A ladder 13 feet long is leaning against the side of a building. If the foot of the ladder is pulled away from the building at a constant rate of 8 inches per second, how fast is the area of the triangle formed by the ladder, the building, and the ground changing in feet squared per second at the instant when the top of the ladder is 12 feet above ground? I can't seem to find anything like this in my notes... Thank you for your help in advance

eh? did you not write down your calculations? How can you possibly not know how you arrived at an answer. I think you mean you are not sure whether your answer is right.

At the moment in question, we have a 5-12-13 right triangle.

x^2 + y^2 = 13^2
2x dx/dt + 2y dy/dt = 0
2(5)(2/3) + 2(12) dy/dt = 0
dy/dt = -5/18 ft/s

the area is 1/2 xy, so

da/dt = 1/2 y dx/dt + 1/2 x dy/dt
= (1/2)((12)(2/3) + (5)(-5/18))
= 119/36

Not sure how you got your minus sign. The ladder has just started to fall, so the area of the triangle will be increasing until x and y are equal.

I got -119/36 ft^2/sec... not really sure how I got that...

To find the rate at which the area of the triangle is changing, we will use the formula for the area of a triangle, which is A = 1/2 * base * height.

Let's denote the base of the triangle as x and the height as y. We are given that the ladder is 13 feet long, so when the top of the ladder is 12 feet above the ground, the base of the triangle is 5 feet (13 - 12) as it forms a right angle with the ground.

We need to find an expression for y in terms of x. We can do this by using the Pythagorean theorem: x^2 + y^2 = 13^2.

Substituting the value of x from above, we get:

5^2 + y^2 = 13^2,
25 + y^2 = 169,
y^2 = 169 - 25,
y^2 = 144,
y = 12.

We have found that when x = 5, y = 12.

Next, we need to find dx/dt (the rate at which x is changing) and dy/dt (the rate at which y is changing).

We are given that the foot of the ladder is pulled away from the building at a constant rate of 8 inches per second. Since we want to find the rate of change in area in feet squared per second, we need to convert inches to feet.

dx/dt = 8 inches per second * (1 foot/12 inches) = 8/12 feet per second = 2/3 feet per second.

To find dy/dt, we can differentiate the equation x^2 + y^2 = 13^2 implicitly with respect to t (time).

2x * dx/dt + 2y * dy/dt = 0,

(2x * dx/dt) + (2y * dy/dt) = 0,

Simplifying, we get:

2x * dx/dt = -2y * dy/dt,

dy/dt = -(x/y) * dx/dt.

Now we can find dy/dt at the given point (x=5, y=12):

dy/dt = -(5/12) * (2/3) = -10/36 = -5/18 feet per second.

Finally, we can find dA/dt (the rate at which the area is changing) by differentiating the formula for the area (A = 1/2 * base * height) with respect to t:

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

At the given point (x=5, y=12), we find:

dA/dt = 1/2 * (5 * (-5/18) + 12 * (2/3)) = 1/2 * (-25/18 + 24/3) = 1/2 * (-25/18 + 144/18) = 1/2 * 119/18 = 119/36 feet squared per second.

Therefore, the area of the triangle is changing at a rate of 119/36 feet squared per second.

To solve this problem, we can use the concepts of similar triangles and the area of a triangle.

Let's denote the distance between the foot of the ladder and the building as x. According to the problem, the foot of the ladder is moving away from the building at a constant rate of 8 inches per second. Since 1 foot is equal to 12 inches, the rate of the foot's movement can be expressed as x'(t) = 8/12 = 2/3 feet per second.

We can apply the Pythagorean theorem to find the height of the triangle formed by the ladder, the building, and the ground. Let h be the height of the triangle. Using the Pythagorean theorem, we have:

h^2 + x^2 = 13^2
h^2 + x^2 = 169

Differentiating both sides of the equation with respect to time t, we have:

2h * h' + 2x * x' = 0

Since x' = 2/3, we can solve for h':

2h * h' + 2x * (2/3) = 0
2h * h' = - (4/3) * x

Now, when the top of the ladder is 12 feet above the ground, we know that x = 13 - 12 = 1 foot. Plugging this value into the equation, we get:

2h * h' = - (4/3) * (1)
2h * h' = -4/3

To find the value of h', we need to know the value of h. Using the Pythagorean theorem again, we can substitute x = 1 into the equation:

h^2 + 1^2 = 13^2
h^2 + 1 = 169
h^2 = 168
h ≈ 12.961

Now, substituting h ≈ 12.961 into the equation 2h * h' = -4/3, we have:

2(12.961) * h' = -4/3
h' ≈ - (4/3) / (2 * 12.961)
h' ≈ -0.05193 feet per second

The negative sign indicates that the height is decreasing over time.

Finally, we can find the rate of change of the area of the triangle by using the formula for the area of a triangle:

A = (1/2) * base * height

In this case, the base of the triangle is the distance that the foot of the ladder is pulled away from the building, x.

The rate of change of the area with respect to time t is given by:

A' = (1/2) * x' * h + (1/2) * x * h'

Substituting the known values, x = 1 and h' ≈ -0.05193, we have:

A' = (1/2) * (2/3) * 12.961 + (1/2) * 1 * (-0.05193)
A' ≈ 4.321 feet squared per second

Therefore, the area of the triangle is changing at a rate of approximately 4.321 feet squared per second at the instant when the top of the ladder is 12 feet above the ground.