A ladder 10ft long rests against a vertical wall. If the bottom on the ladder slides away from the wall at 2 ft/s, how fast is the angle ladder makes with the ground changing when the bottom of the ladder is 5 ft from the wall.

Let the foot be x ft from the wall, let the angle formed by the ladder at the ground be Ø

given: dx/dt = 2 ft/s
find: dØ/dt when x = 5

x/10 = cosØ
x = 10cosØ
dx/dt = -10sinØ dØ/dt

when x=5,
height^2 + 5^2 = 10^2
height = √75

2 = -10(√75/10) dØ/dt
dØ/dt = - 2/√75 radians/second
= appr. - .23 rad/s

100mph

To find the rate at which the angle θ (in radians) between the ladder and the ground is changing, we need to use the concept of related rates.

Let's denote θ as the angle between the ladder and the ground, x as the distance between the bottom of the ladder and the wall, and L as the length of the ladder. In this case, x is changing over time, and we need to find dθ/dt, the rate at which θ is changing with respect to time.

We know that L = 10 ft, and dx/dt = -2 ft/s (negative because the bottom of the ladder slides away from the wall).

Using trigonometry, we can establish a relationship between θ and x:

sin(θ) = x / L

Differentiate both sides of this equation with respect to time (t):

d/dt(sin(θ)) = d/dt(x/L)

cos(θ) * dθ/dt = (1/L) * dx/dt

Since L is a constant (10 ft):

cos(θ) * dθ/dt = (1/10) * (-2)

cos(θ) * dθ/dt = -1/5

Now, we need to find cos(θ) at the specific point when the bottom of the ladder is 5 ft from the wall. Using the Pythagorean theorem:

x^2 + h^2 = L^2
5^2 + h^2 = 10^2
25 + h^2 = 100
h^2 = 75
h = √75 = 5√3 ft

cos(θ) = adjacent / hypotenuse = 5√3 / 10 = √3 / 2

Substituting this value into the equation:

(√3 / 2) * dθ/dt = -1/5

Solving for dθ/dt:

dθ/dt = (-1/5) / (√3 / 2)
dθ/dt = (-2/5) * (√3 / 1)
dθ/dt = -2√3 / 5

Therefore, the rate at which the angle θ is changing when the bottom of the ladder is 5 ft from the wall is -2√3 / 5 radians per second.

To find how fast the angle that the ladder makes with the ground is changing, we can use trigonometry.

Let's assume that the angle the ladder makes with the ground is theta (θ). This angle is changing as the bottom of the ladder moves away from the wall.

We are given the rate at which the bottom of the ladder (let's call it x) is changing. In this case, dx/dt = 2 ft/s.

We need to find dθ/dt, which represents how fast the angle θ is changing with respect to time.

Using trigonometry, we can establish a relationship between the angle θ and the distance x from the wall:

sin(θ) = x / 10

To differentiate both sides of the equation with respect to time, we get:

cos(θ) * dθ/dt = (1/10) * dx/dt

Now, we substitute the known values. Since we are looking for dθ/dt when x = 5 ft, we substitute x = 5 into the equation:

cos(θ) * dθ/dt = (1/10) * 2

cos(θ) * dθ/dt = 1/5

To find dθ/dt, we need to find cos(θ) when x = 5 ft.

Using the pythagorean theorem, we have:

10^2 = 5^2 + h^2
h^2 = 75

Since the ladder is 10 ft long, h must be sqrt(75) = 5√3.

cos(θ) = adj/hypotenuse = 5/10 = 1/2

Now we can substitute cos(θ) = 1/2 into the equation:

(1/2) * dθ/dt = 1/5

dθ/dt = (1/5) / (1/2)

dθ/dt = (1/5) * (2/1)

dθ/dt = 2/5

Therefore, when the bottom of the ladder is 5 ft from the wall, the angle that the ladder makes with the ground is changing at a rate of 2/5 radians per second.