# Related rates problem?

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?

• 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