a ladder of 8 m os leaning against a wall begins to slide. If its upper end slides down the wall at a rate of 0.25 m/sec, at what rate is the angle between the ladder and the ground changing when the foot of the ladder is 5 m from the wall?
Let θ be the angle formed by the ladder and the ground
let the top of the ladder by y m above the ground
sinθ = y/8
8sinθ = y
given: dy/dt = -.25 m/sec, <----- negative, since it is sliding down
find dθ/dt when x = 5
then 25 + y^2 = 64
y = √39 and cosθ = √39/8
8cosθ dθ/dt = dy/dt
dθ/dt = -.25/(8√39/8) = -1/4√39 m/s
To find the rate at which the angle between the ladder and the ground is changing, we need to use related rates. Let's denote the angle between the ladder and the ground as θ and the distance between the foot of the ladder and the wall as x. We are given that dx/dt, the rate at which x is changing, is 0.25 m/sec.
Using trigonometry, we know that sin(θ) = x/8 (where 8 is the length of the ladder). Differentiating both sides with respect to time t, we get:
d/dt (sin(θ)) = d/dt (x/8)
To find the rate at which the angle is changing, we need to solve for dθ/dt, the rate of change of θ with respect to time.
Now, let's differentiate both sides of the equation.
cos(θ) * dθ/dt = (1/8) * dx/dt
We can rearrange this equation to solve for dθ/dt:
dθ/dt = (1/8 * dx/dt) / cos(θ)
We are interested in finding dθ/dt when x = 5 m. To do that, we need to determine the value of cos(θ) when x = 5.
Using trigonometry, cos(θ) = √(1 - sin^2(θ)). Given that sin(θ) = x/8, we can substitute this value and find cos(θ):
cos(θ) = √(1 - (x/8)^2)
cos(θ) = √(1 - (5/8)^2)
cos(θ) = √(1 - (25/64))
cos(θ) = √(64/64 - 25/64)
cos(θ) = √(39/64)
cos(θ) = √39/8
Now, we can substitute the values of dx/dt, x, and cos(θ) into the equation for dθ/dt:
dθ/dt = (1/8 * dx/dt) / cos(θ)
dθ/dt = (1/8 * 0.25) / (√39/8)
dθ/dt = 0.25 / (√39/8)
dθ/dt = 0.25 * 8 / √39
dθ/dt = 2 / √39
So, the rate at which the angle between the ladder and the ground is changing when the foot of the ladder is 5 m from the wall is 2 / √39 radians per second.
To find the rate at which the angle between the ladder and the ground is changing, we can use trigonometry and related rates.
Let's assume the ladder forms a right triangle with the wall and the ground. The ladder represents the hypotenuse, the wall represents the adjacent side, and the distance between the foot of the ladder and the wall represents the opposite side.
Let's denote:
- x as the distance between the foot of the ladder and the wall (opposite side).
- θ as the angle between the ladder and the ground.
We are given:
- dx/dt = 0.25 m/s (the rate at which the upper end of the ladder is sliding down the wall).
- x = 5 m (when the foot of the ladder is 5 m from the wall).
- We need to find dθ/dt (the rate at which the angle is changing).
Using trigonometry, we have the relationship:
cos(θ) = x / 8 (since the ladder is 8 m long)
Differentiating both sides with respect to time (t), we get:
-d(θ)/dt * sin(θ) = (dx/dt) / 8
Rearranging and substituting the known values:
-d(θ)/dt * sin(θ) = 0.25 / 8
Since we want to find dθ/dt, we need to solve for -d(θ)/dt:
-d(θ)/dt = (0.25 / 8) / sin(θ)
To find sin(θ), we can use the Pythagorean identity:
sin²(θ) + cos²(θ) = 1
Substituting the value of cos(θ) from our previous equation:
sin²(θ) + (x/8)² = 1
sin²(θ) + (5/8)² = 1
sin²(θ) + 25/64 = 1
sin²(θ) = 1 - 25/64
sin²(θ) = 39/64
sin(θ) = √(39/64)
Substituting this value of sin(θ) into the equation for -d(θ)/dt:
-d(θ)/dt = (0.25 / 8) / √(39/64)
Now we can solve for -d(θ)/dt:
-d(θ)/dt = (0.25 / 8) / √(39/64)
-d(θ)/dt = 0.25 / (8 * √(39/64))
-d(θ)/dt = 0.25 / (8 * (√39 / √64))
-d(θ)/dt = 0.25 / (8 * (√39 / 8))
-d(θ)/dt = 0.25 / √39
Finally, to find dθ/dt, we need to take the negative value (-d(θ)/dt) since it represents the rate of decrease:
dθ/dt = - 0.25 / √39
Therefore, when the foot of the ladder is 5 m from the wall, the rate at which the angle between the ladder and the ground is changing is approximately -0.0401 radians/s.