A 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?
I need help on both questions. Thanks in advance smart people!!
At a time of t sec, let the height be h ft.
let the angle of elevation be k
let the distance between them be y
1.
h^2 + 50^2 = y^2
2h dh/dt = 2y dy/dt
dy/dt = (h dh/dt)/y
when t = 5, h = 100
y^2 = 50^2 + 100^2 = 12500
y= 50sqrt(5)
dy/dt = 100(20)/sqrt(5)
= 50/sqrt(5) = appr 22.36 ft/sec
2.
then tank = h/50
h = 50tank
dh/dt = 50sec^2 k dk/dt
given: dk/dt = 20 ft/sec
when t =5, h = 100 ft
tank = 200/50 = 2
I sketched a triangle, and
cosk = 1/squr(5)
seck = squr(5)
sec^2 k = 5
dh/dt = 50sec^2 k dk/dt
20 = 50(5) dk/dt
dk/dt = 20/(250) rad/sec
= .08 rads/sec
To find the answers to these questions, we can use the concepts of related rates, which involve identifying relevant variables and their rates of change.
1. How fast is the distance between the observer and the balloon changing?
Let's denote the distance between the observer and the balloon as "d" (in feet) and the time as "t" (in seconds). We are given that at 5 seconds after lift-off, the balloon is rising at a rate of 20 ft/sec.
To find how fast the distance is changing, we need to differentiate the equation that represents this situation with respect to time (t). The equation is:
d^2 = (distance from observer)^2 + height^2
Differentiating both sides with respect to time gives:
2d * dd/dt = 2 * (distance from observer) * (dd/dt) + 2 * height * dh/dt
Since the distance between the observer and the balloon is given as 50 feet, height is not changing (dh/dt = 0), and we know dd/dt (20 ft/sec), we can simplify the equation to find the desired rate:
2d * dd/dt = 2 * 50 * (20) + 2 * 0
Simplifying further:
2d * dd/dt = 2000
Finally, we can solve for the rate of change of the distance (dd/dt):
dd/dt = 2000 / (2d)
At 5 seconds after lift-off, we know that d = 50 feet, so we can substitute this value into the equation:
dd/dt = 2000 / (2 * 50)
= 2000 / 100
= 20 feet/second
Therefore, at 5 seconds after lift-off, the distance between the observer and the balloon is changing at a rate of 20 feet/second.
2. How fast is the angle of elevation changing?
To find the rate at which the angle of elevation is changing, we need to consider the tangent (tan) of the angle. The tangent of an angle is equal to the height divided by the distance from the observer. In this case, the height is not changing (dh/dt = 0), so we can differentiate the equation:
tan(angle) = height / distance
To find the rate of change of the angle (d(angle)/dt), we can differentiate this equation implicitly with respect to time (t):
sec^2(angle) * (d(angle)/dt) = (dh/dt * distance - height * dd/dt) / (distance^2)
Since the height (h) is not changing (dh/dt = 0) and we are given dd/dt (20 ft/sec), we can simplify the equation further:
sec^2(angle) * (d(angle)/dt) = - h * dd/dt / (d^2)
Using the Pythagorean theorem, we can rewrite the equation as:
sec^2(angle) * (d(angle)/dt) = -(h * dd/dt) / (d^2)
At 5 seconds after lift-off, we know that:
h = 0 (the height is not changing)
d = 50 ft (given distance between the observer and the balloon)
Therefore, the equation becomes:
sec^2(angle) * (d(angle)/dt) = 0 / (50^2)
= 0
As a result, at 5 seconds after lift-off, the rate at which the angle of elevation is changing is 0. The angle of elevation remains constant over time.