Find an arc length parametrization of the circle in the plane z = 16 with radius 5 and center (5, 5, 16).

To find an arc length parametrization of the circle, we need to find a parameterization of the form r(t) = (x(t), y(t), z(t)), where t is the parameter.

Given a circle with center (a, b, c) and radius r, we can parameterize the circle using polar coordinates. Let's try to do that.

The equation of a circle in the plane z = 16 with center (5, 5, 16) and radius 5 is given by:

(x - 5)^2 + (y - 5)^2 = 5^2

Expanding this equation, we get:

(x^2 - 10x + 25) + (y^2 - 10y + 25) = 25

Simplifying further, we get:

x^2 - 10x + y^2 - 10y + 25 = 0

Now, let's rewrite this equation in terms of polar coordinates. We know that x = a + r*cos(theta) and y = b + r*sin(theta), where r is the radius and theta is the angle.

Substituting these values into the equation, we get:

(a + r*cos(theta))^2 - 10(a + r*cos(theta)) + (b + r*sin(theta))^2 - 10(b + r*sin(theta)) + 25 = 0

Expanding and simplifying, we have:

r^2*cos(theta)^2 - 10r*cos(theta) + a^2 - 10a + r^2*sin(theta)^2 - 10r*sin(theta) + b^2 - 10b + 25 = 0

Since we are given that a = 5, b = 5, and c = 16, let's substitute these values into the equation:

r^2*cos(theta)^2 - 10r*cos(theta) + 25 - 50 + r^2*sin(theta)^2 - 10r*sin(theta) + 25 - 50 + 25 = 0

Simplifying further, we have:

r^2*cos(theta)^2 + r^2*sin(theta)^2 - 10r*cos(theta) - 10r*sin(theta) + 25 = 0

Since cos(theta)^2 + sin(theta)^2 = 1, we can simplify the equation to:

r^2 - 10r*cos(theta) - 10r*sin(theta) + 25 = 0

Rearranging the terms, we have:

r^2 - 10r*cos(theta) - 10r*sin(theta) = -25

Now, let's solve for r in terms of theta. We have a quadratic equation in terms of r. Using the quadratic formula, we get:

r = (10*cos(theta) + sqrt(100*cos(theta)^2 + 100*sin(theta)^2 + 25)) / 2

Simplifying further, we have:

r = 5*cos(theta) + sqrt(25 + 25) = 5*cos(theta) + 5

Therefore, the arc length parametrization of the circle is:

r(t) = (5*cos(t) + 5, 5*sin(t) + 5, 16), where t is the parameter.