An airplane is flying overhead at a constant elevation of 4090 ft. A person is viewing the plane from a position 3010 ft from the base of a radio tower. The airplane is flying horizontally away from the person. If the plane is flying at the rate of 700 ft/s, at what rate is the distance between the person and the plane increasing when the plane passes over the radio tower
Answers
Answer:
415 ft/s
Step-by-step explanation:
I've attached a diagram showing the triangle formed between the plane, the person and the radio tower.
From the attached diagram, x represents the position between the person and the position on the ground directly below the airplane while s represents the distance between the person and the plane.
The constant elevation remains 4090 ft while the value of x and s varies with time.
Thus, x varies with respect to time as dx/dt while s varies with respect to time as ds/dt.
Now, we are told that the plane is flying at the rate of 700 ft/s. This means that; dx/dt = 700 ft/s
Now we want to find the rate at which the distance between the person and the plane is increasing when the plane passes over the radio tower.
This means that we want to find ds/dt when x = 3010 ft
From pythagoras theorem if x = 3010 ft and the Elevation remains 4090 ft, we can find s as;
s = √(4090² + 3010²)
s = √25788200
s ≈ 5078.21 ft
Since we want to find ds/dt at x = 3010 ft, from the diagram, we can say from pythagoras theorem that;
x² + 4090² = s²
Differentiating x and s with respect to t, we have;
2x(dx/dt) = 2s(ds/dt)
2 will cancel out to give;
x(dx/dt) = s(ds/dt)
ds/dt = (x/s)(dx/dt)
Plugging in 700 for dx/dt, 3010 for x and 5078.21 for s, we have;
ds/dt = (3010/5078.21) × 700
ds/dt ≈ 415 ft/s
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Answers
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