An 8m ladder is leaning against a vertical wall for a repair of crack on the wall of your farm. The bottom of the ladder is pulled horizontally away from the wall at 1.5m/s how fast is the wall when the bottom of the ladder is 5m away from the wall.

x^2 + y^2 = 64

when x=5, y=√39
x dx/dt + y dy/dt = 0
so plug in your numbers and solve for dy/dt

Well, it sounds like this ladder is having quite the adventure! Let's see if we can help it out.

We need to find the rate at which the wall is moving when the bottom of the ladder is 5 meters away. From what we know, the ladder is 8 meters long and is leaning against the wall. The bottom of the ladder is being pulled horizontally at a rate of 1.5 meters per second.

To solve this, we can use similar triangles. The height of the triangle formed by the ladder, the wall, and the ground is the wall's ratem and the base is the horizontal distance between the ladder and the wall.

So, we have a similar triangle with a base of 5 meters and a height of m (the rate at which the wall is moving). The original triangle has a base of 0 meters and a height of 8 meters.

By setting up proportions, we can find the value of m:

8 / 0 = m / 5

Simplifying this equation, we get:

0 = 5m / 8

Since the numerator is 0, the rate of the wall is also 0.

So, when the bottom of the ladder is 5 meters away from the wall, the wall is not moving at all! I guess it's just taking a break from all those cracks.

Hope that brings a little humor to your problem. If you have any more questions, feel free to ask!

To find the rate at which the wall is moving, we need to use related rates and trigonometry.

Let's assume that the distance between the bottom of the ladder and the wall is given by the variable "x" (in meters). We are given that the bottom of the ladder, x, is changing at a rate of 1.5m/s.

We can set up a right triangle to represent the ladder leaning against the wall. The ladder is the hypotenuse, and the wall is the height. The base of the triangle is the distance between the bottom of the ladder and the wall (x). Let's call the height of the wall "h."

Using the Pythagorean theorem, we can relate the sides of the triangle:

h^2 + x^2 = 8^2

Differentiating both sides of this equation with respect to time t gives:

2h * dh/dt + 2x * dx/dt = 0

Since we want to find dh/dt, we can solve for it:

dh/dt = -(x / h) * dx/dt

Now, we are given that x = 5m, and we need to find dh/dt when x = 5m. To do this, we need to find the value of h.

Using the Pythagorean theorem, we have:

h^2 + 5^2 = 8^2

Simplifying, we find:

h^2 = 64 - 25 = 39

Taking the square root of both sides, we get:

h = √39

Now we can substitute the values into our equation for dh/dt:

dh/dt = -(5 / √39) * 1.5

Evaluating this expression, we find:

dh/dt ≈ -1.5 * (5 / √39)

So, the wall is moving at a rate of approximately -1.5 times (5 divided by the square root of 39) meters per second when the bottom of the ladder is 5 meters away from the wall. The negative sign indicates that the height is decreasing.

To determine how fast the wall is moving when the bottom of the ladder is 5m away from it, we need to use related rates. We have the following information:

- The ladder is 8m long and is leaning against the vertical wall.
- The bottom of the ladder is being pulled horizontally away from the wall at a rate of 1.5m/s.

Let's assign some variables to the problem:

- Let "d" represent the distance between the bottom of the ladder and the wall.
- Let "x" represent the distance from the bottom of the ladder to the base of the wall (where the ladder touches the ground).
- Let "y" represent the height of the wall.
- Let "θ" represent the angle between the ladder and the ground.

Since we have a right triangle created by the ladder, wall, and ground, we can use the Pythagorean theorem to relate the variables:

d² + y² = 8² (Equation 1)

Now, we want to find the rate at which the wall is moving when d = 5m. We need to differentiate Equation 1 with respect to time (t) to get the related rates equation:

2d(d/dt) + 2y(dy/dt) = 0

We want to find dy/dt when d = 5m. We already know that dx/dt (the rate at which the distance from the bottom of the ladder to the base of the wall is changing) is 1.5m/s.

Plugging the values into the equation:

2(5)(d/dt) + 2y(dy/dt) = 0
10(d/dt) + 2y(dy/dt) = 0

We can solve this equation for dy/dt:
dy/dt = -(10(d/dt)) / (2y)

Now, we need to find y when d = 5m. Rearranging Equation 1 gives us:
y² = 8² - d²
y² = 8² - (5²)
y² = 64 - 25
y² = 39
y = √(39)

Plugging in the values:
dy/dt = -(10(1.5)) / (2√(39))

Finally, we can solve this to find the rate at which the wall is moving:
dy/dt = -(15) / (√(39))

The rate at which the wall is moving when the bottom of the ladder is 5m away from it is approximately -(15 / √(39)) m/s. The negative sign indicates that the wall is moving downwards.