Respuesta :
Answer:
[tex]\dfrac{d\theta}{dt} =-0.233\ rad/s[/tex]
Explanation:
given,
length of ladder = 10 ft
let x be the distance of the bottom and y be the distance of the top of ladder.
x² + y² = 100
differentiating with respect to time we get
[tex]2 x\dfrac{dx}{dt}+2y\dfrac{dy}{dt} = 0[/tex]..............(1)
when x = 8 and y = 6 and when \dfrac{dx}{dt} = 1.4ft/s
from equation (1)
now,
[tex]16\times 1.4 + 12\dfrac{dy}{dt} = 0[/tex]
[tex]\dfrac{dy}{dt} = -\dfrac{5.6}{3}[/tex]
let the angle between the ladders be θ
[tex]tan\theta = \dfrac{y}{x}[/tex]
y = xtan θ
[tex]\dfrac{dy}{dt} =\dfrac{dy}{dt} tan\theta + x sec^2\theta\dfrac{d\theta}{dt} [/tex]
[tex]-\dfrac{5.6}{3} =1.4\times \dfrac{6}{8} + 8 (1+\dfrac{9}{16})\dfrac{d\theta}{dt}[/tex]
[tex]\dfrac{25}{2} \dfrac{d\theta}{dt} =\dfrac{-17.5}{6}[/tex]
[tex]\dfrac{d\theta}{dt} =-0.233\ rad/s[/tex]
