Question

A ladder 10 ft long leans against a vertical wall. If the bottom of the ladder slides away from the base of the wall at a speed of 2 ft/s, how fast is the angle between the ladder and the wall changing when the bottom of the ladder is 6 ft from the base of the wall?

Answer 1

Answer:

$\frac{dx}{dt}=\frac{-8}{3}\frac{ft}{s}$

Step-by-step explanation:

Be $\frac{dy}{dt}=2\frac{ft}{s}$ to find $\frac{dx}{dt}=?, x=6ft$

$x^{2} +y^{2}=10^{2};6^{2}+y^{2}=100;y^{2}=100-36=64;y=\sqrt{64}=+-8$→ y=8 for being the positive distance, deriving from t, $x^{2} +y^{2}=100$→$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$→$2x\frac{dx}{dt}=-2y\frac{dy}{dt};\frac{dx}{dt}=-\frac{2y}{2x}\frac{dy}{dt}; \frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}$, if x=6 and y=8

$\frac{dx}{dt}=\frac{-8}{6}2$→$\frac{dx}{dt}=\frac{-8}{3}\frac{ft}{s}$

we must find the rate of change in radians over seconds, being the speed 8/3 ft / s = 2.66 ft / s the variation in degrees is determined when traveling 6 ft