A hot air balloon rising vertically is tracked by an observer located 3 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is pi/5 , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment?
______ miles/min
To solve this problem, we'll need to use trigonometry.
Let's denote:
- r as the distance between the observer and the lift-off point (which is 3 miles in this case)
- θ as the angle between the observer's line-of-sight and the horizontal
- h as the height of the balloon above the ground
- dh/dt as the rate at which the balloon is rising (what we are trying to find)
We know that tan(θ) = h/r. Differentiating both sides of this equation with respect to time (t), we get:
sec^2(θ) * dθ/dt = (1/r) * dh/dt
Given that θ = π/5 (as mentioned in the problem), dθ/dt = 0.1 rad/min, and r = 3 miles, we can substitute these values into the equation:
sec^2(π/5) * 0.1 = (1/3) * dh/dt
Next, let's find the value of sec^2(π/5):
sec^2(π/5) = 1/cos^2(π/5)
Using the identity cos^2(θ) + sin^2(θ) = 1, we have:
cos^2(π/5) = 1 - sin^2(π/5)
Since the balloon is rising vertically, the angle between the observer's line-of-sight and the vertical is a right angle. Therefore, sin(θ) = h/r, where h is the height of the balloon. Given that θ = π/5 and r = 3 miles, we can solve for sin(π/5):
sin(π/5) = h/3
Rearranging this equation, we get h = 3 * sin(π/5).
Substituting h = 3 * sin(π/5) into the previous equation, we have:
cos^2(π/5) = 1 - sin^2(π/5) = 1 - (3 * sin(π/5)/3)^2 = 1 - (9 * sin^2(π/5))/9 = (9 - 9* sin^2(π/5))/9 = 9(1 - sin^2(π/5))/9 = 9 cos^2(π/5)/9 = cos^2(π/5)
Therefore, sec^2(π/5) = 1/cos^2(π/5) = 1/cos^2(π/5) = 1.
Substituting sec^2(π/5) = 1 into the aforementioned equation, we have:
1 * 0.1 = (1/3) * dh/dt
0.1 = (dh/dt)/3
Multiplying both sides by 3, we find:
dh/dt = 0.1 * 3 = 0.3 miles/min
Therefore, the balloon is rising at a rate of 0.3 miles/min at the given moment.
To find out how fast the balloon is rising at the moment, we can use trigonometry and rate of change concepts.
Let's assign some variables:
- Let h represent the height of the balloon above the ground (in miles).
- Let θ represent the angle between the observer's line-of-sight and the horizontal (in radians).
- Let dh/dt represent the rate at which the height is changing (in miles per minute).
From the problem, we are given the following information:
- h = ?
- θ = π/5 radians
- dθ/dt = 0.1 rad/min
Now, we need to find a relationship between h, θ, and dh/dt. By drawing a diagram, we can see that tan(θ) = h/3 (since the balloon is rising vertically).
Differentiating both sides of this equation implicitly with respect to time (t), we get:
sec^2(θ) * dθ/dt = dh/dt
Now, let's substitute the given values into the equation:
sec^2(π/5) * 0.1 = dh/dt
Simplifying the equation, we have:
(1/cos^2(π/5)) * 0.1 = dh/dt
To find the value of dh/dt, calculate cos^2(π/5) and substitute it into the equation above. The result will give you the rate at which the balloon is rising at that particular moment.
Now you can solve the equation using a calculator or mathematical software to find the value of dh/dt, which gives you the rate at which the balloon is rising.
tanTheta=h/x
take derivative
sec^2 theta dtheta/dt= 1/x dh/dt
find dh/dt x= 3 miles theta= pi/5,dtheta/dt= .1rad/min