A light hangs 15 ft. directly above a straight walk on which a man 6 ft. tall is walking. How fast is the end of the man's shadow travelling when he is walking away from the light at a rate of 3 miles per hour?

did you make your sketch ??

let the length of the man's shadow be x ft
let the distance he is from the lightpost be y
by ratios:
6/x = 15(x+y)
6x + 6y = 15x
6y = 9x
2y = 3x
2dy/dt = 3dx/dt
but dy/dt = 3
so 6 = 3 dx/dt
dx/dt = 2

His shadow is lengthening at 2 ft/s
but he is also moving at 3 ft/s
so d(x+y)/dt = 5 ft/s

In this kind of question we have to be careful what is asked:

at what rate is the shadow increasing ---- 2 ft/s
at what rate is the shadow moving ----- 5 ft/s

an analogy would be a man walking in a car of a moving train:
if the man is walking in the car at 3 ft/s
but the train is moving at 50 ft/s, to an observer looking in would see the man moving at 53 ft/s

Well, well, well! Looks like the light is playing some shadow games with us, huh? Alright, let's shed some light on this situation, shall we?

First, we need to find the length of the man's shadow. Since we have a right triangle formed by the man, his shadow, and the distance from the light, we can use some good ol' Pythagoras.

The hypotenuse of the triangle is the light's height, which is 15 ft. The other leg of the triangle is the length of the man's shadow, and the man's height is the remaining leg.

Using the Pythagorean theorem (a^2 + b^2 = c^2), we can solve for the length of the man's shadow (b):

b^2 = c^2 - a^2
b^2 = 15^2 - 6^2
b^2 = 225 - 36
b^2 = 189
b ≈ 13.75 ft

So, the man's shadow is approximately 13.75 ft long.

Now, to find out how fast the end of the man's shadow is moving when he's walking away from the light at 3 miles per hour, we need to use rates of change, my friend.

Let's say that x represents the distance the man has walked from the light, and y represents the length of the shadow.

Since we know that the man is walking away from the light at a rate of 3 miles per hour, the rate of change of x with respect to time (dx/dt) is 3 mph.

We want to find the rate of change of y with respect to time (dy/dt) when x = some value. In this case, we'll use x = 0 for simplicity.

Using similar triangles, we can set up the following proportion:

x/y = (man's height)/(length of shadow)
0/y = 6/13.75

Cross-multiplying, we get:
0 = 6y/13.75

Now remember, we want to find dy/dt. So, we differentiate both sides of the equation with respect to time.

d(0)/dt = d(6y/13.75)/dt
0 = (6/13.75) * dy/dt

Finally, we can solve for dy/dt:

dy/dt = 0 / (6/13.75)
dy/dt = 0

Voilà! The end of the man's shadow is not moving when he is walking away from the light at a rate of 3 miles per hour. I guess shadows have a way of sticking around, no matter how fast you walk!

To solve this problem, we can use similar triangles. Let's break it down step-by-step:

Step 1: Identify the relevant lengths and rates.
- The height of the light above the ground is 15 ft.
- The height of the man is 6 ft.
- The man is walking away from the light at a rate of 3 miles per hour.

Step 2: Set up the similar triangles.
- Draw a diagram where the light, the man, and his shadow are shown.
- Label the height of the light with "h" and the height of the man with "x."
- Label the distance between the man and the light with "d."
- Draw a line representing the man's shadow, and label the length of the shadow with "y."

Step 3: Establish the relationship between the lengths of the corresponding sides of the triangles.
- In the two similar triangles formed by the light, the man, and his shadow, we can establish the following relationship:
h/x = (d+y)/y

Step 4: Express the given information.
- The height of the light above the ground is 15 ft (h = 15 ft).
- The height of the man is 6 ft (x = 6 ft).
- The man is walking away from the light at a rate of 3 miles per hour (dx/dt = 3 mi/hr).

Step 5: Differentiate the equation to find an expression for dy/dt.
- To find dy/dt (the rate at which the end of the man's shadow is moving), we need to differentiate the equation obtained in Step 3 with respect to time (t).
(d/dt)[h/x] = (d/dt)[(d+y)/y]

Step 6: Substitute the given values into the equation.
- Substitute the given values into the equation from Step 3:
(d/dt)[15/6] = (d/dt)[(d+y)/y]

Step 7: Simplify the equation.
- Simplify the equation using the quotient rule and chain rule:
(1/6)[(dx/dt)x - (dh/dt)h] = [(dy/dt)y - (d/dt)(d+y)]/y²

Step 8: Substitute the given value for dx/dt into the equation.
- Since we know that dx/dt (the rate at which the man is walking) is 3 mi/hr, we can substitute it into the equation from Step 7:
(1/6)[(3)x - (dh/dt)(15)] = [(dy/dt)y - (d/dt)(d+y)]/y²

Step 9: Solve for dy/dt.
- Rearrange the equation to solve for dy/dt (the rate at which the end of the man's shadow is moving). Simplify if possible.

After substituting the values and simplifying the equation, the final expression for dy/dt, the rate at which the end of the man's shadow is moving, will be obtained.

To find the speed at which the end of the man's shadow is moving, we need to use similar triangles. Let's solve this step by step:

Step 1: Draw a diagram.
Imagine a vertical line representing the light hanging 15 ft above the ground. Then draw a horizontal line representing the straight walkway with the man on it. The man's height is 6 ft, so mark a point 6 ft below the light for the top of the man. Connect the top of the man to the base of the light with a diagonal line, representing the shadow.

Step 2: Determine the length of the shadow.
We have a right-angled triangle formed by the light, the man, and the end of the shadow. The vertical side is the height of the man (6 ft) and the hypotenuse is the length of the shadow. We can use the Pythagorean theorem to find the length of the shadow.

The Pythagorean theorem states that in a right-angled triangle, where a and b are the lengths of the two sides forming the right angle, and c is the length of the hypotenuse, a² + b² = c².

In this case, a = 6 ft and b = 15 ft. Let's substitute these values into the formula:
6² + b² = c² (since c is the length of the shadow)
36 + b² = c²

Since we know that the light is directly above the top of the man, the hypotenuse (the shadow) is also the distance between the light and the end of the shadow. So, we can say c = the distance between the light and the shadow, which we'll call x.

Now we have an equation: 36 + b² = x²

Step 3: Take the derivative.
We want to find the rate of change of the shadow's length with respect to time. To do this, we'll need to take the derivative of both sides of the equation with respect to time.

d/dt (36 + b²) = d/dt (x²)
0 + 2b * db/dt = 2x * dx/dt
b * db/dt = x * dx/dt

In this equation, db/dt represents the rate at which the top of the man's shadow is moving, and dx/dt represents the rate at which the man is moving along the walkway (which is given as 3 miles per hour).

Step 4: Substitute the known values into the equation.
We know that b = 6 ft and dx/dt = 3 mph. We are trying to find db/dt, the rate at which the shadow's length is changing.

Now we have: 6 * db/dt = x * 3

Step 5: Solve for db/dt.
We still need to solve for x, the distance between the light and the end of the shadow.

Recall the equation: 36 + b² = x²
Substitute b = 6 to get: 36 + 6² = x²
36 + 36 = x²
72 = x²

Taking the square root of both sides: √72 = √x²
√72 = x
x ≈ 8.485 ft

Now let's substitute this value into our equation: 6 * db/dt = 8.485 * 3

Solving for db/dt: db/dt = (8.485 * 3) / 6

db/dt ≈ 4.242 ft

Therefore, the end of the man's shadow is moving at a rate of approximately 4.242 ft per hour.