Question

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.4 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall.)

Answer 1

Answer:

$\dfrac{d\theta}{dt} =-0.233\ rad/s$

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

$2 x\dfrac{dx}{dt}+2y\dfrac{dy}{dt} = 0$..............(1)

when x = 8 and y = 6 and when \dfrac{dx}{dt} = 1.4ft/s

from equation (1)

now,

$16\times 1.4 + 12\dfrac{dy}{dt} = 0$

$\dfrac{dy}{dt} = -\dfrac{5.6}{3}$

let the angle between the ladders be θ

$tan\theta = \dfrac{y}{x}$

y = xtan θ

$\dfrac{dy}{dt} =\dfrac{dy}{dt} tan\theta + x sec^2\theta\dfrac{d\theta}{dt}$

$-\dfrac{5.6}{3} =1.4\times \dfrac{6}{8} + 8 (1+\dfrac{9}{16})\dfrac{d\theta}{dt}$

$\dfrac{25}{2} \dfrac{d\theta}{dt} =\dfrac{-17.5}{6}$

$\dfrac{d\theta}{dt} =-0.233\ rad/s$

Answer 2

Answer:

-0.233

Explanation: