The first step is to write down the relevant equations.
For part a we care about the distance from the rocket to the camera: this is the hypotenuse
We use the Pythagorean theorem for this.
a2 + b2 = c2 a = 3000 b = 4000 c = 5000
Because we care about the rate of change we must take the derivative with respect to time of this equation.
When taking the derivative of variables that depend on time follow the chain rule.
d/dt (a2) = 2a * da/dt (2a is the derivative of the outside da/dt is the derivative of the inside)
Solving the full equation we get.
2a * (da/dt) + 2b * (db/dt) = 2c * (dc/dt) Remember da/dt is just a variable
db/dt = 0 because it doesn't change over time.
Simplifying we get.
a * da/dt = c * dc/dt
Plugging in values we finally get.
3000 * 500 = 5000 * dc/dt
Solving we get dc/dt = 300 ft/s
For part b the equation we start with is the tangent of an angle formula.
tan(θ) = a/b I am using a and b from the previous part.
Take the derivative of both sides, and note that θ and a depend on time while b is a constant
sec2(θ) * dθ/dt = (da/dt) /b
Plugging in values sec(θ) = 1/cos(θ) = 1/(adjacent/hypotenuse) = 5000/4000 = 1.25
(1.25)2 * dθ/dt = 500 / 4000
Solve for dθ/dt
dθ/dt = 0.08 radians per second
or
dθ/dt = 4.58 degrees per second when the rocket is 3000 ft above the pad