A child flies a kite at a height of 90 ft, the wind carrying the kite horizontally away from the child at a rate of 30 ft/sec. How fast must the child let out the string when the kite is 150 ft away from the child?
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2 Answers

The string can be considered the hypotenuse of a right triangle that has a constant 90 ft vertical leg (the height of the kite) and a horizontal leg that the length of is unknown but that is increasing at a rate of 30 ft/sec. Calling the length of this leg x, the length of the string L at any given moment is determined by
LÂ² = 90Â² + xÂ².
The rate of release of the string is dL/dt. That is the quantity you are after at the moment when L = 150. When L = 150, x is found by substitution
150Â² = 90Â² + xÂ², so x = âˆš(150Â²  90Â²) = 120 ft.
The rate of change of L is found by differentiation
2L dL/dt = 0 + 2x dx/dt
When L = 150 and x = 120
2(150) dL/dt = 2(120) (30)
dL/dt = 24.
She must let the string out at a rate of 24 ft/sec. That seems kind of fast, but so does the 30 ft/sec wind!

This problem can be modeled by a right triangle with sides x & y and hypotenuse z.
y = 90
z = 150
dx/dt = 30
Find dz/dt:
xÂ² = zÂ²  yÂ²
xÂ² = (150)Â²  (90)Â²
xÂ² = 22500  8100
xÂ² = 14400
x = âˆš14400
x = 120 (Negative value disregarded)
zÂ² = xÂ² + yÂ²
zÂ² = xÂ² + 90Â²
zÂ² = xÂ² + 8100
dz/dt = dx/dt + d(8100)/dt
2z(dz/dt) = 2x(dx/dt) + 0
z(dz/dt) = x(dx/dt)
150(dz/dt) = 120(30)
150(dz/dt) = 3600
dz/dt = 3600 / 150
dz/dt = 24
The child must let out the line at a rate of 24 ft./sec.
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Source(s): 9/9/10