A rocket is fired vertically into the air at the rate of 6 miles/min. An observer on the ground is located 4 miles from the launching pad. When the rocket is 3 miles high, how fast is the angle of elevation between the rocket and the observer changing? Specify units.
at the moment in question, the rocket is 5 miles from the observer.
h/4 = tanθ
dh/dt = 4 sec^2θ dθ/dt
6 = 4(5/4)^2 dθ/dt
24/25 = dθ/dt
I expect you can deduce the units.
yes steve right
To find the rate at which the angle of elevation is changing, we can use trigonometry. The tangent of the angle of elevation can be defined as the ratio of the height of the rocket (opposite side) to the distance from the observer to the launching pad (adjacent side).
Given that the rocket is fired vertically, we can create a right triangle with the height of the rocket (3 miles), the distance from the observer to the launching pad (4 miles), and the hypotenuse representing the distance the rocket has traveled vertically.
Let's denote the angle of elevation as θ, the height of the rocket as h, and the distance from the observer to the launching pad as d.
Using the tangent function, we have:
tan(θ) = h / d
Differentiating implicitly with respect to time (t), we have:
(d/dt) [tan(θ)] = (d/dt) [h / d]
To find the rate at which the angle of elevation is changing, we need to solve for (d/dt) [θ].
Now, let's differentiate each term:
(d/dt) [tan(θ)] = (d/dt) [h / d]
(sec^2(θ)) * (dθ/dt) = (dh/dt * d - h * dd/dt) / d^2
Since the rocket is fired vertically, the horizontal distance (d) remains constant at 4 miles. Thus, dd/dt = 0.
Now, substitute the known values:
2 * (dθ/dt) = (dh/dt) / 4^2
Simplifying:
2 * (dθ/dt) = (dh/dt) / 16
(dθ/dt) = (dh/dt) / (2 * 16)
(dθ/dt) = (dh/dt) / 32
Now, we need to find the rate at which the height of the rocket is changing, (dh/dt). Given that the rocket is fired at a rate of 6 miles/min, (dh/dt) = 6 miles/min.
Substituting this value:
(dθ/dt) = 6 miles/min / 32
(dθ/dt) ≈ 0.1875 miles/min
Therefore, the angle of elevation between the rocket and the observer is changing at a rate of approximately 0.1875 miles per minute.
To find the rate of change of the angle of elevation, we need to analyze the situation and apply trigonometry. Let's start by visualizing the scenario.
We have the observer on the ground, the rocket, and a right triangle formed by the observer, the rocket's position, and the top of the rocket. The angle of elevation is the angle between the ground and the line of sight from the observer to the rocket.
Let's denote:
- The distance from the observer to the launching pad as x (which is 4 miles).
- The height of the rocket as y at a given time.
- The angle of elevation as θ.
Taking the derivative of both sides with respect to time (t), we have:
d/dt(sinθ) = d/dt(y/x)
The derivative of sinθ is cosθ, so we can rewrite the equation as:
cosθ(dθ/dt) = (1/x)(dy/dt) [Differentiating sinθ and using the chain rule]
We are given that x = 4 miles and dy/dt = 6 miles/min. We need to find dθ/dt, the rate of change of the angle of elevation.
Now, let's find cosθ. Using the right triangle, we have:
cosθ = (base/hypotenuse) = (x/y)
By substituting x = 4 and y = 3, we can find cosθ:
cosθ = (4/3)
Returning to our equation:
(cosθ)(dθ/dt) = (1/x)(dy/dt)
Substituting the values we know:
(4/3)(dθ/dt) = (1/4)(6 miles/min) [x = 4 miles, dy/dt = 6 miles/min]
Simplifying:
(4/3)(dθ/dt) = (3/2) miles/min
Now, solve for dθ/dt:
dθ/dt = (3/2)(3/4) miles/min [Cross multiply and divide]
dθ/dt = 9/8 miles/min
Therefore, the rate at which the angle of elevation is changing is 9/8 miles per minute.