A street light is at the top of a 11 ft. tall pole. A man 6.2 ft tall walks away from the pole with a speed of 4.5 feet/sec along a straight path. How fast is the tip of his shadow moving when he is 31 feet from the pole?
To solve this problem, we can use similar triangles. The height of the pole and the height of the man form a set of similar triangles with the shadow.
Let's label the variables:
- h = height of the pole (11 ft)
- x = distance from the man to the pole (31 ft)
- y = length of the man's shadow
Now, let's set up a proportion between the similar triangles:
h / x = (h + y) / y
To solve for y, we can cross-multiply and simplify the equation:
y * h = x * (h + y)
y * h = h * x + x * y
y * h - x * y = h * x
y(h - x) = h * x
y = (h * x) / (h - x)
Now, we can differentiate both sides of the equation with respect to time (t) to find the rate of change:
dy/dt = (d/dt) [(h * x) / (h - x)]
To find the rate at which the tip of the shadow is moving, we need to relate dx/dt (rate at which the man is walking) to dy/dt:
dx/dt = 4.5 ft/sec
Now, we substitute the given values into the equation and solve for dy/dt:
dy/dt = (11 ft * 31 ft/sec) / (11 ft - 31 ft)
dy/dt = -341 ft^2/sec / -20 ft
dy/dt = 17.05 ft/sec
Therefore, the tip of the man's shadow is moving at a rate of 17.05 ft/sec when he is 31 ft from the pole.
To find the speed at which the tip of the man's shadow is moving when he is 31 feet from the pole, we need to concern ourselves with similar triangles and how the length of the shadow is changing over time.
Let's consider the following diagram:
/
/
/ x <--(Man)
/---------
/|| | |\
/ || | h | \
-------------/------- S | | | \
|| | | L |
|| | | |
|| | | |
|| | | |
|| | | |
------------------------------------- (Pole)
Here, "h" is the height of the man, "S" is the length of the shadow, and "L" represents the distance between the man and the pole.
Since the triangles formed by the man, the pole, and his shadow are similar, we can establish the following proportion:
x / (x + L) = h / S
where "x" represents the length of the shadow cast at a given time.
To find the rate at which the tip of the shadow is moving, we need to differentiate this equation with respect to time (t).
Differentiating both sides of the equation gives us:
(d/dt)(x / (x + L)) = (d/dt)(h / S)
To find (d/dt)x, we need to determine the rates at which L and S are changing over time:
L is simply the distance between the man and the pole, and since the man's velocity is given as 4.5 ft/sec, the rate of change of L is -4.5 ft/sec (negative because he is walking away from the pole).
S is the length of the shadow, but we don't have its exact value. However, we can use the given dimensions to establish another relationship.
We have a right-angled triangle formed by the pole, the length of the shadow, and the distance from the tip of the shadow to the pole. We can apply the Pythagorean theorem to determine S:
S^2 = L^2 + (11)^2
Now, we can differentiate both sides of this equation with respect to time:
(d/dt)(S^2) = (d/dt)(L^2 + (11)^2)
2S(dS/dt) = 2L(dL/dt)
Solving for dS/dt:
(dS/dt) = (L * (dL/dt)) / S
Now, substituting both (d/dt)x and (d/dt)S into the differentiated equation from before:
(d/dt)x / (x + L) = h * (L * (dL/dt)) / (S^2)
Now, we can plug in the given values:
h = 6.2 ft (height of the man)
L = 31 ft (distance between the man and the pole)
(dL/dt) = -4.5 ft/sec (rate of change of L)
S = sqrt(31^2 + 11^2) ≈ 32.015 ft (length of the shadow)
Substituting these values in the equation gives us:
(d/dt)x / (x + 31) = (6.2 * (-4.5) * (31)) / (32.015^2)
Now, we can solve for (d/dt)x by multiplying both sides by (x + 31) and dividing by the right-hand side of the equation:
(d/dt)x = ((6.2 * (-4.5) * (31)) / (32.015^2)) * (x + 31)
So, the final expression for (d/dt)x, the rate at which the tip of the shadow is moving, is:
(d/dt)x = ((6.2 * (-4.5) * (31)) / (32.015^2)) * (x + 31)
Substituting x = 31 ft (as given in the question), we can evaluate the expression to find the actual speed.
x = distance man from pole
y = distance tip of shadow from pole
(y-x)/6.2 = y/11
11(y-x) = 6.2 y
4.8 y = 11 x
y = 2.29 x
dy/dt = 2.29 dx/dt = 2.29 * 4.5 = 10.3ft/s