Please help!! A hot-air balloon is rising straight up from a level field at a constant rate. This balloon is tracked by a range finder located on the ground 496feetfrom the lift-off point. At the moment the elevation angle from the range finder to the balloon is 0.875radian, this elevation angle is increasing at the rate of 0.225 radianper minute. At what rate is the distance between the range finder and the balloon changing when the elevation angle is 0.875 radian? Keep 3decimal places in your final answer

46.091 m

Step-by-step explanation:

Given that

Height of the range finder, = 496 feet

Angle of elevation, θ = 0.875 Rad = 50°

Velocity of increment, = 0.225 rad/min

For starters, we try and assume that it all takes place in a triangular shape. Using Pythagoras theorem, we then know that

Tanθ = y / 496

y = 496 tanθ

On differentiating, we have

dy/dt = 496 sec²θ

dy/dt = 496 * 0.413 * de/df

dy/dt = 204.848 * de/df

dy/dt = 204.848 * 0.225

dy/dt = 46.091 m

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

its kind of blurry. can you put a clearer pic?

step-by-step explanation: