An observer is 23m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 15m 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 18 m above the ground? When the elevator is 38m above the ground?
Why are you on every bloody question Steve?
when the elevator is at height x,
tanθ = (x-23)/15
so,
sec^2θ dθ/dt = 1/15 dx/dt
(1+tan^2θ) dθ/dt = 1/15 dx/dt
(1 + ((x-23)/15)^2) dθ/dt = 1/15 dx/dt
Now just plug and chug. You have x, and dx/dt=4 ...
Well, if the elevator is rising at a rate of 4 m/s, we can imagine it as a magical disco elevator that's always going up. Now, let's do some math to figure out the rate of change of the angle of elevation.
First, let's imagine that the elevator is at ground level (0 m above the ground). At this point, the observer is looking straight ahead, so the angle of elevation is 0 degrees.
Now, when the elevator is 18 m above the ground, we need to find the rate of change of the angle of elevation. To do this, we'll use similar triangles and some trigonometry. So grab a slice of pizza and let's dive in!
We have a right triangle with the opposite side being the height of the elevator (18 m) and the adjacent side being the horizontal distance to the elevator shaft (15 m). The hypotenuse is the line of sight from the observer to the elevator. We want to find the rate of change of the angle of elevation, which is the derivative of the angle with respect to time.
Using the trigonometric function tangent, we can define the angle of elevation (let's call it θ) as the inverse tangent (arctan) of the opposite side divided by the adjacent side:
tan(θ) = opposite/adjacent
Now, let's differentiate both sides with respect to time (t):
d/dt(tan(θ)) = d/dt(opposite/adjacent)
Using the chain rule, we get:
sec²(θ) * dθ/dt = (d(opposite)/dt * adjacent - opposite * d(adjacent)/dt) / (adjacent)²
In simpler terms, this equation means that the rate of change of the angle of elevation is equal to (d(opposite)/dt * adjacent - opposite * d(adjacent)/dt) divided by the square of the adjacent side, all multiplied by sec²(θ).
Since we are given that the elevator is rising at a rate of 4 m/s, we have:
d(opposite)/dt = 4 m/s
The horizontal distance to the elevator shaft (adjacent) is always 15 m, so d(adjacent)/dt is 0 m/s.
Plugging these values into the equation, we get:
sec²(θ) * dθ/dt = (4 m/s * 15 m - 18 m * 0 m/s) / (15 m)²
Since 0 m/s * 18 m is 0, this simplifies to:
sec²(θ) * dθ/dt = (4 m/s * 15 m) / (15 m)²
sec²(θ) * dθ/dt = 4 m/s / 15 m
Now, let's solve for dθ/dt by dividing both sides by sec²(θ):
dθ/dt = (4 m/s / 15 m) * (1 / sec²(θ))
Since sec²(θ) is equal to 1 + tan²(θ), we can rewrite the equation as:
dθ/dt = (4 m/s / 15 m) * (1 / (1 + tan²(θ)))
Now, we don't know the value of θ yet, but we can find it using the inverse tangent (arctan) function. By substituting the values we know into the equation, we can find dθ/dt when the elevator is 18 m above the ground.
Phew, that was a wild ride! Let's go get some cotton candy and we'll continue with the calculations for when the elevator is 38 m above the ground, shall we?
To find the rate of change of the angle of elevation, we need to determine how the angle of elevation is related to the elevation of the elevator.
Let's denote the angle of elevation as θ and the elevation of the elevator as h.
From the problem, we know that the observer is 23m above the ground floor and the elevator shaft is 15m horizontally from the observer. This forms a right triangle where the observer is at the base, the elevator shaft is the adjacent side, and the height difference between the observer and elevator is the opposite side.
Using trigonometry, we can express tan(θ) as the opposite side divided by the adjacent side:
tan(θ) = (h - 23) / 15
Now, let's differentiate this expression with respect to time (t represents time):
d/dt (tan(θ)) = d/dt ((h - 23) / 15)
The derivative of tan(θ) is sec^2(θ). To express the rate of change in terms of the rate of change of the elevation, we need to use the chain rule:
(sec^2(θ)) * dθ/dt = (d/dt ((h - 23) / 15))
Now, we need to find the value of dθ/dt when the elevator is at a certain height, specifically 18m and 38m above the ground.
For the first case, when the elevator is 18m above the ground, we substitute h = 18 into our equation:
(sec^2(θ)) * dθ/dt = (d/dt ((18 - 23) / 15))
Simplifying further, we get:
(sec^2(θ)) * dθ/dt = (d/dt (-5 / 15))
(sec^2(θ)) * dθ/dt = (-1/3)
Now, divide both sides by sec^2(θ), which is equivalent to 1 + tan^2(θ):
dθ/dt = (-1/3) / (1 + tan^2(θ))
To find the exact value of dθ/dt, we need to calculate the value of θ. Since θ is the angle of elevation, we should use the inverse tangent function:
θ = arctan((18 - 23) / 15) = arctan(-1/3) ≈ -18.43° (approximately)
Substituting this value into our equation, we can solve for dθ/dt.
Similarly, for the second case when the elevator is 38m above the ground, we follow the same process and substitute h = 38 into our equation:
dθ/dt = ? (To be calculated)
θ = arctan((38 - 23) / 15) = arctan(1) = 45°
Substituting these values into our equation, we can solve for dθ/dt.
Remember, the angle of elevation can be positive or negative depending on the direction the observer is looking. So, the rate of change of the angle of elevation should consider the appropriate sign.