An observer is 36m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 26m horizontally from the observer. The angle of elevation of the elevator is the angle of the observer's line of sight makes with the horizontal (it may be positive or negative). Assuming that the elevator rises at a rate of 4/ms, what is the rate of change of the angle of elevation when the elevator is 23 m above the ground? When the elevator is 49m above the ground?
PLEASE TELL ME THE ANSWERS, I CAN'T SEEM TO UNDERSTAND HOW TO DO IT AND I HAVE A TEST IN A COUPLE OF MINUTES.
never come here expecting help "in a couple of minutes." you can't count on anyone's watching every post as it comes up.
The first related question below is this exact problem, just with different numbers. Always be sure to check the related questions when you do a posting. Your answer may often be already there.
To find the rate of change of the angle of elevation, we can use trigonometry and differentiate the equation. Let's denote the angle of elevation as θ and the height of the elevator above the ground as h.
Given:
Observer's height above the ground floor, y = 36 m
Horizontal distance between observer and elevator, x = 26 m
Rate of change of elevation, dh/dt = 4 m/s
We can form a right triangle with the observer, elevator, and a point on the ground directly below the elevator. The observer's line of sight makes an angle θ with the horizontal.
Using trigonometry, we have the equation:
tan(θ) = (h - y) / x ........(1)
To find the rate of change of the angle of elevation, we need to differentiate equation (1) with respect to time (t).
Differentiating both sides with respect to t:
sec^2(θ) * dθ/dt = [(dh/dt) - 0] / x
sec^2(θ) * dθ/dt = dh/dt / x ........(2)
Since sec^2(θ) = 1 + tan^2(θ), we can rewrite equation (2) as:
(1 + tan^2(θ)) * dθ/dt = dh/dt / x
dθ/dt + tan^2(θ) * dθ/dt = dh/dt / x
Substituting for tan^2(θ) using equation (1):
dθ/dt + ((h - y) / x)^2 * dθ/dt = dh/dt / x
dθ/dt + (h - y)^2 * dθ/dt / x^2 = dh/dt / x
dθ/dt(1 + (h - y)^2 / x^2) = dh/dt / x
dθ/dt = (dh/dt / x) / (1 + (h - y)^2 / x^2)
Now we can substitute the given values to find the rate of change of the angle of elevation at specific heights of the elevator.
When the elevator is 23 m above the ground (h = 23):
dθ/dt = (4 / 26) / (1 + (23 - 36)^2 / 26^2)
dθ/dt = 0.1538 rad/s
When the elevator is 49 m above the ground (h = 49):
dθ/dt = (4 / 26) / (1 + (49 - 36)^2 / 26^2)
dθ/dt = 0.0769 rad/s
Therefore, the rate of change of the angle of elevation when the elevator is 23 m above the ground is approximately 0.1538 rad/s, and when the elevator is 49 m above the ground is approximately 0.0769 rad/s.
To find the rate of change of the angle of elevation, we need to use trigonometry and calculus. Let's break the problem down into steps:
Step 1: Define the variables and assign known values.
Let:
- x be the horizontal distance between the elevator shaft and the observer.
- y be the vertical distance between the elevator and the ground.
- θ be the angle of elevation.
Given:
- x = 26m (horizontal distance)
- y = 36m (vertical distance from the observer to the ground)
Step 2: Calculate the initial angle of elevation.
To find the initial angle of elevation, we can use tangent:
tan(θ) = y / x
θ = atan(y / x)
In this case, θ = atan(36 / 26) = 54.37 degrees.
Step 3: Determine the relationship between x, y, and θ.
Consider a right triangle formed by the observer, elevator, and the ground. The vertical distance y is the opposite side, and the horizontal distance x is the adjacent side. Therefore, we have:
tan(θ) = y / x
Step 4: Determine the rate of change of θ with respect to y.
We need to find dθ/dt, the rate of change of the angle of elevation θ with respect to time t. We're given that the elevator is rising at a rate of 4 m/s, so dy/dt = 4 m/s.
To find dθ/dt, we differentiate both sides of the equation from step 3 with respect to time:
sec^2(θ) * dθ/dt = (1/x) * dy/dt
Step 5: Substitute known values and solve for dθ/dt.
We are asked to find the rate of change of the angle of elevation when the elevator is at two different heights: 23m and 49m.
For y = 23m:
We can determine x by using the Pythagorean theorem:
x^2 + y^2 = (26m)^2
x^2 + (23m)^2 = (26m)^2
x^2 + 529m^2 = 676m^2
x^2 = 676m^2 - 529m^2
x^2 = 147m^2
x ≈ 12.12m
Substituting the values into the equation from step 4:
sec^2(θ) * dθ/dt = (1/x) * dy/dt
sec^2(54.37) * dθ/dt = (1/12.12) * 4
(1.4713)^2 * dθ/dt ≈ 0.3298
dθ/dt ≈ 0.3298 / (1.4713)^2
dθ/dt ≈ 0.1482 degrees per second
For y = 49m:
Using the same process as above, we find:
x ≈ 23.34m
dθ/dt ≈ 0.0472 degrees per second
Therefore, when the elevator is 23m above the ground, the rate of change of the angle of elevation is approximately 0.1482 degrees per second. When the elevator is 49m above the ground, the rate of change is approximately 0.0472 degrees per second.