# 26-27 Time Rates: Kite moving horizontally

**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**

$s^2 = x^2 + 40^2$
when s = 50 ft

50

x = 30 ft

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

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

50

^{2}= x

^{2}+ 40

^{2}

x = 30 ft

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**

Slope

$m = \dfrac{40}{x}$

$m = \dfrac{40}{x}$

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

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*

Subscribe to MATHalino.com on