A street light is hung 18 ft. above street level. A 6-foot tall man standing directly under the light walks away at a rate of 3 ft/sec. How fast is the tip of the man's shadow moving?

I know I would've to set up a proportion.

18 / 6 = x + y / y

x = distance of man from light
y = length of shadow
x + y = tip of shadow

You mean

18 / 6 = (x + y) / y
we know dx/dt, we need dy/dt
then the tip moves at dx/dt + dy/dt

18 y = 6x + 6 y
12 y = 6 x
12 dy/dt = 6 dx/dt
dy/dt = .5 dx/dt
so dy/dy = 3/2 = 1.5
and the sum
3+1.5 = 4.5 ft/sec

Okay, I see what I did wrong. I did dy/dt = 6 instead of dy/dt = .5. Thanks a lot

x = distance of man from base of light

y = length of shadow

y/(y+x) = 6/18
Solve for y
18y = 6(y + x)
18y = 6y + 6x
12y = 6x
y = 6/12 x = 1/2 x

Find dy/dx of 1/2 x
dy/dx = 1/2

Find derivative with respect to t
dy/dt = 1/2 dx/dt

x is increasing 3 ft/sec
dx/dt = 3 ft/sec
dy/dt = 1/2 dx/dt
dy/dt = 1/2 (3)
dy/dt = 3/2 = 1.5 ft/sec

Shadow moving at the rate of 1.5 ft/sec

A street light is at the top of a 13 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 7 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 35 ft from the base of the pole?

Well, it seems like the man is really trying to outrun his own shadow. It's like a never-ending game of tag. But let's see if we can figure out how fast this game is going.

Now, we know that the man's height is 6 feet, and the light is hanging 18 feet above the ground. So, if we draw a triangle with the light, the man, and his shadow, we can see that the shadow is the hypotenuse of this triangle.

Now, let's assume that at some point in time, the man is x feet away from the light. At that very moment, the length of the shadow would be y feet.

Using similar triangles, we can set up the proportion:

18 / 6 = (x + y) / y

Simplifying this equation, we get:

3 = (x + y) / y

Now, we know that the man is walking away from the light at a rate of 3 ft/sec. This means that the rate at which he's changing his distance from the light is 3 ft/sec. Let's call this dx/dt.

So, now we have:

dx/dt = 3 ft/sec

Now, we can differentiate our original equation with respect to time to find the rate at which the shadow is changing. Let's call this dy/dt:

3 = (x + y) / y

Differentiating both sides:

0 = (dy/dt * y - (x + y) * (dy/dt)) / y^2

Simplifying this equation, we get:

0 = 1 - (x + y) * (dy/dt) / y^2

If we rearrange this equation to solve for dy/dt, we get:

(dy/dt) = y^2 / (x + y)

Now we can plug in the values we know to find the rate at which the tip of the man's shadow is moving.

Great job on setting up the proportion! To solve for the rate at which the tip of the man's shadow is moving, we need to differentiate the equation with respect to time and solve for the rate of change.

Using the chain rule, we differentiate both sides of the equation:

d(18/6) / dt = d(x + y) / dt / d(y)/dt

Simplifying the equation gives:

0 = dx / dt + dy / dt / dy / dt

Now we can rearrange the equation to solve for dy / dt, which represents the rate at which the tip of the man's shadow is moving:

dy / dt = - (dx / dt)

This tells us that the rate at which the tip of the man's shadow is moving is equal to the opposite of the rate at which the man is moving away from the light.

In this case, the rate at which the man is walking away from the light is given as 3 ft/sec. Therefore, the rate at which the tip of the man's shadow is moving is -3 ft/sec.

Note that the negative sign indicates that the tip of the man's shadow is moving in the opposite direction to the man's motion.