A television camera is positioned 2500 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 that its speed is 800 ft/s when it has risen 4000 ft.
How fast is the distance from the camera to the rocket changing in ft/s at that moment?
If the televison camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing in radians/s at that moment?
To find the rate at which the distance from the camera to the rocket is changing, we can differentiate the equation that relates the distance and time. Let's represent the distance from the camera to the rocket as "d" and the time as "t".
Given that the rocket rises vertically, we can use the equation for distance traveled with constant acceleration:
d = 1/2 * at^2
We are told that the speed of the rocket is 800 ft/s when it has risen 4000 ft. This gives us the initial velocity:
v = 800 ft/s
We also know that the distance from the camera to the rocket at the beginning (when the rocket is on the launching pad) is 2500 ft. So we have:
d = 2500 ft
Now, let's find the acceleration "a" using the information given. We can use the equation for velocity:
v = at
Since we know the velocity (800 ft/s) and the time at which the velocity is achieved (4000 ft), we can substitute those values:
800 ft/s = a * 4000 ft
Solving for "a":
a = 800 ft/s / 4000 ft
a = 0.2 ft/s^2
Now, we can differentiate the equation for distance with respect to time to find how fast the distance is changing:
d/dt (d) = d/dt (1/2 * at^2)
Using the power rule of differentiation:
d/dt (d) = 1/2 * d/dt (at^2)
Differentiating, we have:
d/dt (d) = 1/2 * 2at * dt/dt
Since dt/dt is simply 1:
d/dt (d) = at
Plugging in the values we know, we have:
d/dt (d) = (0.2 ft/s^2) * t
At the moment when the rocket has risen 4000 ft, we can substitute t = 4000 ft / 800 ft/s:
d/dt (d) = (0.2 ft/s^2) * (4000 ft / 800 ft/s)
Simplifying:
d/dt (d) = 1 ft/s
So, at that moment, the distance from the camera to the rocket is changing at a rate of 1 ft/s.
Now let's find the rate at which the angle of elevation of the camera is changing in radians/s. To do this, we need to use the concept of similar triangles.
Let's represent the angle of elevation as "θ" and the distance between the camera and the rocket as "d".
From the problem, we know that the rocket rises vertically, so the angle of elevation of the camera remains constant.
Now, let's consider a small change in distance "dd" and a small change in angle "dθ". The relationship between these changes can be represented using the concept of tangent:
tan(dθ) = dd / d
We can rearrange this equation to solve for dθ:
dθ = tan^(-1)(dd / d)
To find the rate at which the angle of elevation is changing, we need to differentiate this equation with respect to time:
dθ/dt = d/dt (tan^(-1)(dd / d))
Using the chain rule of differentiation, we get:
dθ/dt = (1 / (1 + (dd / d)^2)) * (d/dd (dd / d)) * (dd / dt)
Now, since the rocket is rising vertically, we have:
dd / dt = 1 ft/s (as found earlier)
Also, since the angle of elevation remains constant (the camera is always aimed at the rocket), we can assume dd is very small in comparison to d, and hence (dd / d)^2 is negligible.
With that in mind, the equation becomes:
dθ/dt = (1 / (1 + (dd / d)^2)) * (d/dd (dd / d)) * (dd / dt)
Simplifying:
dθ/dt ≈ (1 / (1 + 0)) * (d/dd (dd / d)) * (dd / dt)
dθ/dt ≈ d/dd (dd / d)
Now, let's differentiate dd / d with respect to d:
d/dd (dd / d) = d/dd (d / d^(-1))
Using the quotient rule:
d/dd (dd / d) = (d*(-1) - d / d^(-1)^2) = (-dd / d^2)
Finally, plugging this result into the equation:
dθ/dt ≈ (-dd / d^2)
At the moment when the rocket has risen 4000 ft, let's substitute dd = 1 ft (as we found earlier) and d = 2500 ft:
dθ/dt ≈ (-1 ft / (2500 ft)^2)
Calculating:
dθ/dt ≈ -1 ft / 6250000 ft^2
Converting to radians:
dθ/dt ≈ -1 / 6250000 rad/s
So, at that moment, the camera's angle of elevation is changing at a rate of approximately -1 / 6250000 rad/s.