A lamp is placed on the ground 100 feet away from a wall. A man six feet tall is walking at a speed of 10 ft/sec from the lamp to the nearest point on the wall. When he is midway between the lamp and the wall, the rate of change in the length of his shadow is (in ft/ sec)?

Discussion:

Let the length of the shadow at any point of time me y and distance of the man from Lamp be x.

By similarity of triangles we can say \(\frac {y}{100} = \frac {6}{x} \).

Differentiating both sides with respect to time t we get \(\frac {1}{100} \frac {dy}{dt} = 6 \cdot \frac {-1}{x^2} \frac {dx}{dt} \). Replace dx/dt by 10, x by 50 we get dy/dt = -2.4 ft/sec where the negative sign implies reduction in the length of the shadow.

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