This problem can be solved using related rates in calculus.

Step 1: Identify what you know and what you need to find out.

We know that the ladder is 13 feet long (the hypotenuse of the triangle), the rate at which the bottom of the ladder is moving away from the wall (dx/dt = 4 ft/sec), and we need to find out the rate at which the area of the triangle is changing (dA/dt) when the bottom of the ladder is 12 feet from the wall.

Step 2: Set up the equation for the area of the triangle.

The area of a triangle is given by A = 1/2 * base * height. In this case, the base is x (the distance from the wall to the bottom of the ladder) and the height is y (the height up the wall to where the ladder touches). So, A = 1/2 * x * y.

Step 3: Use the Pythagorean theorem to find y.

The ladder, wall, and ground form a right triangle, so we can use the Pythagorean theorem (a^2 + b^2 = c^2) to find y. The ladder is the hypotenuse (c), x is one leg (a), and y is the other leg (b). So, y = sqrt(c^2 - a^2) = sqrt(13^2 - 12^2) = sqrt(169 - 144) = sqrt(25) = 5 feet.

Step 4: Differentiate the area equation with respect to time (t).

dA/dt = 1/2 * (x * dy/dt + y * dx/dt)

Step 5: Find dy/dt.

Again using the Pythagorean theorem, differentiate both sides with respect to time to get 2x * dx/dt + 2y * dy/dt = 0 (since the ladder length is constant and does not change with time). Solve for dy/dt to get dy/dt = -x/y * dx/dt = -12/5 * 4 = -48/5 ft/sec.

Step 6: Substitute x, y, dx/dt, and dy/dt into the differentiated area equation to find dA/dt.

dA/dt = 1/2 * (12 * -48/5 + 5 * 4) = -288/5 + 10 = -58/5 = -11.6 square feet per second.

So, the area of the triangle is decreasing at a rate of 11.6 square feet per second when the bottom of the ladder is 12 feet from the wall.

Question

This problem can be solved using related rates in calculus.

Step 1: Identify what you know and what you need to find out.

We know that the ladder is 13 feet long (the hypotenuse of the triangle), the rate at which the bottom of the ladder is moving away from the wall (dx/dt = 4 ft/sec), and we need to find out the rate at which the area of the triangle is changing (dA/dt) when the bottom of the ladder is 12 feet from the wall.

Step 2: Set up the equation for the area of the triangle.

The area of a triangle is given by A = 1/2 * base * height. In this case, the base is x (the distance from the wall to the bottom of the ladder) and the height is y (the height up the wall to where the ladder touches). So, A = 1/2 * x * y.

Step 3: Use the Pythagorean theorem to find y.

The ladder, wall, and ground form a right triangle, so we can use the Pythagorean theorem (a^2 + b^2 = c^2) to find y. The ladder is the hypotenuse (c), x is one leg (a), and y is the other leg (b). So, y = sqrt(c^2 - a^2) = sqrt(13^2 - 12^2) = sqrt(169 - 144) = sqrt(25) = 5 feet.

Step 4: Differentiate the area equation with respect to time (t).

dA/dt = 1/2 * (x * dy/dt + y * dx/dt)

Step 5: Find dy/dt.

Again using the Pythagorean theorem, differentiate both sides with respect to time to get 2x * dx/dt + 2y * dy/dt = 0 (since the ladder length is constant and does not change with time). Solve for dy/dt to get dy/dt = -x/y * dx/dt = -12/5 * 4 = -48/5 ft/sec.

Step 6: Substitute x, y, dx/dt, and dy/dt into the differentiated area equation to find dA/dt.

dA/dt = 1/2 * (12 * -48/5 + 5 * 4) = -288/5 + 10 = -58/5 = -11.6 square feet per second.

So, the area of the triangle is decreasing at a rate of 11.6 square feet per second when the bottom of the ladder is 12 feet from the wall.

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