A hot air balloon, 50 feet from an observer, is rising at 20 ft/sec. At 5 seconds after lift off
1. How fast is the distance between the observer and the balloon changing?
2. How fast is the angle of elevation changing?
the distance z at height x is
z^2 = 50^2 + x^2
so,
2z dz/dt = 2x dx/dt
find z when x = 5*20=100, and plug and chug.
tanθ = x/50
sec^2 θ dθ/dt = 1/50 dx/dt
To answer these questions, we can use the principles of calculus, specifically related rates.
1. How fast is the distance between the observer and the balloon changing?
Let's start by assigning some variables to the quantities involved:
- Let t represent time in seconds.
- Let d represent the distance between the observer and the balloon in feet.
We are given that the hot air balloon is rising at a rate of 20 ft/sec. This means that the rate of change of the distance d (with respect to time t) is 20 ft/sec.
Now, we want to find the rate of change of d when t = 5 seconds, which can be denoted as dd/dt (the derivative of d with respect to t).
To find dd/dt, we can use the chain rule of calculus. The chain rule states that if we have a function y = f(g(t)), then dy/dt = (dy/dg) * (dg/dt).
In our case, d is a function of t, so we can write d = f(t). Therefore, dd/dt = (dd/df) * (df/dt).
Given that df/dt = 20 ft/sec (the rate at which the balloon is rising), we need to find dd/df (the derivative of d with respect to f).
To do this, we can use the Pythagorean theorem, which states that for a right triangle, the square of the hypotenuse (d^2) is equal to the sum of the squares of the other two sides.
In this case, the hypotenuse represents the distance between the observer and the balloon (d), and the other two sides represent x (horizontal distance) and y (vertical distance).
We can write the equation using the Pythagorean theorem as d^2 = x^2 + y^2.
Taking the derivative of both sides of this equation with respect to f, we get:
2d * dd/df = 2x * dx/df + 2y * dy/df.
Since the observer is not moving horizontally (dx/df = 0), we can simplify the equation to:
2d * dd/df = 2y * dy/df.
Now, we need to find the value of y when t = 5 seconds.
Given that the balloon is rising at a rate of 20 ft/sec, the value of y at t = 5 can be calculated as follows:
y = 20 * (5 seconds) = 100 feet.
Substituting the values into the equation, we have:
2(50 feet) * dd/df = 2(100 feet) * (dy/df).
Cancelling out the common factors, we get:
50 * dd/df = 100 * (dy/df).
Now, we can solve for dd/df:
dd/df = (100 * (dy/df)) / 50
Simplifying, we get:
dd/df = 2 * (dy/df).
Since the rate of change of y (dy/df) is given as 20 ft/sec, we can substitute this value into the equation:
dd/df = 2 * (20 ft/sec).
Therefore, when t = 5 seconds, dd/dt (the rate of change of the distance between the observer and the balloon) is:
dd/dt = 2 * 20 ft/sec = 40 ft/sec.
Hence, the distance between the observer and the balloon is changing at a rate of 40 ft/sec at 5 seconds after lift-off.
2. How fast is the angle of elevation changing?
To find the rate of change of the angle of elevation, we need to consider the right triangle formed by the observer, the balloon, and a point on the ground directly below the balloon.
The angle of elevation can be defined as the angle between the horizontal line (represented by x) and the line of sight from the observer to the balloon.
Let's denote the angle of elevation as θ and the horizontal distance as x.
Now, we want to find the rate of change of θ with respect to time (dθ/dt).
To do this, we can use trigonometry. Based on the right triangle formed, we can write the tangent of the angle of elevation as:
tan(θ) = y / x.
Differentiating both sides of this equation with respect to time t, we get:
sec^2(θ) * (dθ/dt) = (dy/dt) / x.
We know that dy/dt (the rate of change of y) is given as 20 ft/sec, and the horizontal distance x is a constant value of 50 feet.
Substituting these values into the equation, we have:
sec^2(θ) * (dθ/dt) = (20 ft/sec) / 50 ft.
Simplifying, we get:
sec^2(θ) * (dθ/dt) = 0.4 ft/sec.
To find dθ/dt, we need to determine the value of sec^2(θ) at t = 5 seconds.
Therefore, we need to find the value of θ when t = 5 seconds.
Given that the balloon is 50 feet from the observer, at a height of 100 feet, we can calculate the value of θ using the inverse tangent (arctan) function:
θ = arctan(100/50) = arctan(2).
Substituting the value of θ into the equation, we have:
sec^2(arctan(2)) * (dθ/dt) = 0.4 ft/sec.
Now, we can solve for dθ/dt:
(dθ/dt) = (0.4 ft/sec) / sec^2(arctan(2)).
Using a calculator, we find that sec^2(arctan(2)) ≈ 2.4037.
Substituting this value into the equation, we have:
(dθ/dt) ≈ (0.4 ft/sec) / 2.4037.
Hence, the rate at which the angle of elevation is changing is approximately:
(dθ/dt) ≈ 0.166 ft/sec.