A ladder 10 feet long rests against a vertical wall. If the bottom of the
ladder slides away from the wall at a speed of 2 ft/s, how fast is the angle
between the top of the ladder and the wall changing when the angle is �/4
radians?
What kind of a number is �/4 ?
I believe it's suppose to be Pi/4 radians
To solve this problem, we need to use derivatives and apply trigonometric functions.
Let's consider a right triangle formed by the ladder, the wall, and the ground. The ladder acts as the hypotenuse of the triangle, and the bottom of the ladder sliding away from the wall creates a right angle at the bottom of the triangle.
Let's assign some variables:
- Let x be the distance from the wall to the bottom of the ladder.
- Let θ be the angle between the ladder and the wall.
- Let L be the length of the ladder.
We're given that dx/dt = 2 ft/s (the speed at which the bottom of the ladder slides away from the wall).
Using trigonometry, we know that cos(θ) = x / L.
Differentiating both sides of the equation with respect to time (t), we get:
-d/dt(sin(θ)) = (d/dt(x / L))
-d/dt(sin(θ)) = (dx/dt / L)
We are looking for the rate of change of the angle, so we need to find dθ/dt.
Next, let's use the Pythagorean theorem to relate x, L, and θ:
x^2 + y^2 = L^2
Differentiating both sides of the equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 2L(dL/dt)
Substituting dx/dt = 2 ft/s and dy/dt = 0 since the height doesn't change, we simplify the equation to:
2x(2) = 2L(dL/dt)
4x = 2L(dL/dt)
2x = L(dL/dt)
We know that cos(θ) = x / L, so we can substitute 2x = L(dL/dt):
2(cos(θ)) = (dL/dt)
dL/dt = 2cos(θ)
Now we can plug this into our equation for d/dt(sin(θ)):
-d/dt(sin(θ)) = (dx/dt / L)
-d/dt(sin(θ)) = (2 ft/s / L)
-d/dt(sin(θ)) = 2 / (2cos(θ))
-d/dt(sin(θ)) = 1 / cos(θ)
Finally, we can find dθ/dt by using the identity d/dt(sin(θ)) = cos(θ) * dθ/dt:
cos(θ) * dθ/dt = 1 / cos(θ)
Simplifying the equation, we get:
dθ/dt = 1 / (cos(θ))^2
Given that θ = π/4 radians, we can substitute this value into the equation to find dθ/dt:
dθ/dt = 1 / (cos(π/4))^2
dθ/dt = 1 / (sqrt(2)/2)^2
dθ/dt = 1 / (2/4)
dθ/dt = 2
Therefore, the angle between the top of the ladder and the wall is changing at a rate of 2 radians per second when the angle is π/4 radians.