a spotlight on the ground shines on a wall 12 m away. if a man 2m tall walks from the spotlight toward the building at a speed of 1.6m/s, how fast is the length of his shadow on the building decreasing when he is 4m from the building?
0.6
To solve this problem, we can use similar triangles and the Chain Rule from calculus.
Let's define the following variables:
- Let x represent the distance of the man from the spotlight (along the ground) at any given time.
- Let y represent the length of the man's shadow on the building at any given time.
- Let D represent the distance from the spotlight to the building, which is 12 m.
We are given:
- The man's height is 2 m.
- The man's speed is 1.6 m/s.
- We need to find out how fast the length of his shadow is decreasing when he is 4 m from the building, which means we need to find dy/dt when x = 4.
We can start by setting up a proportion of similar triangles:
(y + 2) / x = y / D
Now, let's differentiate both sides of this equation with respect to time t:
(d/dt) [(y + 2) / x] = (d/dt) (y / D)
To differentiate the left side of the equation, we can use the Quotient Rule:
[(d/dt) (y + 2) * x - (d/dt) (x) * (y + 2)] / (x^2) = (d/dt) (y) / D
Since the man's speed is constant, the derivative of x with respect to t is simply dx/dt = 1.6 m/s.
Also, we are given that dy/dt represents the rate of change of y with respect to t, which is what we want to find.
Plug in the given values:
[(dy/dt + 0) * 4 - 1.6 * (2 + 0)] / (4^2) = dy/dt / 12
Simplifying the equation gives us:
4(dy/dt) - 3.2 = (dy/dt) / 12
Multiply both sides of the equation by 12 to eliminate the denominator:
48(dy/dt) - 38.4 = dy/dt
Rearranging the equation gives us:
48(dy/dt) - dy/dt = 38.4
Simplifying further:
47(dy/dt) = 38.4
Finally, solving for dy/dt:
dy/dt = 38.4 / 47
Therefore, the length of the man's shadow on the building is decreasing at a rate of approximately 0.817 m/s when he is 4 m from the building.