A light is placed on the ground 30 ft from a building. A man 6 ft tall walks from the light toward the building at the rate of 5 ft/sec. Find the rate at which the length of his shadow is changing when he is 15 ft from the building.
using similar triangles, when the man is x feet from the building,
s/30 = 6/(30-x)
1/30 ds/dt = -6/(30-x)^2 dx/dt
Now just plug in your numbers to find ds/dt
To find the rate at which the length of his shadow is changing, we can use related rates and the concept of similar triangles.
Let's assume that the length of the man's shadow is represented by s and the distance between the man and the building is represented by x.
Since the man is walking towards the building, x is decreasing with time.
Given:
Rate at which the man is walking (dx/dt) = -5 ft/sec (negative since x is decreasing)
Length of the man (h) = 6 ft
Distance between the light source and the building (d) = 30 ft
We need to find:
Rate at which the length of his shadow is changing (ds/dt) when the man is 15 ft from the building.
We can set up a proportion using the similar triangles formed by the man, his shadow, and the building:
h/s = d/(d + x)
Substitute the known values:
6/s = 30/(30 + x)
Cross-multiply:
6(30 + x) = 30s
Expand:
180 + 6x = 30s
Rearrange to solve for s:
s = (180 + 6x)/30
s = (6x + 180)/30
s = (x + 30)/5
Differentiate both sides of the equation with respect to time t:
ds/dt = (d/dt)((x + 30)/5)
Using the quotient rule:
ds/dt = (1/5)(dx/dt)
Substitute the given value for dx/dt:
ds/dt = (1/5)(-5 ft/sec)
Simplify:
ds/dt = -1 ft/sec
Therefore, the rate at which the length of his shadow is changing when he is 15 ft from the building is -1 ft/sec (negative indicates that the shadow is decreasing in length).
To find the rate at which the length of the man's shadow is changing, we can use similar triangles. Let's call the length of the man's shadow 'x' and the distance of the man from the light 'y'.
According to the problem, the man is walking towards the building at a rate of 5 ft/sec, so dy/dt (the rate at which 'y' is changing) is equal to -5 ft/sec because 'y' decreases as the man moves towards the building.
We want to find dx/dt (the rate at which 'x' is changing) when 'y' is equal to 15 ft.
Using similar triangles, we can set up the following proportion:
x/y = (height of the building) / (distance from the building to the light)
Since the height of the building is 6 ft, and the distance from the building to the light is fixed at 30 ft, we can rewrite the proportion as:
x/15 = 6/30
Cross-multiplying, we have:
x = (6/30) * 15
Now, we differentiate both sides of the equation with respect to time 't' to find the rate at which 'x' is changing (dx/dt) when 'y' is equal to 15 ft:
dx/dt = d/dt ((6/30) * 15)
dx/dt = (6/30) * d/dt (15)
dx/dt = (6/30) * 0
Since 0 multiplied by any number is 0, we get:
dx/dt = 0
Therefore, the rate at which the length of the man's shadow is changing when he is 15 ft from the building is 0 ft/sec. This means that the length of his shadow is not changing at this point.