an observer stands 200 m from the launch site of a hot-air balloon. the balloon rises vertically at a constant rate of 4 m/s. how fast is the angle of elevation increasing 30 s after the launch.
To find the rate at which the angle of elevation is increasing, we need to differentiate the equation that relates the angle of elevation to the distance between the observer and the hot-air balloon.
Let's denote the angle of elevation as θ and the distance between the observer and the hot-air balloon as x. From the problem, we know that the balloon rises vertically at a constant rate of 4 m/s. Therefore, the rate of change of x with respect to time (dx/dt) is also 4 m/s.
We want to find dθ/dt, the rate at which the angle of elevation is changing with respect to time, 30 seconds after the launch. To find this, we need to differentiate the equation tan(θ) = x/200, with respect to time.
Differentiating both sides of the equation:
d/dt(tan(θ)) = d/dt(x/200)
Using the chain rule on the left side:
sec^2(θ) * dθ/dt = (1/200) * dx/dt
Substituting the given values:
sec^2(θ) * dθ/dt = (1/200) * 4
Since this equation involves the secant function, we need to find the value of θ. We can use the right-angled triangle formed by the observer, the balloon, and the ground.
In this triangle, the vertical side represents the height of the balloon, which is given by h(t) = 4t, where t is the time after launch. The horizontal side represents the distance between the observer and the balloon, which is x. The hypotenuse represents the distance from the observer to the balloon and remains constant at 200 m.
Using Pythagoras' theorem:
x^2 + (4t)^2 = 200^2
x^2 + 16t^2 = 40000
From this equation, we can solve for x in terms of t:
x^2 = 40000 - 16t^2
x = sqrt(40000 - 16t^2)
Now we can substitute the expression for x in terms of t into the equation involving θ:
tan(θ) = sqrt(40000 - 16t^2) / 200
To find the angle θ, we can take the inverse tangent (arctan) of both sides:
θ = arctan(sqrt(40000 - 16t^2) / 200)
Now we can substitute θ into the equation involving the rate of change of the angle of elevation:
sec^2(θ) * dθ/dt = (1/200) * 4
Using the identity sec^2(θ) = 1 + tan^2(θ):
(1 + tan^2(θ)) * dθ/dt = (1/200) * 4
Substituting θ = arctan(sqrt(40000 - 16t^2) / 200):
(1 + (sqrt(40000 - 16t^2) / 200)^2) * dθ/dt = (1/200) * 4
Simplifying the equation:
(1 + (40000 - 16t^2) / 20000) * dθ/dt = 1/50
Simplifying further:
(1 + (40000 - 16t^2)) / 20000 * dθ/dt = 1/50
(40000 - 16t^2 + 20000) / 20000 * dθ/dt = 1/50
(60000 - 16t^2) / 20000 * dθ/dt = 1/50
Now we can solve for dθ/dt by substituting t=30 into the equation:
(60000 - 16(30)^2) / 20000 * dθ/dt = 1/50
(60000 - 16(900)) / 20000 * dθ/dt = 1/50
(60000 - 14400) / 20000 * dθ/dt = 1/50
45600 / 20000 * dθ/dt = 1/50
2.28 * dθ/dt = 1/50
dθ/dt = (1/50) / 2.28
Simplifying further:
dθ/dt ≈ 0.0219 radians per second
Therefore, the angle of elevation is increasing at a rate of approximately 0.0219 radians per second, 30 seconds after the launch.
To find the rate at which the angle of elevation is increasing, we can use trigonometry.
Let's assume that the observer is standing at point O, the launch site is point L, and the balloon is at point B.
Given:
Distance OL = 200 m
Rate of ascent of the balloon (vertical distance) = 4 m/s
We are asked to find how fast the angle of elevation is increasing, so we need to find dθ/dt (the derivative of the angle of elevation with respect to time).
Now, we can set up a right triangle with the balloon's position at time t (30 seconds after launch). Let the height of the balloon at time t be h(t).
Using trigonometry, we have:
tan(θ) = h(t) / OL
Differentiating both sides of the equation with respect to time t:
sec²(θ) * (dθ/dt) = (dh/dt) / OL
Since we are interested in finding dθ/dt, we can isolate it on one side of the equation:
(dθ/dt) = (dh/dt) / (OL * sec²(θ))
Now, let's find the values needed to calculate dθ/dt.
At time t = 30 seconds, the height of the balloon is:
h(30) = rate of ascent * time
h(30) = 4 m/s * 30 s
h(30) = 120 m
We can determine the angle using the right triangle:
tan(θ) = h(30) / OL
tan(θ) = 120 m / 200 m
tan(θ) = 0.6
To find θ, we can take the arctan of both sides of the equation:
θ = arctan(0.6)
θ ≈ 30.96°
Now, substitute the values into the equation for (dθ/dt):
(dθ/dt) = (dh/dt) / (OL * sec²(θ))
(dθ/dt) = 4 m/s / (200 m * sec²(30.96°))
Using a calculator:
(dθ/dt) ≈ 0.029 degrees per second
Therefore, the angle of elevation is increasing at a rate of approximately 0.029 degrees per second 30 seconds after the launch.