A street light is at the top of a 19 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 4 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 50 ft from the base of the pole?
Well, this sounds like an illuminating problem! Let's light it up with some math and humor.
First, we need to find the rate at which the tip of her shadow is moving. To do that, we can use similar triangles. The height of the pole and the woman's height form one set of parallel sides, while the length of the shadow and the distance from the woman to the pole form another set.
Since these triangles are similar, we can set up the following proportion:
(Height of the pole)/(Length of shadow) = (Height of the woman)/(Distance from the woman to the pole)
So, we have:
19 ft/x = 6 ft/(50 ft)
To find x (the length of the shadow), we can cross multiply and solve for x:
x = (19 ft * 50 ft) / 6 ft
Now, let's calculate the rate at which the tip of her shadow is moving. We can differentiate both sides of the equation with respect to time:
d(x) / dt = (d(19 ft * 50 ft) / dt) / 6 ft
d(x) / dt = (0 ft^2/sec) / 6 ft
d(x) / dt = 0 ft/sec
Surprisingly, the tip of her shadow is not moving at all! I guess it's too cool to follow her as she walks away. Maybe her shadow is just enjoying some downtime and decided to stay put. Shadow, you da real MVP!
I hope this illuminating and humorous answer brightened your day! If you have any more questions, feel free to ask!
To find the rate at which the tip of her shadow is moving, we need to use similar triangles and the chain rule from calculus.
Let's set up the following diagram:
woman
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We have the following information:
- The height of the pole is 19 ft.
- The height of the woman is 6 ft.
- The woman is walking 50 ft away from the base of the pole at a rate of 4 ft/sec.
Let's define some variables:
- Let y be the distance from the tip of the shadow to the base of the pole.
- Let x be the distance from the tip of the shadow to the woman.
We need to find dy/dt, the rate at which y is changing with respect to time when x = 50 ft.
Now, let's look at the similar triangles formed by the pole, the woman, and the shadow. The ratio of the woman's height to her distance from the pole is equal to the ratio of the shadow's height to its distance from the pole.
These ratios can be written as:
6 / x = (19 + 6 - y) / y
Cross-multiplying, we get:
6y = x(19 + 6 - y)
Expanding the equation:
6y = 25x - xy
To solve for y, we rearrange the equation:
xy + 6y = 25x
Factoring out y:
y(x + 6) = 25x
Now, we can solve for y:
y = (25x) / (x + 6)
To find dy/dt, the rate of change of y with respect to time, we differentiate both sides of the equation with respect to t:
dy/dt = (25(dx/dt)(x + 6) - (25x)(dx/dt)) / (x + 6)^2
Substituting the given values:
dx/dt = 4 ft/sec (rate at which the woman is walking)
x = 50 ft (distance from the tip of the shadow to the woman)
x + 6 = 56 ft (distance from the tip of the shadow to the base of the pole)
dy/dt = (25(4)(50 + 6) - (25)(50)) / (56)^2
Simplifying the expression:
dy/dt = (25(4)(56) - (25)(50)) / (56)^2
Calculating the numerator:
= (5600 - 1250) / (56)^2
= 4350 / (56)^2
Calculating the denominator:
= 4350 / 3136
= 1.386 ft/sec
Therefore, the tip of her shadow is moving at a rate of approximately 1.386 ft/sec when she is 50 ft from the base of the pole.