Sure, to solve this problem we need to first develop a few key vectors.
Given the vector
r(t) = (3t*sin(t) + 3cos(t))i + 3j + (3t*cos(t) - 3sin(t))k,
Let's first calculate the first derivative r'(t):
r'(t) = 3t*cos(t)i + 0j -3t*sin(t)k.
Following that, let's determine the second derivative - r''(t):
r''(t) = (-3t*sin(t) + 3cos(t))i + 0j - (3t*cos(t) + 3sin(t))k.
To find the curvature, we first need to find the magnitude of the first derivative, which is:
|r'(t)| = sqrt[9(t*sin(t))^2 + 0^2 + (-3t*sin(t))^2]
= sqrt[9(t*sin(t))^2 + 9(t*cos(t))^2]
Curvature of a curve is given by |r'(t) ✕ r''(t)| / |r'(t)|^3, where ✕ denotes the cross product.
Calculating cross product of r'(t) and r''(t) we get:
r'(t) ✕ r''(t) = 3t*(3t*sin(t) - 3cos(t))sin(t)i - 0j + 3t*(3t*cos(t) + 3sin(t))cos(t)k.
Taking the magnitude of the cross product is:
|r'(t) ✕ r''(t)| = Abs(3t*(3t*sin(t) - 3cos(t))*sin(t) + 3t*(3t*cos(t) + 3sin(t))*cos(t))
Now we can substitute |r'(t)| and |r'(t) ✕ r''(t)| into the formula for curvature, getting:
curvature = Abs(3t*(3t*sin(t) - 3cos(t))*sin(t) + 3t*(3t*cos(t) + 3sin(t))*cos(t))/(9(t*sin(t))^2 + 9(t*cos(t))^2)^(3/2)
So the curvature of the given space curve r(t) is **Abs(3t*(3t*sin(t) - 3cos(t))*sin(t) + 3t*(3t*cos(t) + 3sin(t))*cos(t))/(9(t*sin(t))^2 + 9(t*cos(t))^2)^(3/2)**.