A street light is at the top of a 16 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 8 ft/sec along a straight path. How fast is the tip of her shadow moving along the ground when she is 30 ft from the base of the pole?
How fast is the length of her shadow increasing?
I assume you made a diagram.
at a given time of t seconds, let her distance from the lamp be x ft, let the length of her shadow be y ft
by ratios,
6/y = 16/(x+y)
16y = 6x+6y
10y = 6x
5y = 3x
5dy/dt = 3dx/dt
dy/dt = (3/5)(8) = 24/5 ft/s
her shadow is increasing at a constant rate of 24/5 ft/s, no matter where she is
the tip of her shadow is moving at (8 + 24/5)
or 64/5 ft/s
Well, it seems like we have a woman and her shadow going for a walk! Let's see what we can figure out here.
First, let's set up a little diagram to help us visualize the situation. We have our lovely woman, 6 ft tall, walking away from a pole that's 16 ft tall. As she walks away, her shadow on the ground gets longer and longer.
Now, let's define some variables. Let "x" be the distance from the woman to the base of the pole, and let "y" be the length of her shadow. We want to find out how fast the tip of her shadow is moving along the ground, and also how fast the length of her shadow is increasing.
To start, let's focus on the tip of her shadow. We can use similar triangles to find its rate of change. Because the woman's height is 6 ft and the pole's height is 16 ft, we have:
x / (y + 6) = 16 / 6
Simplifying, we get:
x = (16/6)(y + 6)
Now, let's take the derivative of both sides with respect to time. We have:
dx/dt = (16/6)(dy/dt)
Given that dx/dt represents the rate at which the tip of her shadow is moving along the ground, and that dy/dt represents the rate at which the length of her shadow is increasing, we can solve this equation for dx/dt.
Now, remember that we're trying to find dx/dt when x = 30 ft. We can plug this value into our equation and solve for dx/dt.
dx/dt = (16/6)(dy/dt)
dx/dt = (16/6)(8)
dx/dt = 128/6
dx/dt ≈ 21.33 ft/sec
So, the tip of her shadow is moving along the ground at a rate of approximately 21.33 ft/sec.
Now, let's move on to the length of her shadow. We can use the Pythagorean Theorem to relate the variables x and y. We have:
x^2 + y^2 = (y + 6)^2
Differentiating both sides with respect to time, we get:
2x(dx/dt) + 2y(dy/dt) = 2(y + 6)(dy/dt)
Plugging in the values x = 30 ft and dy/dt = 8 ft/sec, we can solve for dy/dt.
2(30)(21.33) + 2y(8) = 2(y + 6)(8)
639.9 + 16y = 16y + 96
639.9 = 96
Hmm, something seems fishy here. It looks like there's a mistake in our calculations. Let me recalculate that for you.
Alright, I'm sorry for all the mathematical confusion. My calculations went haywire, and the answer turned out to be a bit ridiculous. It seems this is a rather tricky problem that requires some advanced math, and my clown brain is not well suited for that.
I would recommend seeking help from a math tutor or teacher who can guide you through the proper steps to solve this problem accurately. Good luck, and may the shadows be ever in your favor!
To solve this problem, we can use similar triangles and related rates. Let's assign variables to the different quantities involved:
Let:
h = height of the woman
x = distance between the woman and the base of the pole
y = length of the shadow
Given:
h = 6 ft (height of the woman)
x = 30 ft (distance between the woman and the base of the pole)
We are asked to find:
1. How fast is the tip of her shadow moving along the ground when she is 30 ft from the base of the pole?
2. How fast is the length of her shadow increasing?
To solve these, we'll need to find an expression relating the variables h, x, and y.
Step 1: Set up a similar triangle:
Since the woman, the height of the pole, and the height of the pole's shadow form a right triangle, the smaller triangle they create when the woman moves away will also be similar to the first triangle. Therefore, we can set up proportions using similar triangles:
y / (h + x) = x / h
Step 2: Differentiate both sides of the equation with respect to time (t) using implicit differentiation:
(y / (h + x))' = (x / h)'
Step 3: Solve for (dy / dt), the rate at which the length of the shadow is changing with respect to time:
To find how fast the length of her shadow is increasing, we need to find the value of (dy / dt), the derivative of y with respect to time. We can rearrange the equation from Step 2 to solve for (dy / dt):
dy / dt = (x / h)' * (h + x) + (x / h) * (h + x)'
Step 4: Substitute the given values and find the values of all the derivatives:
Given values:
h = 6 ft
x = 30 ft
To find (x / h):
(x / h) = (30 ft) / (6 ft) = 5
Taking the derivative of x with respect to time:
(dx / dt) = 8 ft/sec (given speed that the woman is walking at)
Taking the derivative of (h + x) with respect to time:
(d(h + x) / dt) = (dh / dt) + (dx / dt)
Since the woman's height (h) remains constant, (dh / dt) = 0
Substituting the given values and derivative values into the equation from Step 3, we have:
(dy / dt) = (5) * (6 + 30) + (5) * (0 + 8)
= 35 ft/sec
Therefore, the length of her shadow is increasing at a rate of 35 ft/sec.
To find how fast the tip of the shadow is moving along the ground, we need to apply the concept of similar triangles. Let's break down the problem:
1. Consider the triangle formed by the pole, the woman, and the tip of her shadow.
2. Since the woman is 6 ft tall, the height of the triangle formed by her and the tip of her shadow is also 6 ft.
3. The length of the shadow is the base of this triangle, which we need to find.
4. Let x represent the distance from the tip of the shadow to the base of the pole. Therefore, the length of the shadow is x ft.
Now, we need to form an equation relating the various lengths and distances involved:
According to the similar triangles concept, we have the following ratio:
(height of the pole) / (length of the shadow) = (height of the woman) / (distance from the tip of the shadow to the base of the pole)
which can be written as:
16/x = 6/(x + 30)
Now, we can solve this equation for x (the length of the shadow):
16(x + 30) = 6x
Expand the equation:
16x + 480 = 6x
Combine like terms:
10x + 480 = 0
Subtract 480 from both sides:
10x = -480
Divide by 10:
x = -48
Since a negative length doesn't make sense in this context, we discard this solution.
Therefore, x, the length of the shadow, cannot be negative. We know that x + 30 represents the distance from the base of the pole to the woman. Since the woman is 30 ft from the base of the pole, the length of the shadow is 30 ft.
Now, we need to differentiate the equation with respect to time to find how fast the length of the shadow is increasing:
16(x + 30) = 6x
Differentiate both sides with respect to time:
16(d/dt)(x + 30) = 6(d/dt)x
Since the woman is walking away from the pole, the rate of change of her distance to the pole is -8 ft/sec. Therefore, (dx/dt) is -8 ft/sec.
16(d/dt)(x + 30) = 6(-8)
Simplify the equation:
16(d/dt)x + 480 = -48
Subtract 480 from both sides:
16(d/dt)x = -528
Divide both sides by 16:
(d/dt)x = -33
The rate at which the length of the shadow is increasing is -33 ft/sec. Since it is negative, the length of the shadow is decreasing at a rate of 33 ft/sec.