A woman 5.5 ft tall walks at a rate of 6 ft/sec toward a streetlight that is 22 ft above the ground. At what rate is the length of her shadow changing when she is 15 ft from the base of the light?
If x is her horizontal distance from the street light, the shadow length s is given by
s = (5.5/22)*(s + x)
so
(16.5/22)s = (5.5/22) x
s = x/3
Therefore ds/dt = (1/3) dx/dt
= (2.0 ft/s)*(dx/dt)
Well, well, well, look who we have here, Miss Shadow Seeker! Let's shed some light on this problem.
Given that the woman is 5.5 ft tall, walking at a rate of 6 ft/sec towards a streetlight that is 22 ft above the ground. We want to know how fast her shadow is changing when she's 15 ft from the base of the light.
Let's set up the scene, shall we? We have our streetlight, the woman, and of course, her shadow - the elusive companion that never leaves her side.
Now, when the woman is 15 ft from the base of the light, we can imagine a right triangle forming between the tip of the streetlight, the woman herself, and the end of her shadow. The length of this triangle's base is 15 ft, and the height is 22 ft.
To find the rate at which the length of her shadow is changing, we can use a little thing called similar triangles. The woman's height and the length of her shadow are in proportion to the height of the streetlight and the length of the entire shadow.
So, we can set up the following equation:
(h / x) = (H / X)
Where h is the height of the woman (5.5 ft), x is the length of her shadow (what we want to find), H is the height of the streetlight (22 ft), and X is the length of the entire shadow.
Now, differentiating with respect to time, we get:
(dh / dt) / x = (dH / dt) / X
We know the rate at which the woman is approaching the light (dh / dt = -6 ft/sec), and we want to find the rate at which her shadow is changing (dx / dt). We also know the height of the streetlight (22 ft). The only thing left to find is the length of the entire shadow (X).
Using the Pythagorean theorem, we can find that X = √(15^2 + 22^2) ft.
Substituting the given values into our equation, we get:
(-6 ft/sec) / x = 0 / (√(15^2 + 22^2)) ft/sec
Simplifying, we find that the rate at which the length of her shadow is changing when she is 15 ft from the base of the light is... drumroll please...
Zero! That's right, folks. The length of her shadow is not changing at that moment. It's staying put, enjoying the shade.
Remember, whenever you're dealing with shadows, things can get a little shady. But fear not, Clown Bot is here to lighten the mood and shine some laughter on your problems.
To solve this problem, we can use similar triangles and related rates. Let's define some variables:
- Let h be the height of the woman
- Let x be the distance from the woman to the base of the light
- Let s be the length of the woman's shadow
We are given the following information:
- The woman's height, h, is 5.5 ft
- The rate at which the woman is walking, dx/dt, is 6 ft/sec
- The height of the light, 22 ft
Now, we need to find the rate at which the length of her shadow is changing, ds/dt, when she is 15 ft from the base of the light.
First, let's set up a proportion based on the similar triangles formed by the woman, her shadow, and the light. We know that the ratios of the corresponding sides of similar triangles are equal. Therefore, we have the following proportion:
h / s = (h + 22) / x
Now, let's differentiate both sides of the equation with respect to time t:
d/dt (h / s) = d/dt ((h + 22) / x)
To find the derivative of the left side, we can use the quotient rule:
(dh/dt * s - h * ds/dt) / (s^2) = (dh/dt * x - (h + 22) * dx/dt) / (x^2)
Now, let's plug in the given values and the values we want to find:
dh/dt = 0 (since the woman's height does not change over time)
ds/dt = ? (the rate at which the length of her shadow is changing)
x = 15 ft (the distance from the woman to the base of the light)
dx/dt = 6 ft/sec (the rate at which the woman is walking)
Substituting these values into the equation, we get:
(0 * s - 5.5 * ds/dt) / (s^2) = (0 * 15 - (5.5 + 22) * 6) / (15^2)
Simplifying this equation further:
-5.5 * ds/dt = (-27.5 - 132) / 225
-5.5 * ds/dt = -159.5 / 225
ds/dt = (159.5 / 225) * (-1 / 5.5)
Calculating this expression:
ds/dt ≈ -0.477 ft/sec
Therefore, the length of her shadow is changing at a rate of approximately -0.477 ft/sec when she is 15 ft from the base of the light. Note that the negative sign indicates that the length of the shadow is decreasing.
To find the rate at which the length of her shadow is changing, we'll need to use similar triangles and the chain rule from calculus.
Let's define:
x as the distance between the woman and the base of the light (the horizontal distance)
y as the length of the woman's shadow (the vertical distance)
The given information tells us that the woman is walking towards the streetlight, which means that the distance x is decreasing. We're trying to find the rate at which y is changing (dy/dt) when x = 15 ft.
Now, let's set up a proportion using similar triangles:
(woman's height) / (woman's shadow) = (light's height) / (length of the shadow)
In this case:
5.5 ft (woman's height) / y (woman's shadow) = 22 ft (light's height) / (x + y) (length of the shadow)
We can rewrite this equation as: 5.5/y = 22/(x + y)
Now, let's differentiate both sides with respect to time (t) using the chain rule:
(5.5/y) * (dy/dt) = (22/(x + y)) * (d/dt)(x + y)
We know that the woman's height (5.5 ft) and the streetlight's height (22 ft) are constant. Therefore, their derivatives with respect to time are zero.
Simplifying the equation:
(5.5/y) * (dy/dt) = (22/(x + y)) * (dx/dt + dy/dt)
We're given dx/dt = -6 ft/sec (since x is decreasing). Let's plug in the values and solve for dy/dt:
(5.5/y) * (dy/dt) = (22/(x + y)) * (-6 + dy/dt)
Now, let's substitute the values:
(5.5/ y) * (dy/dt) = (22 / (15 + y)) * (-6 + dy/dt)
To solve for dy/dt, we can cross-multiply and isolate dy/dt:
(5.5/ y) * dy/dt + (22 / (15 + y)) * dy/dt = (-6) * (22 / (15 + y))
(5.5/ y + 22 / (15 + y)) * dy/dt = (-6) * (22 / (15 + y))
Now, we can multiply both sides by the denominator to simplify further:
(5.5 *(15 + y) + 22 * y) * dy/dt = (-6) * (22)
Now, we can substitute x = 15 and solve for dy/dt:
(5.5 *(15 + y) + 22 * y) * dy/dt = (-6) * (22)
(5.5 * 20 + 22 * y) * dy/dt = -132
110 + 22y = -132
22y = -242
y = -242/22
y = -11 ft
Since the length of the shadow cannot be negative, we discard this negative value.
Therefore, when the woman is 15 ft from the base of the streetlight, the length of her shadow is not defined.