A street lamp is 6m above a straight road.A
Man 2m tall.walks along the road away from
the lamp at a constant speed of 1.5m/s.At what
rate is his shadow lengthening?
Well, it depends where he is I think
at the beginning
x = man from light pole= Xi
then
x = Xi + 1.5 t
if he starts at the pole at t = 0 then Xi = 0 but we can not assume that
y = shadow length=(2/6)(x+y)
or
y = x/2
so
y=(1/2)(Xi + 1.5 t)
and
dy/dt = (1/2)(1.5) = 3/4 m/s
Draw a diagram. Using similar triangles, if
x = the distance from the lamp post
s = length of shadow
s/1.5 = (x+s)/6
s = x/3
so,
ds/dt = 1/3 dx/dt
...
Extra credit: how fast is the tip of the shadow moving?
Using Steve's definitions
by similar triangles:
2/s = 6/(s+x)
6s = 2s + 2x
4s = 2x
2s = x
2 ds/dt = dx/dt
ds/dt = 1.5/2 = 3/4 or .75 m/s
which is what Damon had. I think Steve misread the man's speed as his distance.
Notice the speed of the shadow is independent of where the man is walking.
Reiny is correct.
Can you explain me about similar triangle concept.?
To find the rate at which the shadow lengthens, we need to consider similar triangles.
Let's call the length of the man's shadow at any given time x. At the same time, the height of the street lamp is 6m, and the man's height is 2m. Since the man is walking away from the lamp, the distance between the man and the lamp is increasing.
Now, let's consider two situations: one at the current moment and another after a very short period of time Δt.
In the current situation, we have a right triangle formed between the lamp, the man, and his shadow. The height of the triangle is 6m (the height of the lamp), the base is x (the length of the shadow), and the hypotenuse is x + d (the total distance between the lamp and the man).
In the second situation, after a short period of time Δt, the man has moved a distance of 1.5m/s * Δt away from the lamp. Therefore, the distance between the lamp and the man has increased by Δt.
Now, let's look at the similar triangles formed in the two situations.
In the current situation:
(height of the lamp) / (length of the shadow) = (distance between the lamp and the man) / (total distance between the lamp and the man)
6 / x = (x + 0) / (x + d)
In the situation after a short period of time Δt:
(height of the lamp) / (length of the shadow after Δt) = (distance between the lamp and the man + Δt) / (total distance between the lamp and the man + Δt)
6 / (x + Δx) = (x + Δx + Δt) / (x + d + Δt)
To find the rate at which the shadow lengthens, we can take the derivative with respect to time of the equation:
6 / x = (x + 0) / (x + d)
Differentiating both sides with respect to time, we get:
-6 / x^2 * dx/dt = (dx/dt) / (x + d)
Now, we can solve for dx/dt, which represents the rate at which the shadow lengthens:
dx/dt = -6 * (x + d) / (x^2)
Substitute the given values:
dx/dt = -6 * (x + 0) / (x^2)
Since we know that the man's height is 2m, we can substitute x = 2 into the equation:
dx/dt = -6 * (2 + d) / (4)
Simplifying the equation, we have:
dx/dt = -3 * (2 + d)
Therefore, the rate at which the man's shadow lengthens is given by the expression -3 * (2 + d), where d represents the distance between the lamp and the man.