Introduction

The problem presented involves a man who is 6 meters tall, standing away from a light source that is 10 meters above the ground. The objective is to calculate the rate at which the tip of his shadow is moving, given that his shadow is lengthening at a rate of 2 meters per second, and the man is walking at a rate of 4/3 meters per second. This problem involves the principles of geometry and the concept of related rates in calculus.

Understanding the Geometry

Let's visualize the scenario. The light source is at the top of a 10-meter pole, and the man is standing at a certain distance from the pole. The height of the man is 6 meters, and we need to find the rate at which the tip of his shadow is moving. We can approximate the light source as a point source, which simplifies the problem using similar triangles.

Let's denote the following:

h 6 m (height of the man) H 10 m (height of the light source) x(t) is the horizontal distance of the man from the base of the light source s(t) is the length of the shadow V(x) is the length of the shadow as a function of the man's distance from the light source V'(x) is the rate of change of the shadow length with respect to time V'(x) 2 m/s (given that the shadow is lengthening at 2 meters per second) x'(t) 4/3 m/s (given that the man is walking at 4/3 meters per second)

By similar triangles, we can establish the following relationship:

[frac{H}{s x} frac{h}{s}]

Substituting the given values:

[frac{10}{s x} frac{6}{s}]

Rearranging the equation:

[frac{10s}{6(s x)} 1 Rightarrow 10s 6(s x) Rightarrow 10s 6s 6x Rightarrow 4s 6x Rightarrow s frac{3}{2}x]

This means the shadow length is (frac{3}{2}) times the distance of the man from the light source. Therefore:

[frac{dS}{dt} 3 frac{dx}{dt}]

We know that (frac{dx}{dt} frac{4}{3} m/s), so:

[frac{dS}{dt} 3 cdot frac{4}{3} 4 m/s]

The rate at which the tip of the shadow is moving is the sum of the rate at which the man is walking and the rate at which the shadow is lengthening:

[frac{dS_{tip}}{dt} frac{dx}{dt} V'(x) frac{4}{3} 2 frac{4 6}{3} frac{10}{3} m/s]

Conclusion

The tip of the shadow is moving at a rate of (frac{10}{3}) meters per second. This problem demonstrates the application of related rates in calculus and how the concept of similar triangles can be used to solve real-world problems.

Additional Insights

Understanding the principles of related rates can help in various scenarios where rates of change are involved, such as in physics, engineering, and economics. It is crucial to visualize the problem and set up the correct equations before solving.

Keywords: Shadow movement, man walking, light source, rate of change, geometry.