a 4m ladder is leaning against a wall when its base starts to slide away. By the time the base is 2m from the wall, the base is moving at the rate of 1m/sec.

find
a) the speed at which the top of the ladder is sliding down the wall

b) the rate at which the area of the triangle formed by the ladder, wall and ground is changing.

c) the rate at which the angle theta between the ladder and the ground is changing.

Thanks!!

with x=distance of base from wall, y the height,

x^2 + y^2 = 16
2x dx/dt + 2y dy/dt = 0
when x=2, dx/dt=1
y = √12, so
2(2) + 2√12 dy/dt = 0
dy/dt = -4/2√12 = -1/√3

a = xy/2
2da/dt = y dx/dt + x dy/dt
= √12*1 - 2/√3
= 4/√3
so da/dt = 2/√3

tanθ = y/x
sec^2θ dθ/dt = (xy' - yx')/x^2
when x=2, tanθ = √12/2 = √3, so θ=π/6
sec^2θ = 4/3, so
4/3 dθ/dt = (2(-1/√3)-(√12)(1))/4
dθ/dt = -√3 / 2

as always, double-check my algebra

To solve this problem, we can use related rates, a branch of calculus that involves finding the rates of change of two or more related variables. Let's walk through each part of the problem step by step.

a) The speed at which the top of the ladder is sliding down the wall:
Let's denote the height of the ladder as "h" and the distance of the base from the wall as "x". We are given that dx/dt = 1 m/sec when x = 2 m.

We can use the Pythagorean theorem to relate h and x. According to the theorem, h^2 + x^2 = 4^2 (since a 4m ladder is used). We can differentiate both sides of the equation with respect to time "t" to find the related rates:

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

Since we are interested in finding dh/dt (the rate at which the top of the ladder is sliding down the wall), we can solve the equation for dh/dt:

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

Substituting the given values, x = 2 m, h = 4 m, and dx/dt = 1 m/sec, we get:

dh/dt = (-2/4) * 1 m/sec = -0.5 m/sec

Therefore, the speed at which the top of the ladder is sliding down the wall is -0.5 m/sec. The negative sign indicates that the height is decreasing.

b) The rate at which the area of the triangle formed by the ladder, wall, and ground is changing:
The area of a triangle can be calculated using the formula: A = (1/2) * b * h

In this case, the base of the triangle is the distance between the wall and the ladder (x), and the height is the ladder's length (4 m). Therefore, the area can be written as A = (1/2) * x * 4.

We need to find dA/dt, the rate at which the area is changing. The derivative of the area is given by:

dA/dt = (1/2) * (4) * (dx/dt)

Substituting the given value dx/dt = 1 m/sec, we get:

dA/dt = (1/2) * (4) * (1 m/sec) = 2 m²/sec

Therefore, the rate at which the area of the triangle is changing is 2 m²/sec.

c) The rate at which the angle theta between the ladder and the ground is changing:
To find the rate at which the angle theta is changing, we can use trigonometry. Let's denote the angle theta as θ. According to the problem, the ladder's base is moving away from the wall at a rate of 1 m/sec.

The trigonometric relationship between the angle, the height, and the base of the triangle is given by the tangent function:

tan(θ) = h / x

Differentiating both sides with respect to time "t" gives:

sec²(θ) * dθ/dt = (dh/dt * x - h * dx/dt) / x²

We can solve the equation for dθ/dt, the rate at which the angle theta is changing:

dθ/dt = (dh/dt * x - h * dx/dt) / (x² * sec²(θ))

Substituting the given values, dh/dt = -0.5 m/sec, h = 4 m, x = 2 m, and dx/dt = 1 m/sec, we can evaluate the equation:

dθ/dt = (-0.5 * 2 - 4 * 1) / (2² * sec²(θ))

Simplifying further, we get:

dθ/dt = (-1 - 4) / (4 * sec²(θ))
= -5 / (4 * sec²(θ))

Therefore, the rate at which the angle theta is changing can be determined by knowing the value of sec²(θ). Without any information about θ, we cannot find the exact value for dθ/dt.

These are the solutions to the three parts of the problem. If you have any more questions, feel free to ask!