Calculus

A spotlight on the ground is shining on a wall 24\text{m} away. If a woman 2\text{m} tall walks from the spotlight toward the building at a speed of 0.6\text{m/s}, how fast is the length of her shadow on the building decreasing when she is 2\text{m} from the building?

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  1. when she is x meters from the light, and her shadow has height y, then

    2/x = y/24
    -2/x^2 dx/dt = 1/24 dy/dt
    When she is 2m from the building (22 m from the light),

    -2/22^2 * 0.6 = 1/24 dy/dt
    dy/dt = -0.06 m/s

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