A street light is mounted at the top of a 6-meter-tall pole. A man 2 m tall walks away from the pole with a speed of 1.4 m/s along a straight path. How fast is the tip of his shadow moving when he is 16 m from the pole?
same question ....
http://www.jiskha.com/display.cgi?id=1333489636
just change the necessary numbers
since you have a calculus class, you probably ought to do it Reiny's way, but sometimes a little simplification can save you some work:
If the man is x away from pole, ad his shadow length is s, then using similar triangles,
s/2 = (x+s)/6
s = 1/2 x
so, ds/dt = 1/2 dx/dt
since the tip of the shadow is x+s from the pole, it is moving at
dx/dt + ds/dt = dx/dt + 1/2 dx/dt = 3/2 dx/dt = 3/2(1.4) = 2.1 m/s
To solve this problem, we can use similar triangles and related rates.
Let's denote the distance between the man and the base of the pole as "x" and the length of his shadow as "s". We are given that the man's height is 2 m and the pole's height is 6 m.
By using similar triangles, we can write the proportion:
(Length of shadow + Height of pole) / Height of man = Length of shadow / Height of man
Using the values given, we get:
(s + 6) / 2 = s / 2
Simplifying this equation, we find:
s + 6 = s / 2
Multiplying both sides of the equation by 2, we have:
2s + 12 = s
Subtracting "s" from both sides, we get:
s = -12
This negative value for s doesn't make sense in this context, so we discard it. Therefore, the length of the shadow is 12 m.
Now, let's differentiate the equation with respect to time (t):
(s + 6) / 2 = s / 2
Differentiating both sides with respect to t, we have:
[(ds/dt) + 0] / 2 = (ds/dt)/2
Simplifying this equation, we find:
(ds/dt) / 2 = (ds/dt) / 2
This equation tells us that the rate at which the tip of the shadow is moving (ds/dt) is equal to the rate at which the length of the shadow is changing (ds/dt).
We are given that the man is walking away from the pole with a speed of 1.4 m/s. This means that dx/dt = -1.4 m/s (negative sign because the man is moving away from the pole).
We want to find the value of ds/dt when x = 16 m. From the similar triangles, we can deduce that:
x / s = (x + 6) / (s + 6)
Substituting the values, we get:
16 / s = (16 + 6) / (s + 6)
Multiplying both sides by s and expanding, we find:
16s + 96 = 22s + 132
Rearranging this equation, we have:
6s = 36
Dividing both sides by 6, we get:
s = 6
So, when x = 16 m, the length of the shadow is 6 m.
Now, substituting dx/dt = -1.4 m/s and ds/dt = ? in the equation:
(ds/dt) / 2 = (ds/dt) / 2
(-1.4) / 2 = (ds/dt) / 2
Multiplying both sides by 2, we find:
-1.4 = ds/dt
Therefore, the tip of the man's shadow is moving at a speed of -1.4 m/s when he is 16 m from the pole.
Note: The negative sign indicates that the tip of the shadow is moving in the opposite direction of the man.
To find the rate at which the tip of the man's shadow is moving, we need to use similar triangles and the concept of rates.
Let's consider the situation at any given time when the man is 16m from the pole.
First, let's determine the length of the man's shadow. We can set up a proportion using the similar triangles formed by the man, the pole, and the shadow:
(man's height) / (length of man's shadow) = (height of pole) / (distance between man and pole)
Substituting the given values into the equation, we have:
2m / x = 6m / 16m
Now, we can cross-multiply and solve for x, which represents the length of the man's shadow:
2m * 16m = 6m * x
32m^2 = 6m * x
Next, let's differentiate both sides of the equation with respect to time (t) to find the rate of change:
d(32m^2) / dt = d(6m * x) / dt
Now, we need to find the rate at which the tip of the man's shadow is moving, so we differentiate each term with respect to time:
64m * (dm / dt) = 6m * (dx / dt)
We know that the man's height is constant, so dm / dt is zero. Therefore, we can simplify the equation to:
64m * 0 = 6m * (dx / dt)
0 = 6m * (dx / dt)
Now, we can solve for dx / dt, which represents the rate at which the tip of the man's shadow is moving:
dx / dt = 0 / 6m
Since anything divided by zero is undefined, the rate at which the tip of the man's shadow is moving when he is 16m from the pole is undefined.