A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. To 2 decimal places in radians per second, find the rate of change of the angle of elevation when the firework is 40 feet above the ground. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).
To find the rate of change of the angle of elevation, we need to use trigonometry. Let's denote the angle of elevation as θ, the distance from the observer to the firework as x, and the height of the firework as y.
Given:
dx/dt = 10 ft/s (the rate of change of x)
x = 50 ft (the initial distance from the observer)
y = 40 ft (the height of the firework)
We need to find dy/dt, the rate of change of the height of the firework.
Since we have a right triangle, we can use the tangent function to relate the height (y) and the distance (x):
tan(θ) = y / x
To find dy/dt, we will differentiate this equation with respect to time t:
sec^2(θ) * dθ/dt = (dy/dt * x - y * dx/dt) / x^2
We can rearrange this equation to solve for dθ/dt:
dθ/dt = (sec^2(θ) * dx/dt * x - (dy/dt * x)) / (x^2 * sec^2(θ))
Now, let's plug in the values we have:
dx/dt = 10 ft/s
x = 50 ft
y = 40 ft
We also need the value of θ, which we can find using the Pythagorean theorem:
x^2 + y^2 = h^2, where h is the distance between the observer and the firework
50^2 + 40^2 = h^2
2500 + 1600 = h^2
4100 = h^2
h ≈ 64.03 ft
Using the tangent function, we can find θ:
tan(θ) = y / x
tan(θ) = 40 / 50
θ ≈ 38.66 degrees
Now we can calculate dθ/dt:
dθ/dt = (sec^2(θ) * dx/dt * x - (dy/dt * x)) / (x^2 * sec^2(θ))
sec^2(θ) = 1 + tan^2(θ) = 1 + (40/50)^2 = 1 + 0.64 = 1.64
dθ/dt = (1.64 * 10 * 50 - (dy/dt * 50)) / (50^2 * 1.64)
Finally, we need to find dy/dt. Let's use the Pythagorean theorem:
h^2 = x^2 + y^2
64.03^2 = 50^2 + y^2
4096.80 = 2500 + y^2
1596.80 = y^2
y ≈ 39.96 ft
Now, we can plug in the values to calculate dθ/dt:
dθ/dt = (1.64 * 10 * 50 - (dy/dt * 50)) / (50^2 * 1.64)
dθ/dt = (82 * 50 - (dy/dt * 50)) / (2500 * 1.64)
dθ/dt = (4100 - (dy/dt * 50)) / (4100)
We can further simplify this equation by substituting dy/dt with dx/dt since the height and distance are related:
dθ/dt = (4100 - (dx/dt * 50)) / (4100)
Now we can substitute the given value for dx/dt:
dθ/dt = (4100 - (10 * 50)) / (4100)
⇒ dθ/dt = (4100 - 500) / (4100)
⇒ dθ/dt = 3600 / 4100
⇒ dθ/dt ≈ 0.8780 radians per second (rounded to two decimal places)
Therefore, the rate of change of the angle of elevation when the firework is 40 feet above the ground is approximately 0.88 radians per second.
To find the rate of change of the angle of elevation, we need to differentiate the equation that relates the angle of elevation to the height of the firework.
Let:
- θ = angle of elevation (in radians)
- h = height of the firework above the ground (in feet)
From trigonometry, we know that tan(θ) = h / 50. Rearranging this equation, we have h = 50tan(θ).
Now, we need to differentiate this equation with respect to time to find the rate of change of the angle of elevation (dθ/dt) when h = 40.
First, differentiate both sides of the equation with respect to time:
d(h) / dt = d(50tan(θ)) / dt
Next, since θ is a function of time, we can use the chain rule to differentiate both sides:
d(50tan(θ)) / dt = 50sec^2(θ) * d(θ) / dt
Now, we have an equation involving the rate of change of the angle of elevation (d(θ) / dt) and the angle of elevation (θ). We need to find the value of d(θ) / dt when h = 40.
From the given information, we know that d(h) / dt = 10 feet per second. Thus, we substitute this value into the equation:
10 = 50sec^2(θ) * d(θ) / dt
To find sec^2(θ), we can use the fact that tan^2(θ) + 1 = sec^2(θ). Substituting h = 40 into the equation h = 50tan(θ), we can find θ:
40 = 50tan(θ)
tan(θ) = 40/50
tan(θ) = 0.8
θ ≈ arctan(0.8)
Using a calculator, we can find that θ ≈ 0.6745 radians.
Now, substitute sec^2(θ) = tan^2(θ) + 1 into the rate of change equation:
10 = 50(tan^2(θ) + 1) * d(θ) / dt
Simplifying further:
10 = 50(0.8^2 + 1) * d(θ) / dt
10 = 50(1.64) * d(θ) / dt
Now, solve for d(θ) / dt:
d(θ) / dt = 10 / (50 * 1.64)
Using a calculator, we can find that d(θ) / dt ≈ 0.122 radians per second.
Therefore, to 2 decimal places, the rate of change of the angle of elevation when the firework is 40 feet above the ground is approximately 0.12 radians per second.
at height h,
tanθ = h/50
sec^2θ dθ/dt = 1/50
so, plug in your numbers and evaluate dθ/dt