A lamp is located on the ground 10m from a building. A person 2m tall walks from thelighttowardthebuildingatarateof2m/s. Findtherateatwhichtheperson’sshadow on the wall is shortening when the person is 5m from the building.
x distance from light
2/x = h/10
at t = 0, x = 5 so h = 4
dx/dt = 2 given
so
x = 5 + 2 t
dx = 2 dt
h = 20/x = 20/(2t+5)
dh/dt = -20(2)/(2t+5)^2
=-20/(25) = -4/5 = -0.8 m/s
To find the rate at which the person's shadow on the wall is shortening, we can use the concept of similar triangles.
Let's imagine a right-angled triangle formed by the person, the lamp, and the top of the building. The person's height (2m) corresponds to the length of one side of the triangle, and the distance between the lamp and the building (10m) corresponds to the length of another side.
We need to find the rate at which the shadow is shortening when the person is 5m from the building, which means we need to find the rate at which the third side of the triangle (the length of the shadow) is changing with respect to time.
Let's denote the length of the shadow as x and the distance traveled by the person as d. Since the person is walking towards the building at a rate of 2m/s, we have d/dt = 2.
Using similar triangles, we can write the equation:
2m / x = (2m+2d) / (10m)
Cross-multiplying, we get:
x = 20m / (2m+2d)
Now, we can differentiate both sides of the equation with respect to time (t) to find the rate at which x is changing:
dx/dt = d/dt (20m / (2m+2d))
To simplify, let's substitute d = 5m (since we're interested in when the person is 5m from the building):
dx/dt = d/dt (20m / (2m+2*5m))
= d/dt (20m / (2m+10m))
= d/dt (20m / 12m)
= d/dt (5/3)m/s
Since we previously found that d/dt = 2, we can substitute this value:
dx/dt = 2 * (5/3)m/s
= (10/3)m/s
Therefore, the rate at which the person's shadow on the wall is shortening when the person is 5m from the building is 10/3 meters per second.