A child flies a kite at a height of 80 ft , the wind carrying the kite horizontally away from the child at a rate of \( 34 \mathrm{ft} / \mathrm{sec} \). How fast must the child let out the string when the kite is 170 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 170 ft away from the child. (Simplify your answer.)

To solve this problem, we can use the Pythagorean theorem. Let \( x \) be the horizontal distance between the kite and the child, \( h \) be the height of the kite (80 ft), and \( s \) be the length of the kite string. Given that the kite is 170 ft away, we can set up the relationship: \[ s^2 = x^2 + h^2. \] Since \( h = 80 \), and at the instant when the kite is 170 ft away, we can find \( x \) using \( \sqrt{170^2 - 80^2} \). Differentiate both sides of the equation to find \( \frac{ds}{dt} \), knowing \( \frac{dx}{dt} = 34 \, \text{ft/sec} \) and the height remains constant. Now, the relationship is: - \( s = 170 \, \text{ft} \) - \( h = 80 \, \text{ft} \) - \( x = \sqrt{170^2 - 80^2} \) Calculating this gives \( x = \sqrt{28900 - 6400} = \sqrt{22500} = 150 \, \text{ft} \). Now, differentiate: \[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt}. \] Reducing gives: \[ \frac{ds}{dt} = \frac{x}{s} \frac{dx}{dt}. \] Substituting the values: \[ \frac{ds}{dt} = \frac{150}{170} \times 34 = 30 \, \text{ft/sec}. \] Thus, the child must let out the string at a rate of \( 30 \, \text{ft/sec} \).