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?

The distance y when the rocket has reached height h is given by

y^2 = h^2 + 2500^2
So, when h=4000, y = 500√89

2y dy/dt = 2h dh/dt
2(500√89) dy/dt = 2(4000)(800)
dy/dt = 6400/√89 = 678.4 ft/s

To find how fast the distance from the camera to the rocket is changing, we can use the concept of related rates.

Let's establish some variables:
- Let D be the distance from the camera to the rocket.
- Let t be the time elapsed since the rocket started rising.
- Let x be the height of the rocket.

Given that the rocket's speed is 800 ft/s when it has risen 4000 ft, we can use this information to find the rocket's height as a function of time:
x = 800t

Since the camera is positioned 2500 ft from the base of the rocket launching pad, the distance from the camera to the rocket is the hypotenuse of a right triangle with sides x (height of the rocket) and 2500 ft (distance from the camera to the base of the launching pad):

D = sqrt(x^2 + 2500^2)

Differentiating both sides of the equation with respect to time (t) using the chain rule, we get:

dD/dt = (dD/dx) * (dx/dt)

To find dD/dt, we need to find dD/dx and dx/dt.

Derivative of D with respect to x (dD/dx):
dD/dx = (1/2) (x^2 + 2500^2)^(-1/2) * 2x
dD/dx = x / sqrt(x^2 + 2500^2)

Derivative of x with respect to t (dx/dt):
dx/dt = 800 ft/s

Therefore, substituting dD/dx and dx/dt back into the equation, we get:

dD/dt = (x / sqrt(x^2 + 2500^2)) * (800 ft/s)

Now, we need to find dD/dt at the moment when x = 4000 ft. Substituting x = 4000 in the equation, we have:

dD/dt = (4000 / sqrt(4000^2 + 2500^2)) * (800 ft/s)
dD/dt = (4000 / sqrt(16000000 + 6250000)) * (800 ft/s)
dD/dt = (4000 / sqrt(22250000)) * (800 ft/s)
dD/dt = (4000 / 4714.91) * (800 ft/s)
dD/dt ≈ 678.18 ft/s

Therefore, at that moment, the distance from the camera to the rocket is changing at a rate of approximately 678.18 ft/s.

Now let's determine how fast the camera's angle of elevation is changing in radians/s at that moment.

The angle of elevation of the camera can be represented by the arctan of the ratio of the height of the rocket (x) to the distance from the camera to the rocket (D):

θ = arctan(x / D)

Taking the derivative of both sides of the equation with respect to time (t) using the chain rule, we get:

dθ/dt = (dθ/dx) * (dx/dt) + (dθ/dD) * (dD/dt)

To find dθ/dt, we need to find dθ/dx, dθ/dD, and dD/dt (which we have already calculated above as 678.18 ft/s).

Derivative of θ with respect to x (dθ/dx):
dθ/dx = 1 / (1 + (x / D)^2) * (1 / D)

Derivative of θ with respect to D (dθ/dD):
dθ/dD = -x / (D * (1 + (x / D)^2))

Now, substituting the values of dθ/dx, dθ/dD, and dD/dt, we have:

dθ/dt = (1 / (1 + (x / D)^2) * (1 / D)) * (dx/dt) + (-x / (D * (1 + (x / D)^2))) * (dD/dt)

Substituting the values of x = 4000 ft, D ≈ 5000 ft (since D ≈ sqrt(4000^2 + 2500^2)), and dx/dt = 800 ft/s, we get:

dθ/dt = (1 / (1 + (4000 / 5000)^2) * (1 / 5000)) * (800 ft/s) + (-4000 / (5000 * (1 + (4000 / 5000)^2))) * (678.18 ft/s)

Simplifying the equation, we get:

dθ/dt ≈ 0.153 radians/s

Therefore, the camera's angle of elevation is changing at a rate of approximately 0.153 radians/s at that moment.