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 1.7 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building? (Round your answer to one decimal place.)
Question
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 1.7 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building? (Round your answer to one decimal place.)
Solution
This problem can be solved using similar triangles and the concept of related rates in calculus.
Step 1: Set up the problem We can set up two similar triangles: one with the man and his shadow, and one with the spotlight, the man, and the top of his shadow on the wall.
Step 2: Set up the equation Let's denote:
- x as the distance between the man and the spotlight
- y as the length of the shadow on the wall
From the similar triangles, we have the relationship: 2/(12-x) = (2+y)/12
Step 3: Differentiate both sides with respect to time (t) Differentiating both sides with respect to time gives us: 2/(12-x)^2 * dx/dt = 1/12 * dy/dt
Step 4: Substitute the given values We know that dx/dt = -1.7 m/s (the man is moving towards the spotlight), and we want to find dy/dt when x = 4 m. Substituting these values into the equation gives us: 2/(12-4)^2 * -1.7 = 1/12 * dy/dt
Step 5: Solve for dy/dt Solving for dy/dt gives us: dy/dt = -2.55 m/s
So, the length of his shadow on the building is decreasing at a rate of 2.55 m/s when he is 4 m from the building.
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