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A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.6 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

  • Calculus -

    make a sketch, place the man x m from the light
    let the length of the shadow on the wall me y m

    by similar triangles : 2/x = y/12
    xy = 24

    x dy/dt + y dx/dt = 0
    dy/dt = -y dx/dt / x

    when he is 4 m from the wall x = 8 (look at our diagram)
    so in xy=24 --> 8y = 24 or y = 3
    and dx/dt = 1.6

    dy/dt = -3(1.6)/8
    = - .6 m

    at that moment his shadow is decreasing at .6 m /s

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