A ladder 10 ft 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?
sorry this one is already done i forgot i posted it already
To find how fast the angle the ladder makes with the ground is changing, we need to use the concept of related rates.
Let's start by labeling the given information:
- The ladder is 10 ft long.
- The bottom of the ladder slides away from the wall at a rate of 2 ft/s.
- We want to find the rate at which the angle the ladder makes with the ground is changing.
Now let's establish some notation:
- Let x represent the distance the bottom of the ladder is from the wall.
- Let θ represent the angle the ladder makes with the ground.
To solve this problem, we can use trigonometry. The relationship between x, θ, and the ladder length (10 ft) is given by the equation:
sin(θ) = x / 10
To find how fast the angle θ is changing, we can take the derivative of both sides with respect to time (t):
d(sin(θ)) / dt = d(x / 10) / dt
Now, let's differentiate both sides of the equation:
cos(θ) * dθ / dt = (1/10) * dx / dt
Rearranging the equation, we can isolate dθ / dt (the rate of change of the angle θ):
dθ / dt = [(1/10) * dx / dt] / cos(θ)
We have dx / dt given as 2 ft/s, and we want to find dθ / dt when x = 5 ft.
Let's substitute the given values into the equation:
dθ / dt = [(1/10) * 2 ft/s] / cos(θ)
To find cos(θ), we can use the fact that sin(θ) = x / 10:
sin(θ) = x / 10
θ = arcsin(x / 10)
Now we can substitute the value of x = 5 ft into this equation:
θ = arcsin(5 / 10)
θ ≈ arcsin(0.5)
Using a calculator, we can find that θ ≈ 30°.
Substituting this value into the equation for dθ / dt:
dθ / dt = [(1/10) * 2 ft/s] / cos(30°)
Calculating cos(30°) ≈ 0.866:
dθ / dt ≈ (1/10) * 2 ft/s / 0.866
dθ / dt ≈ 0.2 ft/s / 0.866
dθ / dt ≈ 0.231 ft/s
Therefore, when the bottom of the ladder is 5 ft from the wall, the angle the ladder makes with the ground is changing at a rate of approximately 0.231 ft/s.