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.
tan(a) = h/200
sec^2(a) da/dt = 1/200 dh/dt
da/dt = 1/200 cos^2(a) dh/dt
When t=30, h = 4*30 = 120m
tan(a) = 120/200 = 0.6
so, cos(a) = 0.857
da/dt = .005 * 0.857^2 * 4 = .0147 rad/s
That's roughly pi/68 rad/s
Were you drunk when you posted this Steve?
To solve this problem, we can use trigonometry. Let's consider the triangle formed by the observer, the balloon, and the vertical distance between them at time t.
Let:
- d be the horizontal distance from the observer to the balloon at time t.
- h be the vertical distance from the observer to the balloon at time t.
- θ be the angle of elevation from the observer to the balloon at time t.
Since the balloon is rising vertically, the horizontal distance d remains constant at 200m throughout.
According to trigonometry, we can relate the angle of elevation θ to the vertical distance h and the horizontal distance d through the equation: tan(θ) = h/d.
To solve for how fast the angle of elevation is changing (dθ/dt), we need to differentiate this equation with respect to time t, and then solve for dθ/dt.
Differentiating both sides of the equation with respect to t:
(d/dt) tan(θ) = (d/dt) (h/d)
Now, let's find the individual derivatives:
1. Derivative of tan(θ) with respect to t:
(d/dt) tan(θ) = sec^2(θ) * (dθ/dt)
2. Derivative of h/d with respect to t:
(d/dt) (h/d) = (d/dt) h * (1/d) + h * (d/dt) (1/d)
Given that the rate of change of horizontal distance (dd/dt) is 0 (since the horizontal distance remains constant), the second term in the above equation becomes 0.
So we are left with:
(d/dt) (h/d) = (d/dt) h * (1/d)
Substituting these derivatives back into the original equation:
sec^2(θ) * (dθ/dt) = (d/dt) h * (1/d)
Since tan(θ) = h/d, we can rewrite this as:
sec^2(θ) * (dθ/dt) = (d/dt) (h/d) = (d/dt) (h/d) * (d/d)
Now, let's evaluate the given information:
- The balloon rises vertically at a constant rate of 4 m/s. This means (d/dt) h = 4 m/s.
We also know that at time t = 30s, the observer stands 200 m away from the launch site.
Now, substituting the given values and solving for (dθ/dt):
sec^2(θ) * (dθ/dt) = (4 m/s) * (1/200 m)
To find sec^2(θ), we can use the Pythagorean theorem:
d^2 = h^2 + d^2
200^2 = h^2 + 200^2
h^2 = 200^2 - 200^2 = 40000
So, h = √(40000) = 200√2
Now, we can substitute h in the equation:
sec^2(θ) * (dθ/dt) = (4 m/s) * (1/200 m)
sec^2(θ) * (dθ/dt) = 1/50 s^(-1) (since m/m cancel out)
To solve for (dθ/dt), we need to find sec^2(θ). We can use the equation:
tan^2(θ) + 1 = sec^2(θ)
Substituting h and d, we get:
(tan^2(θ) + 1) * (dθ/dt) = 1/50 s^(-1)
tan^2(θ) * (dθ/dt) + (dθ/dt) = 1/50 s^(-1)
Since tan(θ) = h/d = (200√2 m)/(200 m) = √2, we can substitute this value:
√2 * √2 * (dθ/dt) + (dθ/dt) = 1/50 s^(-1)
2 *(dθ/dt) + (dθ/dt) = 1/50 s^(-1)
Now, simplifying the equation:
3 *(dθ/dt) = 1/50 s^(-1)
Finally, solving for (dθ/dt):
(dθ/dt) = (1/50 s^(-1)) / 3
(dθ/dt) = 1/150 s^(-1)
Therefore, the angle of elevation is increasing at a rate of 1/150 s^(-1) 30 seconds after the launch.
To find the rate of change of the angle of elevation, we need to use trigonometry. Let's assume that the observer is located at point O, the launch site is at point L, and the position of the hot-air balloon at time t is point B.
We can create a right triangle OLB, where OL is the horizontal distance from the observer to the launch site (200 m) and BL is the height of the balloon. The angle of elevation, θ, is the angle that OL makes with OB, as shown in the diagram below:
```
O
/|
/ |
L/ |
/ |
/ |
/θ | B
```
Using trigonometry, we can find the relationship between the rate of change of the angle of elevation (dθ/dt) and the rate of change of the height of the balloon (dh/dt):
tan(θ) = BL / OL
Differentiating both sides of this equation with respect to time (t), we get:
sec^2(θ) * dθ/dt = (dBL/dt) / OL
We are given that the rate of change of the height of the balloon (dBL/dt) is 4 m/s, and OL is 200 m.
Now, we need to find the value of sec(θ). By drawing a new triangle, we can see that:
cos(θ) = OL / OB
Therefore, sec(θ) = 1 / cos(θ) = 1 / (OL / OB) = OB / OL.
Since the balloon is rising vertically, OB is equal to BL. Thus, sec(θ) = BL / OL.
Plugging this value into our differentiation equation, we have:
(sec(θ))^2 * dθ/dt = (dBL/dt) / OL
BL / OL) ^2 * dθ/dt = 4 / 200
Substituting in the given values, we have:
(4 / 200) * (200 / OL)^2 = dθ/dt
Simplifying and evaluating at t = 30 s, we can determine the rate at which the angle of elevation is changing:
(4 / 200) * (200 / 200)^2 = dθ/dt
(4 / 200) * 1^2 = dθ/dt
(4 / 200) = dθ/dt
dθ/dt = 0.02 radians/s
Therefore, the angle of elevation is increasing at a rate of 0.02 radians per second.