A hot air balloon rising vertically is tracked by an observer located 2 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is π5 , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment?
nope
What was your starting equation?
To find the rate at which the balloon is rising at a certain moment, we can use trigonometry and differentiate with respect to time.
Let's denote the distance between the observer and the point directly below the balloon as "x" (in miles) and the height of the balloon as "h" (in miles). We are given that the observer is located 2 miles from the lift-off point, so x = 2.
The tangent of the angle between the observer's line-of-sight and the horizontal is equal to the height of the balloon divided by the distance between the observer and the balloon:
tan(π/5) = h/x
Now, let's differentiate both sides of the equation with respect to time, t:
d/dt(tan(π/5)) = d/dt(h/x)
The derivative of the tangent function with respect to time is the rate of change of the angle, which is given as 0.1 rad/min:
sec^2(π/5) * d(π/5)/dt = dh/dt * (1/x)
Simplifying the equation:
(1/2) * sec^2(π/5) * d(π/5)/dt = dh/dt * (1/2)
Now we can solve for dh/dt, the rate at which the balloon is rising:
dh/dt = sec^2(π/5) * d(π/5)/dt
To find the value of sec^2(π/5), we can use a calculator or reference table.
Substitute the value of d(π/5)/dt, which is given as 0.1 rad/min:
dh/dt = sec^2(π/5) * 0.1
Now, compute the value of sec^2(π/5) and multiply it by 0.1 to find the rate at which the balloon is rising.