A person whose height is 6 feet is walking away from the base of a streetlight along a straight path at 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?
To find the rate at which the person's shadow is lengthening, we can use similar triangles. Let's assume that at a certain moment, the person is "x" feet away from the base of the streetlight, and the length of their shadow is "y" feet.
Since the person's height and the length of their shadow create similar triangles, we can set up a ratio:
Height of the person / Length of the person's shadow = Height of the streetlight / Length from the person to the streetlight.
Using the given values, we have:
6 feet / y = 15 feet / x.
Cross-multiplying, we get:
6x = 15y.
To solve for the rate at which the person's shadow is lengthening, we need to find dy/dt, the rate at which y is changing with respect to time. We know that dx/dt (the rate at which x is changing with respect to time) is 4 feet per second.
Differentiating both sides of the equation with respect to time, we get:
6(dx/dt) = 15(dy/dt).
Now we can substitute dx/dt with the given value:
6(4) = 15(dy/dt).
Simplifying the equation, we have:
24 = 15(dy/dt).
Dividing both sides by 15, we find:
dy/dt = 24/15 = 8/5 feet per second.
Therefore, the rate at which the person's shadow is lengthening is 8/5 feet per second.
To find the rate at which the person's shadow is lengthening, we need to use similar triangles. Let's consider the two triangles formed: the person's shadow triangle and the person's height triangle.
In the person's shadow triangle, we have the height of the streetlight as the object's height and the length of the person's shadow as the base. In the person's height triangle, we have the person's height as the object's height and the length of the person's shadow as the base. Both triangles have the same corresponding angles because they are similar.
Let's denote the length of the person's shadow as "s" at any given time t, and the length of the person's height as "h." The height of the streetlight remains constant at 15 feet.
From the similar triangles, we can set up the following proportion:
(h + 15) / s = h / 6
To find the rate at which the person's shadow is lengthening, we need to find ds/dt, the derivative of the length of the shadow with respect to time. To do this, we differentiate both sides of the equation with respect to time t:
[(d/dt) (h + 15)] / s = (d/dt) (h / 6)
Next, we need to differentiate the equation using the chain rule:
(dh/dt) / s - (h + 15) (ds/dt) / s^2 = (1/6) (dh/dt)
Rearranging the equation and substituting the given values:
(ds/dt) = (6 / s^2)(dh/dt) - (h + 15) (ds/dt) / s^2
We know that dh/dt represents the rate at which the person's height is changing, which is given as 4 feet per second. Now we can substitute the given values:
(ds/dt) = (6 / s^2)(4) - (6 + 15) (ds/dt) / s^2
We can simplify this equation to find ds/dt:
(ds/dt) + (21s/s^2)(ds/dt) = 24/s
Rewriting the equation:
(ds/dt) + (21/s)(ds/dt) = 24/s
(ds/dt) (1 + 21/s) = 24/s
(ds/dt) = (24/s) / (1 + 21/s)
(ds/dt) = 24/(s + 21)
Therefore, the rate at which the person's shadow is lengthening is 24/(s + 21) feet per second.
utr
draw a diagram. Let the distance from the pole be x, and the shadow's length be s. Using similar triangles, we have
s/6 = (x+s)/15
or,
5s = 2x+2s
3s = 2x
so,
ds/dt = 2/3 dx/dt
plug and chug