A man 6 feet tall walks at a rate of 5 feet/sec toward a street lamp that is 16 feet above the ground. At what rate is the tip ,of his shadow moving? At what rate is the length of his shadow changing when he is 10 feet from the base of the lamp post?
To find the rate at which the tip of the man's shadow is moving, we need to use similar triangles.
Let's denote the height of the man as H, the height of the lamp post as L, and the length of the man's shadow as S.
The height of the shadow, denoted as h, can be determined using the following proportion:
H/L = h/S
Now, we can differentiate both sides of the above equation with respect to time (t), as we want to determine the rates of change:
d(H/L)/dt = d(h/S)/dt
Since the height of the man is a constant, its rate of change is 0. Also, since the length of the shadow is changing, we can denote its rate of change as ds/dt. Hence, the equation becomes:
0 = (dh/dt * S - h * ds/dt) / S^2
Next, let's focus on the second part of the question. When the man is 10 feet from the base of the lamp post, we need to find the rate at which the length of his shadow is changing, ds/dt.
To solve this, we can use the Pythagorean theorem:
S^2 + h^2 = L^2
Differentiating both sides with respect to time, we get:
2S(ds/dt) + 2h(dh/dt) = 2L(dL/dt)
Now, we substitute the values from the previous equation into this one:
0 = (dh/dt * S - h * ds/dt) / S^2 = (ds/dt * S + h * dh/dt) / S
Since we are interested in finding ds/dt, we can isolate it:
ds/dt = -h * dh/dt / S
To find the values for dh/dt, h, and S at the given position of the man, we can use the similar triangles we mentioned earlier.
Since the man is 6 feet tall, his height (H) is 6. The height of the lamp post (L) is given as 16. The distance of the man from the base of the lamp post (x) is 10.
Using the similar triangles, we can say that h/x = H/L --> h = H * x / L
Plugging in the known values, we get:
h = 6 * 10 / 16 = 3.75
Also, using the similar triangles, we can say that S/x = (S+H)/L --> S = (H * x) / (L - H)
Plugging in the known values, we get:
S = (6 * 10) / (16 - 6) = 6
Now, we can substitute these values into the equation we derived earlier to find ds/dt:
ds/dt = -h * dh/dt / S = -3.75 * dh/dt / 6
To find the value of dh/dt, we need to use the equation we derived from the Pythagorean theorem:
2S(ds/dt) + 2h(dh/dt) = 2L(dL/dt)
Since we're interested in finding dh/dt when the man is 10 feet from the base of the lamp post, we can plug in the known values:
2(6)(ds/dt) + 2(3.75)(dh/dt) = 2(16)(0)
Simplifying this equation, we get:
12(ds/dt) + 7.5(dh/dt) = 0
Since ds/dt = -3.75 * dh/dt / 6, we can substitute this into the above equation:
12(-3.75 * dh/dt / 6) + 7.5(dh/dt) = 0
Simplifying further, we have:
-6.25(dh/dt) + 7.5(dh/dt) = 0
Since we now have an equation with only one variable, we can solve for dh/dt:
1.25(dh/dt) = 0
dh/dt = 0
Therefore, when the man is 10 feet from the base of the lamp post, the length of his shadow is not changing (dh/dt = 0).
To find the rate at which the tip of the man's shadow is moving, we need to use similar triangles.
Let's denote the length of the man's shadow as "x" and the distance between the man and the lamp post as "y". We know that the height of the lamp post is 16 feet.
The height of the man and the length of his shadow create a set of similar triangles with the lamp post and its shadow. The ratio of the height of the man to the length of his shadow is equal to the ratio of the height of the lamp post to the length of its shadow.
In other words, we have the following proportion:
(6 feet / x) = (16 feet / y)
Now, let's differentiate both sides of the equation with respect to time (t):
d(6 feet / x)/dt = d(16 feet / y)/dt
To find d(6 feet / x)/dt, we use the chain rule:
d(6 feet / x)/dt = (-6 feet / x^2) * dx/dt
We know dx/dt is given as 5 feet/sec since the man is walking at a constant rate of 5 feet/sec. We can substitute these values into the equation:
(-6 feet / x^2) * (5 feet/sec) = (-30 feet / x^2) feet/sec
Therefore, the rate at which the tip of the man's shadow is moving is equal to (-30/x^2) feet/sec.
Next, we need to find the rate at which the length of his shadow is changing when he is 10 feet from the base of the lamp post.
To do this, we need to find dy/dt, which represents the rate at which y is changing with respect to time (t).
We can use the Pythagorean theorem to relate x, y, and the distance from the man to the lamp post (which is 10 feet):
x^2 + y^2 = 10^2
Differentiating this equation with respect to time (t) gives us:
2x * dx/dt + 2y * dy/dt = 0
Substituting the known values, we have:
2x * (5 feet/sec) + 2y * dy/dt = 0
Since we are interested in dy/dt, we can solve for it:
dy/dt = (-2x * dx/dt) / (2y)
Substituting the known values again:
dy/dt = (-2x * 5 feet/sec) / (2y)
dy/dt = (-10x feet/sec) / y
When the man is 10 feet from the base of the lamp post, we can substitute x = 10 into the formula:
dy/dt = (-10 * 10 feet/sec) / y
Since we want to find the rate at which the length of his shadow is changing, we now have -100/y feet/sec.
Therefore, when the man is 10 feet from the base of the lamp post, the length of his shadow is changing at a rate of -100/y feet/sec.
draw a diagram. using similar triangles, if the length of the shadow is x, and the distance to the lamp is y,
x/6 = (x+y)/16
16x = 6x + 6y
10x = 6y
x = 3/5 y
since y is changing at -5ft/sec
x is changing at -3ft/sec
no matter where the man is.