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 \( 30 \mathrm{ft} / \mathrm{sec} \). How fast must the child let out the string when the kite is 150 ft away from the child? The child must let out the string at a rate of \( \square \mathrm{ft} / \mathrm{sec} \) when the kite is 150 ft away from the child. (Simplify your answer.)

To solve this problem, we can use the Pythagorean theorem to relate the height of the kite, the horizontal distance from the child to the kite, and the length of the string. Let: - \( y = 90 \) ft (height of the kite) - \( x \) = horizontal distance from the child to the kite - \( s \) = length of the string From the Pythagorean theorem, we have: \[ s^2 = x^2 + y^2 \] Taking the derivative with respect to time, we get: \[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \] Since the height \( y \) is constant (\( \frac{dy}{dt} = 0 \)), the equation simplifies to: \[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} \] or \[ s \frac{ds}{dt} = x \frac{dx}{dt} \] First, we need to find the lengths and the rates: 1. At the moment when the kite is 150 ft away from the child horizontally, we have \( x = 150 \text{ ft} \) and \( y = 90 \text{ ft} \). Thus, we can calculate \( s \): \[ s = \sqrt{x^2 + y^2} = \sqrt{150^2 + 90^2} = \sqrt{22500 + 8100} = \sqrt{30600} = 30\sqrt{34} \text{ ft} \] 2. The kite is moving away horizontally at \( \frac{dx}{dt} = 30 \) ft/sec. Now inserting our known values into the simplified equation: \[ 30\sqrt{34} \frac{ds}{dt} = 150 \cdot 30 \] Now, solve for \( \frac{ds}{dt} \): \[ \frac{ds}{dt} = \frac{150 \cdot 30}{30\sqrt{34}} = \frac{150}{\sqrt{34}} \] To simplify, we can leave the final answer as: \[ \frac{ds}{dt} = \frac{150}{\sqrt{34}} \text{ ft/sec} \] This can be rationalized if needed, but in terms of the problem, the child must let out the string at a rate of: \[ \frac{150}{\sqrt{34}} \text{ ft/sec} \] That's your answer!