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Nookie1986 [14]
3 years ago
9

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.)
Physics
1 answer:
Art [367]3 years ago
6 0

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

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__________________________________________________________

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