An airplane is flying above an observer standing on the ground. At the moment it passes overhead, the observer judges by its apparent size that the airplane’s altitude is 30,000 feet. It takes 50 seconds for the plane, moving in a straight, horizontal line, to move through an angle of 0.94 radians as seen by the observer. The angle described is swept out by the line segment joining the observer to the plane. At this moment, the angle is increasing at a rate of 34.3 radians per hour. What is the speed of the plane at this time, in miles per hour?
To find the speed of the plane, we need to use the concepts of trigonometry and differentiation.
Let's define some variables:
- h: altitude of the airplane (in feet)
- θ: angle described by the line segment joining the observer to the plane (in radians)
- t: time (in seconds)
- v: speed of the plane (in miles per hour)
We are given the following information:
- h = 30,000 feet
- θ = 0.94 radians
- dθ/dt = 34.3 radians per hour
We need to find the speed of the plane, v.
To relate the altitude, angle, and time, we can use the concept of similar triangles. The altitude, h, can be related to the observed apparent size of the airplane. Since the angle θ is small, we can approximate the tangent of the angle as the angle itself.
Therefore, we have the equation:
tan(θ) = h / d
where d is the distance between the observer and the airplane. Note that d will change with time.
Now we need to relate the angle θ with time. The rate of change of θ with respect to time is given: dθ/dt = 34.3 radians per hour.
Differentiating the equation tan(θ) = h / d with respect to time (t), we get:
sec²(θ) * dθ/dt = -h / d² * dd/dt
Since sec²(θ) = 1 + tan²(θ), we can simplify the equation to:
(1 + tan²(θ)) * dθ/dt = -h / d² * dd/dt
Substituting the values we have, we get:
(1 + tan²(0.94)) * 34.3 = -30,000 / d² * dd/dt
Now we can solve for dd/dt, the rate of change of distance with respect to time:
dd/dt = [(1 + tan²(0.94)) * 34.3 / -30,000] * d²
Integrating both sides of the equation with respect to time, we get:
∫ dd/dt dt = ∫ [(1 + tan²(0.94)) * 34.3 / -30,000] * d² dt
Simplifying, we have:
d = [(1 + tan²(0.94)) * 34.3 / -30,000] * t + C
Where C is the constant of integration. Since at t = 0, d = 0 (the observer and the airplane are at the same point), we can find the value of C:
0 = [(1 + tan²(0.94)) * 34.3 / -30,000] * 0 + C
C = 0
Now we can find d in terms of t:
d = [(1 + tan²(0.94)) * 34.3 / -30,000] * t
To find v, we need to differentiate d with respect to t:
dv/dt = [(1 + tan²(0.94)) * 34.3 / -30,000]
dv/dt = 0.0359 radians per hour
Finally, we need to convert dv/dt to miles per hour. Since 1 hour is equal to 3600 seconds and there are 2π radians in a full circle, we have:
dv/dt in miles per hour = (0.0359 radians per hour) * (2π * 3960 miles) / (2π)
dv/dt = 71.52 miles per hour
Therefore, the speed of the plane at this time is approximately 71.52 miles per hour.