Given a 20 m ladder making an angle θ with the ground and the tip of the ladder falling at a rate of 0.4 m/min, we need to find dθ/dt when θ = π/6 radians.
Let x be the distance from the base of the ladder to the point where the ladder touches the ground, and let y be the height of the ladder from the ground. As the ladder slides, we have a right triangle formed by the ladder, the ground, and the wall.
Using the Pythagorean theorem, we have:
x^2 + y^2 = 20^2 x^2 + y^2 = 400
Now, we differentiate both sides with respect to time (t):
2x(dx/dt) + 2y(dy/dt) = 0
Given that the tip of the ladder is falling at a rate of 0.4 m/min, we have:
dy/dt = -0.4 m/min
Since we are interested in the rate of change of the angle θ, we can use the trigonometric relationship:
sin(θ) = y / 20
When θ = π/6, we have:
sin(π/6) = y / 20
y = 20 * (1/2) = 10 m
Now, we can find the value of x when y = 10 m:
x^2 + 10^2 = 400 x^2 = 300 x = √300
Now we can substitute x, y, dy/dt into the equation we derived from the Pythagorean theorem:
2(√300)(dx/dt) + 2(10)(-0.4) = 0
Solving for dx/dt:
(√300)(dx/dt) = 4 dx/dt = 4 / √300
Now, let's find dθ/dt using the relationship:
cos(θ) = x / 20
Differentiating both sides with respect to time (t):
-sin(θ)(dθ/dt) = (1 / 20)(dx/dt)
Substituting θ = π/6, x = √300, and dx/dt = 4 / √300:
-sin(π/6)(dθ/dt) = (1 / 20)(4 / √300)
Solving for dθ/dt:
-(1/2)(dθ/dt) = (1 / 20)(4 / √300) dθ/dt = -[(1/2) * (20) * (4 / √300)]
dθ/dt ≈ -0.0230 rads/min
Thus, the rate of change of the angle θ with respect to time when θ = π/6 is approximately -0.0230 radians per minute.
