differential calculus

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a man 6 ft tall is walking toward a building at the rate of 4 ft/sec. if there is a light on the ground 40 ft from the building, how fast is the man's shadow on the building growing shorter when he is 30 ft from the building?

  • differential calculus -

    Using similar triangles, if the man is x from the light, and his shadow is h tall,

    h/40 = 6/x
    1/40 dh/dt = -6/x^2 dx/dt
    So, when the man is 10 ft from the light,

    1/40 dh/dt = -6/100 * 4
    dh/dt = -9.6 ft/s

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