a man 6ft tall walks at the rate of 5ft/sec toward a streetlight that is 16ft above the ground. At what rate is the length of his shadow changing when he is 10ft from the base of the light?

If the shadow is length y when the man is x feet from the pole, then using similar triangles,

y/6 = (x+y)/16
y = 3x/5

dy/dt = 3/5 dx/dt = 3/5 (-5) = -3

so, the shadow is shrinking at 3 ft/sec regardless of the distance from the pole.

Well, let's shed some light on this problem (pun intended)! So we have a 6ft tall man walking towards a streetlight. The length of his shadow is changing, and we want to find out the rate at which it is changing when he is 10ft from the base of the light.

Now, we can treat this as a similar triangles problem. We have two similar triangles here: the man's shadow triangle and the man's actual triangle. The height of the streetlight (16ft) corresponds to the height of the man's shadow, and the distance from the base of the streetlight to the man (10ft) corresponds to the length of the man's shadow.

The ratio of the lengths of corresponding sides in similar triangles is always equal. So we can set up a proportion:

Length of the man's shadow / Distance from the base of the streetlight = Height of the man's shadow / Height of the streetlight

Let's call the length of the man's shadow "x". We want to find the rate at which x is changing when the man is 10ft from the base of the light. So we need to find dx/dt.

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

(d/dt) (Length of the man's shadow) / Distance from the base of the streetlight = (d/dt) (Height of the man's shadow) / Height of the streetlight

Since the man's height remains constant at 6ft, (d/dt) (Height of the man's shadow) = 0 (no change in height).

Now, let's focus on the left side of the equation. The man's shadow length (x) is changing with respect to time, so (d/dt) (Length of the man's shadow) = dx/dt.

Substituting the values we know, we have:

dx/dt / 10ft = 0 / 16ft

Don't worry, I'm not trying to divide by zero! The height of the man's shadow is constant, so the rate of change is zero.

Therefore, the rate at which the length of the man's shadow is changing when he is 10ft from the base of the light is 0 ft/sec. It looks like his shadow will stay the same length. Not much of a shadow show, eh?

To solve this problem, we can use similar triangles and the concept of rates.

Let's consider the man, the streetlight, and his shadow. We have a right triangle formed between the man, his shadow, and the streetlight.

Let's assign variables to the different lengths:
- Height of the man: h = 6 ft
- Height of the streetlight: s = 16 ft
- Distance between the man and the base of the streetlight: x = 10 ft

We want to find the rate at which the length of his shadow is changing with respect to time. Let's call this rate ds/dt.

We can use similar triangles to relate the lengths of the sides:

h / s = (h + ds) / (s + x)

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

(dh/dt * (s + x) - ds/dt * h) / (s + x)^2 = dh/dt

Rearranging the equation:

ds/dt = (dh/dt * (s + x))/(s - h)

Now, let's substitute the given values into the equation:

dh/dt = -5 ft/sec (since the man is walking towards the streetlight)
h = 6 ft
s = 16 ft
x = 10 ft

ds/dt = (-5 * (16 + 10))/(16 - 6)
ds/dt = (-5 * 26)/10
ds/dt = -13 ft/sec

Therefore, the rate at which the length of his shadow is changing when he is 10 ft from the base of the light is -13 ft/sec.

To find the rate at which the length of his shadow is changing, we can use similar triangles and related rates.

Let's define the variables:
h = height of the man (6 ft)
x = distance of the man from the base of the light (10 ft)
y = length of the man's shadow
z = height of the man's shadow

We want to find the rate at which y is changing when x = 10 ft.

Using similar triangles, we can set up the following proportion:
h/y = (h+z)/y.

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

d(h/y)/dt = d((h+z)/y)/dt.

To get the desired rate, we need to find the derivative dy/dt with respect to time.

Differentiating the left side:
d(h/y)/dt = d(6/y)/dt.

Before differentiating the right side, we need to relate z and x.

Since we know that x and z are changing with respect to time, we have:
x^2 + z^2 = y^2.

Differentiating both sides with respect to time (t), we get:
2x(dx/dt) + 2z(dz/dt) = 2y(dy/dt).

Now we can substitute h = 6, x = 10, and dz/dt = 0 (since the height of the light is constant), into the equation derived from the similar triangles.

6/y = (6+0)/y.

Simplifying the equation, we find:
6/y = 6/y.

Next, we differentiate both sides of the equation with respect to time (t).

d(6/y)/dt = d(6/y)/dt.

Now, we substitute the known values to solve for dy/dt.

-6/y^2(dy/dt) = -6/y^2.

Canceling out the common factors, we have:
dy/dt = dy/dt.

Since both sides of the equation are the same, dy/dt cancels out, giving us no specific rate at this point.

However, we can substitute x = 10 into the equation derived from the similar triangles:
2(10)(dx/dt) + 2z(dz/dt) = 2y(dy/dt).

Since we know dx/dt = 5 ft/sec, x = 10 ft, and dy/dt is what we're trying to find, we can solve for dz/dt.

2(10)(5) + 2(16)(0) = 2y(dy/dt).

100 = 2y(dy/dt).

Simplifying further:
dy/dt = 100/(2y).

Now, we can find y by using the Pythagorean theorem:
x^2 + z^2 = y^2,
10^2 + 16^2 = y^2,
100 + 256 = y^2,
356 = y^2,
y ≈ √356 ≈ 18.87 ft.

Finally, we can substitute y = 18.87 ft into the equation to find dy/dt:
dy/dt = 100/(2 * 18.87).

Calculating the value, we find:
dy/dt ≈ 2.65 ft/sec.

Therefore, the length of his shadow is changing at a rate of approximately 2.65 ft/sec when he is 10 ft from the base of the light.