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 1.6 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?
make a sketch, place the man x m from the light
let the length of the shadow on the wall me y m
by similar triangles : 2/x = y/12
xy = 24
x dy/dt + y dx/dt = 0
dy/dt = -y dx/dt / x
when he is 4 m from the wall x = 8 (look at our diagram)
so in xy=24 --> 8y = 24 or y = 3
and dx/dt = 1.6
dy/dt = -3(1.6)/8
= - .6 m
at that moment his shadow is decreasing at .6 m /s
To find the rate at which the length of the man's shadow is decreasing, we need to use related rates. Let's call the length of the man's shadow "x" and the distance between the man and the building "y".
We are given that the man's height is 2 m and he is walking towards the building at a speed of 1.6 m/s. This means that the rate at which the distance between the man and the building is changing is -1.6 m/s (negative because he is moving towards the building).
We need to find the rate at which the length of the shadow is changing (dx/dt) when the man is 4 m from the building. To do this, we can use similar triangles.
Using similar triangles, we can set up the following proportion:
x/y = 2/12
Cross-multiplying, we get:
2y = 12x
Differentiating both sides of the equation with respect to time (t), we get:
2(dy/dt) = 12(dx/dt)
Rearranging the equation, we get:
dx/dt = (2/12)(dy/dt)
Substituting the given values, when the man is 4 m from the building, dy/dt = -1.6 m/s, we have:
dx/dt = (2/12)(-1.6) = -0.267 m/s
Therefore, the length of the man's shadow on the building is decreasing at a rate of approximately 0.267 m/s when he is 4 m from the building.