A 13-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away at a rate of 2 feet per second, how fast is the area of the triangle formed by the wall, the ground, and the ladder changing when the bottom is 12 feet away from the wall?

Well, well, well, it seems like the ladder wants to do the limbo! Let's see what we can figure out here.

First things first, we need to find the rate of change of the ladder's height. As the bottom of the ladder is being pulled away at a rate of 2 feet per second, the height of the ladder is decreasing at the same rate. So, the height of the ladder can be given by the function h(t) = 13 - 2t, where t represents time in seconds.

Now, we want to find how fast the area of the triangle formed by the ladder, the wall, and the ground is changing when the bottom is 12 feet away from the wall. We'll call this area A.

The area of a triangle is given by the formula A = (1/2) * base * height. In this case, the base of the triangle is always 12 feet since that's how far away the bottom is from the wall.

To find the rate of change of A, we differentiate the area function with respect to time t and substitute the value of t when the bottom is 12 feet away (t = 6 seconds). So, A'(6) represents the required rate of change.

Differentiating A = (1/2) * 12 * (13 - 2t) with respect to t, we get:

A' = (1/2) * 12 * (-2) = -12

Therefore, the area of the triangle is changing at a rate of -12 square feet per second when the bottom of the ladder is 12 feet away from the wall.

I hope that wasn't too much of a balancing act for you!

To solve this problem, we need to apply the concept of related rates. We are given that the bottom of the ladder is being pulled away at a rate of 2 feet per second. We need to find how fast the area of the triangle formed by the wall, the ground, and the ladder is changing when the bottom is 12 feet away from the wall.

Let's start by labeling the triangle:
The vertical wall forms the height of the triangle, and the distance from the bottom of the ladder to the wall represents the base. The ladder itself represents the hypotenuse.

Let's label:
- The length of the ladder (hypotenuse) as L
- The distance from the bottom of the ladder to the wall (base) as x
- The height of the triangle (vertical wall) as h
- The area of the triangle as A (which we are looking for)
- The rate at which the base is changing as dx/dt (given as 2 ft/s in this question)

We know that the area of a triangle can be calculated using the formula:
A = (1/2) * base * height

So, in this case, we have:
A = (1/2) * x * h

To find the relationship between the variables, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In our case:
L² = x² + h²

Differentiating both sides with respect to time:
(dL/dt) * 2L = 2x(dx/dt) + 2h(dh/dt)

We are given dL/dt = 0 because the ladder length is constant.
Also, at the moment in question, the bottom is 12 feet away from the wall, so x = 12.

Substituting these values into our equation:
0 = 2(12)(dx/dt) + 2h(dh/dt)

Simplifying the equation further:
0 = 24(dx/dt) + 2h(dh/dt) ... (1)

We need to find (dh/dt), which represents how fast the height of the triangle is changing.
To solve for (dh/dt), we need to know the value of h. We can use the right triangle relationship again:

L² = x² + h²

Substituting the known values:
13² = 12² + h²
169 = 144 + h²
h² = 169 - 144
h² = 25
h = 5

Now, we have the value of h, which is 5.

Substituting the values of x = 12 and h = 5 into equation (1):
0 = 24(2) + 2(5)(dh/dt)

Simplifying:
0 = 48 + 10(dh/dt)
48 = -10(dh/dt)
(dh/dt) = -48/10
(dh/dt) = -4.8 ft/s

Therefore, the area of the triangle is changing at a rate of -4.8 square feet per second when the bottom is 12 feet away from the wall.

To find how fast the area of the triangle is changing, we can use the chain rule to differentiate the area formula with respect to time. The formula for the area of a triangle is A = (1/2) * base * height.

First, let's assign variables to the quantities in the problem:
- Let x represent the distance between the bottom of the ladder and the wall.
- Let A represent the area of the triangle.
- Let t represent time.

Given:
- The ladder is 13 feet long, so the height of the triangle is 13 feet.
- The bottom of the ladder is being pulled away at a rate of 2 feet per second, so dx/dt = 2 ft/s.

We need to find dA/dt, the rate at which the area is changing with respect to time.

Now, let's find an equation connecting the three variables: x, A, and t.

Using the formula for the area of a triangle, we have A = (1/2) * x * 13.

Differentiating both sides of the equation with respect to time (t), we get:

dA/dt = (1/2) * (d/dt)(x * 13)

Now, to find (d/dt)(x * 13), we can apply the product rule:

(d/dt)(x * 13) = (dx/dt) * 13 + x * (d/dt)(13)

Since dx/dt = 2 ft/s and (d/dt)(13) = 0 (as 13 is a constant), we simplify further:

(d/dt)(x * 13) = (2) * 13 + x * 0
(d/dt)(x * 13) = 26

Substituting this value back into the equation for dA/dt, we get:

dA/dt = (1/2) * (26)

Simplifying, we have:

dA/dt = 13 ft^2/s

Therefore, when the bottom of the ladder is 12 feet away from the wall, the area of the triangle is changing at a rate of 13 square feet per second.