Problem 26 A kite is 40 ft high with 50 ft cord out. If the kite moves horizontally at 5 miles per hour directly away from the boy flying it, how fast is the cord being paid out?

Solution 26

$2s \dfrac{ds}{dt} = 2x \dfrac{dx}{dt}$

$s \dfrac{ds}{dt} = x \dfrac{dx}{dt}$

Thus, $50 \dfrac{ds}{dt} = 30(\frac{22}{3})$

$\dfrac{ds}{dt} = 4.4 \, \text{ ft/sec}$ answer

Problem 27 In Problem 26, find the rate at which the slope of the cord is decreasing.

Solution 27

$\dfrac{dm}{dt} = \dfrac{-40}{x^2}\,\dfrac{dx}{dt}$

From Solution 26, x = 30 ft when s = 50 ft $\dfrac{dm}{dt} = \dfrac{-40}{30^2}\,(\frac{22}{3})$

$\dfrac{dm}{dt} = -\frac{44}{135} \, \text{ rad/sec}$ answer

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