A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and its speed is 600 ft/s when it has risen 3000 ft.
How fast is the distance from the television camera to the rocket changing at that moment? (ft/s)
If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment? (rad/s)
as always, draw a diagram. The distance z when the rocket is at height h is
z^2 = 4000^2 + h^2
z dz/dt = h dh/dt
Now plug in your numbers, as with the baseball problem below.
tanθ = h/4000
sec^2θ dθ/dt = 1/4000 dh/dt
(5/4)^2 dθ/dt = 600/4000
dθ/dt = 0.096 rad/s
To find the rate at which the distance from the television camera to the rocket is changing at a given moment, we can use related rates.
Let's denote:
- x as the horizontal distance from the camera to the rocket
- y as the vertical distance from the camera to the rocket
- d as the distance from the camera to the rocket
We are given that x = 4000 ft, and we need to find dx/dt when y = 3000 ft. We are also given that dy/dt = 600 ft/s when y = 3000 ft.
Using the Pythagorean theorem, we have:
d^2 = x^2 + y^2
Differentiating both sides with respect to time (t):
2d(dd/dt) = 2x(dx/dt) + 2y(dy/dt)
At the given moment when y = 3000 ft, x = 4000 ft, and dy/dt = 600 ft/s, we can substitute these values into the equation:
2d(dd/dt) = 2(4000)(dx/dt) + 2(3000)(600)
Simplifying the expression:
2d(dd/dt) = 8000(dx/dt) + 3600000
Since the camera is focused on the rocket, the rate of change of the distance between them is the same as the rate of change of y. Therefore, we have:
dd/dt = dy/dt
Substituting this back into the equation:
2(dy/dt) = 8000(dx/dt) + 3600000
Now, we can substitute the given values:
2(600) = 8000(dx/dt) + 3600000
Simplifying further:
1200 = 8000(dx/dt) + 3600000
Rearranging the equation to solve for dx/dt:
8000(dx/dt) = 1200 - 3600000
8000(dx/dt) = -3598800
dx/dt = (-3598800) / 8000
dx/dt = -449.85 ft/s
Therefore, the distance from the television camera to the rocket is changing at a rate of approximately -449.85 ft/s at that moment.
To find the rate at which the camera's angle of elevation is changing at the same moment, we can use trigonometry.
The angle of elevation can be defined as the inverse tangent of y/x:
θ = arctan(y/x)
Differentiating with respect to time (t):
(dθ/dt) = (1 / (1 + (y/x)^2)) * ((dx/dt * y - dy/dt * x) / x^2)
Substituting the values we know:
(dθ/dt) = (1 / (1 + (3000/4000)^2)) * (((-449.85) * 3000) - (600 * 4000)) / (4000^2)
(dθ/dt) = (1 / (1 + (3/4)^2)) * (((-449.85) * 3000) - (600 * 4000)) / (4000^2)
Simplifying the expression:
(dθ/dt) = (1 / (1 + 9/16)) * (((-449.85) * 3000) - (600 * 4000)) / (4000^2)
(dθ/dt) = (1 / (25/16)) * (((-449.85) * 3000) - (600 * 4000)) / (4000^2)
(dθ/dt) = (16/25) * (((-449.85) * 3000) - (600 * 4000)) / (4000^2)
(dθ/dt) ≈ (-4.064) rad/s
Therefore, the camera's angle of elevation is changing at a rate of approximately -4.064 rad/s at that moment.
To find the rate of change of distance between the television camera and the rocket, we can use the concept of related rates.
Let's define:
x = horizontal distance between the television camera and the rocket
y = vertical distance between the television camera and the rocket
We are given that x = 4000 ft and dy/dt (the rate at which the rocket is rising vertically) = 600 ft/s. We need to find dx/dt (the rate at which the distance between the camera and the rocket is changing) when y = 3000 ft.
Using the Pythagorean theorem, we know that:
x^2 + y^2 = 4000^2
Differentiating both sides of this equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since the rocket is rising vertically, dx/dt = 0. We can solve for dy/dt using the given values:
2(4000)(0) + 2(3000)(600) = 0
2(3000)(600) = -2(4000)(dx/dt)
(3000)(600) = -(4000)(dx/dt)
dx/dt = -(3000)(600) / (4000)
dx/dt = -450 ft/s
Hence, the distance from the television camera to the rocket is changing at a rate of -450 ft/s at that moment.
Now, let's find how fast the camera's angle of elevation is changing at that same moment.
The angle of elevation (θ) can be defined as the arctangent of y/x. So, we have:
θ = arctan(y/x)
Differentiating both sides of this equation with respect to time (t), we get:
dθ/dt = [1 / (1 + (y/x)^2)] * [(1/x) * dy/dt - (y/x^2) * dx/dt]
Substituting the given values y = 3000 ft, x = 4000 ft, and dy/dt = 600 ft/s, we can find dθ/dt:
dθ/dt = [1 / (1 + (3000/4000)^2)] * [(1/4000) * 600 - (3000 / (4000^2)) * (-450)]
dθ/dt = [1 / (1 + 0.5625)] * [0.0375 + 0.05625]
dθ/dt = [1 / 1.5625] * 0.09375
dθ/dt = 0.06 rad/s
So, the camera's angle of elevation is changing at a rate of 0.06 rad/s at that moment.