A ladder 20 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per se

Question

A ladder 20 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 16 feet from the wall.

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Ngọc Khuê 5 months 2021-08-23T20:19:46+00:00 1 Answers 0 views 0

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    2021-08-23T20:21:37+00:00

    Answer:

    0.17 °/s

    Step-by-step explanation:

    Since the ladder is leaning against the wall and has a length, L and is at a distance, D from the wall. If θ is the angle between the ladder and the wall, then sinθ = D/L.

    We differentiate the above expression with respect to time to have

    dsinθ/dt = d(D/L)/dt

    cosθdθ/dt = (1/L)dD/dt

    dθ/dt = (1/Lcosθ)dD/dt where dD/dt = rate at which the ladder is being pulled away from the wall = 2 ft/s and dθ/dt = rate at which angle between wall and ladder is increasing.

    We now find dθ/dt when D = 16 ft, dD/dt = + 2 ft/s, and L = 20 ft

    We know from trigonometric ratios, sin²θ + cos²θ = 1. So, cosθ = √(1 – sin²θ) = √[1 – (D/L)²]

    dθ/dt = (1/Lcosθ)dD/dt

    dθ/dt = (1/L√[1 – (D/L)²])dD/dt

    dθ/dt = (1/√[L² – D²])dD/dt

    Substituting the values of the variables, we have

    dθ/dt = (1/√[20² – 16²]) 2 ft/s

    dθ/dt = (1/√[400 – 256]) 2 ft/s

    dθ/dt = (1/√144) 2 ft/s

    dθ/dt = (1/12) 2 ft/s

    dθ/dt = 1/6 °/s

    dθ/dt = 0.17 °/s

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