tan(elevation)=4000/x
d/dx tanE =-4000/x^2 dx/dt
but d/dttanE= sec^2E * dE/dt
finally...
sec^2 E * dE/dt=-4000/x^2 dx/dt
solve for dE/dt when x=2000, You can figure secE from the triangle. You are given dx/dt as 500ft/sec
d/dx tanE =-4000/x^2 dx/dt
but d/dttanE= sec^2E * dE/dt
finally...
sec^2 E * dE/dt=-4000/x^2 dx/dt
solve for dE/dt when x=2000, You can figure secE from the triangle. You are given dx/dt as 500ft/sec
First off, we have a plane flying west at a speed of 500 ft/sec. That's pretty speedy! Meanwhile, the searchlight on the ground is trying to maintain its focus on the plane. Tricky, tricky.
We want to find the change in the angle of elevation of the searchlight at a horizontal distance of 2000 ft. Alright, buckle up!
To figure this out, we can use a little bit of trigonometry. We'll need to consider the altitude of the plane and the horizontal distance it's covering.
At a horizontal distance of 2000 ft, we can create a right triangle with the searchlight at one end, the plane at the other end, and the hypotenuse connecting the two.
The altitude of the plane is 4000 ft. Since the triangle is a right triangle, we can use the tangent function to find the angle of elevation.
The tangent of an angle is equal to the opposite side (altitude) divided by the adjacent side (horizontal distance). So, we have:
tan(angle of elevation) = 4000 ft / 2000 ft
Now, let's simplify that:
tan(angle of elevation) = 2
To find the angle of elevation, we'll have to take the inverse tangent (or arctan) of both sides. So:
angle of elevation = arctan(2)
Now go grab your calculator and find that juicy angle!
And there you have it! The change in the angle of elevation of the searchlight at a horizontal distance of 2000 ft. I hope this answer flies with you!
Let's assume that the angle of elevation of the searchlight is θ at time t. Since the plane is flying west at a constant speed and altitude, we can say that the horizontal distance between the searchlight and the plane is changing at a rate of 500 ft/sec.
The height of the plane remains constant at 4000 ft.
We are given that the horizontal distance between the searchlight and the plane is 2000 ft. To find the change in angle of elevation, we need to find the rate of change of the angle with respect to time, dθ/dt.
To do this, we can use the tangent function:
tan(θ) = opposite/adjacent
Since the opposite side is the altitude of the plane (4000 ft) and the adjacent side is the horizontal distance (2000 ft), we can rewrite the equation:
tan(θ) = 4000/2000 = 2
Taking the inverse tangent of both sides:
θ = arctan(2)
Using a calculator, we find that the angle θ is approximately 63.43 degrees.
Now, we need to find the rate of change of the angle with respect to time, dθ/dt. Since the horizontal distance is changing at a rate of 500 ft/sec, we can say that dθ/dt = -500/2000 = -1/4.
Therefore, the change in the angle of elevation of the searchlight at a horizontal distance of 2000 ft is -1/4 radians per second.
Let's denote the horizontal distance from the searchlight to the plane as "x". In this case, x = 2000 ft.
The altitude of the plane is given as 4000 ft, and it is flying horizontally at a speed of 500 ft/sec. This means that the vertical distance between the plane and the searchlight is constant.
Now, let's consider a small time interval, Δt. During this time interval, the plane will move horizontally by 500 * Δt ft (we know the plane's horizontal speed is 500 ft/sec) and the searchlight will also move horizontally by the same distance.
So, after this time interval Δt, the horizontal distance between the searchlight and the plane will be x + 500 * Δt ft.
Since the vertical distance between the plane and the searchlight is constant, we have a right triangle formed where the vertical distance is 4000 ft, the horizontal distance is x + 500 * Δt ft, and the hypotenuse (the distance between the searchlight and the plane) is the diagonal connecting these two distances.
Applying the Pythagorean Theorem, we have:
(4000)^2 + (x + 500 * Δt)^2 = (distance)^2
Simplifying the equation:
16000000 + x^2 + 1000xΔt + 250000Δt^2 = (distance)^2
Since Δt is very small, the term 250000Δt^2 becomes negligible.
So, the equation becomes:
16000000 + x^2 + 1000xΔt ≈ (distance)^2
Now, if we subtract 16000000 from both sides, we have:
x^2 + 1000xΔt ≈ (distance)^2 - 16000000
Next, divide both sides by Δt:
(x^2 + 1000xΔt) / Δt ≈ [(distance)^2 - 16000000] / Δt
Since Δt is very small, the term x^2Δt can be neglected:
x + 1000 ≈ [(distance)^2 - 16000000] / Δt
Now, the right-hand side of the equation represents the derivative of the distance with respect to time (distance rate).
So, we can rewrite the equation as:
x + 1000 ≈ d(distance) / dt
Now, let's find the distance rate d(distance) / dt. Since the horizontal distance is x + 500 * Δt, we can differentiate it:
d(distance) / dt = d(x + 500 * Δt) / dt
The derivative of x with respect to t is 0 (since x is constant).
The derivative of 500 * Δt with respect to t is 500 (since the derivative of Δt with respect to itself is 1).
Therefore, we have:
d(distance) / dt = d(x + 500 * Δt) / dt = 0 + 500 = 500 ft/sec
Since the distance rate is 500 ft/sec, we can write the equation as:
x + 1000 ≈ 500
Finally, solve for x:
x ≈ 500 - 1000
x ≈ -500
Since the horizontal distance cannot be negative in this context, we can conclude that the searchlight should not track the plane, based on the given information.
Hence, the change in the angle of elevation of the searchlight will be zero.