a man 6 feet tall is walking away from a lamp post at the rate of 60 feet per minute. when the person is 6 feet from the lamp post, his shadow is 12 feet long. find the rate at which the length of his shadow is increasing when he is 145 feet from the lamp post
If the light is at height h, we know that
6/12 = h/18
h = 9
So, if the shadow has length s when the man is x feet from the pole, then we have
s/6 = (x+s)/9
s = 2x
so, ds/dt = 2 dx/dt
Now just plug in your numbers.
To find the rate at which the length of the man's shadow is increasing, we can use similar triangles.
Let's assume that the height of the lamp post is H and the length of the man's shadow is S.
From the given information, when the man is 6 feet from the lamp post, his shadow is 12 feet long. This gives us the following proportion:
(H + S) / S = 6 / 12
Simplifying, we get:
(H + S) / S = 1 / 2
Next, let's find an expression for S in terms of H. We can use the fact that the man is 6 feet tall:
(H + 6) / 6 = 1 / 2
Cross-multiplying, we have:
2(H + 6) = 6
2H + 12 = 6
2H = -6
H = -3
Since the height of the lamp post cannot be negative, we can discard this solution.
Therefore, there must be an error in the given problem or information. The height of the lamp post cannot be determined, so we cannot find the rate at which the length of the man's shadow is increasing when he is 145 feet from the lamp post.
To find the rate at which the length of the shadow is increasing, we can use similar triangles.
Let's denote the length of the shadow as "s" and the distance of the man from the lamp post as "x". We know that when the person is 6 feet from the lamp post, his shadow is 12 feet long. This gives us the following proportion:
(s + 6) / x = 12 / 6
Simplifying the above equation, we can write:
(s + 6) / x = 2
Now, we can differentiate both sides of the equation with respect to time, assuming that "s" and "x" are functions of time:
d(s + 6) / dt = d(2x) / dt
The derivative of "s" with respect to time (ds/dt) represents the rate at which the length of the shadow is increasing. The derivative of "x" with respect to time (dx/dt) represents the rate at which the distance from the lamp post is changing. Since the man is walking away from the lamp post, dx/dt is given as 60 feet per minute. Therefore, we have:
ds/dt = 2(dx/dt)
Now, we need to find dx/dt when the person is 145 feet from the lamp post. Since we know that the rate at which the person is moving away from the lamp post is 60 feet per minute, we have:
dx/dt = 60
Finally, we can substitute the value of dx/dt into the equation to find ds/dt:
ds/dt = 2(dx/dt)
ds/dt = 2(60)
ds/dt = 120
Therefore, the rate at which the length of the man's shadow is increasing when he is 145 feet from the lamp post is 120 feet per minute.