A ladder 10 ft leans against a vertical wall. The upper end slips down the wall at 5 ft/sec. How fast is the ladder turning when it takes an angle of 20 degrees with the ground.
let the top of the ladder by y ft up on the wall
let the angle made at the foot be Ø
sinØ = y/10
10 sinØ = y
10 cosØ dØ/dt = dy/dt
given: when Ø = 20° , dy/dt = -5 ( y is decreasing)
10 cos20° dØ/dt = -5
dØ/dt = -1/(2cos20°) = -.532 radians/sec
or angle is decreasing at .532 radians / sec
which is appr. 30.5° / sec
(notice the angle is changing very rapidly at that moment, since the ladder has almost fallen)
To find the rate at which the ladder is turning when it takes an angle of 20 degrees with the ground, we need to use trigonometry and related rates.
Let's start by drawing a diagram to better visualize the situation. We have a ladder leaning against a vertical wall forming a right triangle with the ground. The ladder is 10 ft long, and its upper end is slipping down at a rate of 5 ft/sec.
```
|
|
| |
10ft |
| |
| |
| θ | x
| |
----------------------- -------------
Wall Ground
```
We are interested in finding how fast the angle θ is changing when it reaches 20 degrees.
Now, let's assign some variables:
- Let x represent the distance from the bottom of the ladder to the wall (the horizontal distance).
- Let θ represent the angle between the ladder and the ground.
- Let y represent the distance from the top of the ladder to the ground (the height or length of the ladder against the wall).
Based on the given information, we know that dx/dt = -5 ft/sec (negative because x is decreasing).
Using trigonometry, we can relate these variables:
- tan(θ) = y / x
To relate the variables using differentiation, we can take the derivative of both sides with respect to time (t) using the chain rule:
- sec^2(θ) * dθ/dt = (dy/dt * x - y * dx/dt) / x^2
Since we are interested in finding dθ/dt, the rate at which the angle is changing, we can rearrange the equation:
- dθ/dt = (dy/dt * x - y * dx/dt) / (x^2 * sec^2(θ))
Now, we can substitute the given values into the equation:
- dx/dt = -5 ft/sec (as given)
- x = 10 ft
- θ = 20 degrees (convert to radians: θ = 20 * π/180)
Substituting these values, we can solve for dθ/dt when θ = 20 degrees.
dθ/dt = (dy/dt * 10 - y * (-5)) / (10^2 * sec^2(20°))
To find dy/dt and y, we need to consider the right triangle formed by the ladder and the wall:
sin(θ) = y / 10
=> y = 10 * sin(θ)
dy/dt = d(10 * sin(θ))/dt
=> dy/dt = 10 * cos(θ) * dθ/dt
Now, substitute y and dy/dt into the previous equation:
dθ/dt = (10 * cos(θ) * dθ/dt * 10 - 10 * sin(θ) * (-5)) / (10^2 * sec^2(20°))
Simplifying further:
dθ/dt = 100 * cos(θ) * dθ/dt + 50 * sin(θ) / (100 * sec^2(20°))
Now, substitute the remaining values:
θ = 20 * π/180
cos(θ) = cos(20°) = 0.9397
sec(θ) = sec(20°) = 1.0472
dθ/dt = 100 * (0.9397) * dθ/dt + 50 * sin(20°) / (100 * (1.0472)^2)
Now, we can solve for dθ/dt by isolating it on one side of the equation:
dθ/dt - 0.9397 * dθ/dt = 50 * sin(20°) / (100 * (1.0472)^2)
dθ/dt * (1 - 0.9397) = 50 * sin(20°) / (100 * (1.0472)^2)
dθ/dt = (50 * sin(20°) / (100 * (1.0472)^2)) / (1 - 0.9397)
Evaluating this expression will give us the rate at which the ladder is turning when it takes an angle of 20 degrees with the ground.