A person whose height is 6feet is walking away from the base of a streetlight along a straight path at a rate of 4 feet per second. If the height of the streetlight is 15 feet, what is the rate at which the person's shadow is lengthening?

Ah, the classic shadow conundrum! Well, let's shed some light on this, shall we? We know that the person's height is 6 feet and they're walking away from the streetlight at 4 feet per second. So, we can assume that the length of the person's shadow is changing over time.

Now, imagine this scenario: as the person walks away, they cast a shadow behind them. As they move, the shadow gets longer. So, the rate at which the person's shadow is lengthening can be determined by using similar triangles!

Since the height of the streetlight is 15 feet and the person's height is 6 feet, the ratio between the person's height and the length of their shadow will always be the same. This means that we can use this ratio to find the rate at which the shadow is lengthening.

So, let's set up a proportion:

Person's height / Length of shadow = 6 / x

Streetlight's height / Distance from streetlight = 15 / (x + 4)

Now, let's cross-multiply and solve for x:

6 * (x + 4) = 15 * x

6x + 24 = 15x

24 = 9x

x = 24 / 9

x ≈ 2.67 feet

Therefore, the rate at which the person's shadow is lengthening is approximately 2.67 feet per second. Hope that shines some light on the situation for you!

To find the rate at which the person's shadow is lengthening, we can use similar triangles. Let's assume the length of the person's shadow at a given time is x feet.

Since the person's height is 6 feet and the height of the streetlight is 15 feet, we can set up the following proportion:

(person's height) / (person's shadow length) = (streetlight's height) / (streetlight's shadow length)

6 / x = 15 / (x + 6).

To find x, we can cross-multiply:

6 * (x + 6) = 15 * x
6x + 36 = 15x
9x = 36
x = 4.

Therefore, the person's shadow is currently 4 feet long.

To find the rate at which the person's shadow is lengthening, we can take the derivative of the equation we found: 6 / (x + 6).

By taking the derivative, we find the rate at which the person's shadow is lengthening:

d/dt (6 / (x + 6)) = -6 / (x + 6)^2 * dx/dt.

Since the person is walking away from the streetlight at a rate of 4 feet per second (dx/dt = 4), we can substitute this value into the equation:

-6 / (4 + 6)^2 * 4 = -6 / 100 * 4 = -24 / 100 = -0.24.

Therefore, the rate at which the person's shadow is lengthening is -0.24 feet per second. Note that the negative sign indicates that the shadow is getting shorter.

To find the rate at which the person's shadow is lengthening, we can use similar triangles.

Let's define a few variables:

- Let x be the distance between the person and the base of the streetlight.
- Let y be the length of the person's shadow.
- Let h be the height of the streetlight.
- Let d be the rate at which the person is walking, which is 4 feet per second.

We can create a proportion using the similar triangle formed by the person, their shadow, and the streetlight:

(x + y) / y = (h + y) / h

Now, let's differentiate both sides of the equation with respect to time (t):

d/dt[(x + y) / y] = d/dt[(h + y) / h]

To find the rate at which the person's shadow is lengthening, we need to evaluate the derivatives on both sides.

The derivative of (x + y) / y with respect to time represents the rate at which the distance between the person and the base of the streetlight is changing. This will be equal to the person's walking speed, d, which is given as 4 feet per second.

The derivative of (h + y) / h represents the rate at which the height of the person's shadow is changing. We want to find this rate.

Let's calculate the derivative on the right side of the equation:

d/dt[(h + y) / h] = (1/h) * d/dt (h + y) = (1/h) * dy/dt

Now we can substitute the given values:

d = 4 feet per second
h = 15 feet

Plugging these values into our equation:

4 = (1/15) * dy/dt

To find the rate at which the person's shadow is lengthening, we can solve for dy/dt:

dy/dt = 4 * 15 = 60 feet per second

Therefore, the person's shadow is lengthening at a rate of 60 feet per second.

draw a diagram. using similar triangles, when the person is x ft from the light pole,

6/s = 15/(x+s)
6(x+s) = 15s
6x = 9s
s = 2/3 x
so
ds/dt = 2/3 dx/dt
since dx/dt = 4, ds/dt = 8/3 ft/s
regardless of the distance from the light