The Pythagora´s Theorem, tell us that in a right triangle, the square of the hypothenuse (L²) is equal to :
L² = x² + y²
x and y are the legs.
The solution is
dy/dt = 48.9 ft/s
We know that in a right triangle:
L² = x² + y² (1)
If the helicopter lifts off vertically, that means x ( the horizontal distance between the observing man and the helicopter, maintains constant).
According to that, if we get derivatives with respect to time on both sides of equation (1) we get:
2×L× dL/dt = 0 + 2×y×dy/dt (2)
And
L = √ ( 58)² + ( 113)² = √12769 + 3364 = 127.02 ft
L = 127.02 ft when y = 113 ft
Now in equation (2) we know:
y = 113 ft dy/dt = 55 ft/s L = 127.02 we look for dL/dt
Then by substitution
2×127.02 × dL/dt = 2×113×55
254.04× dL/dt = 12430
dL/dt = 48.92
dL/dt = 48.9 ft/s
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