasing, which is 6 meters per minute.

The height h is not changing instantaneously (since we're not told the rate at which it's changing), so dh/dt is 0.

The rate of change of the area is therefore:

dA/dt = (1/2) * (6 * 3 + 4 * 0) = (1/2) * 18 = 9 square meters per minute