Final answer:
The length of the curve defined by the vector function r(t) = 3t, 3 cos(t), 3 sin(t) for -5 ≤ t ≤ 5 is 60√(2).
Step-by-step explanation:
To find the length of the curve defined by the vector function r(t) = 3t, 3 cos(t), 3 sin(t) for -5 ≤ t ≤ 5, we can use the arc length formula. The arc length of a curve defined by a vector function r(t) = x(t), y(t), z(t) is given by the integral of the magnitude of the derivative of r with respect to t over the interval [a, b]. In this case, the derivative is r'(t) = 3, -3 sin(t), 3 cos(t), and the magnitude of the derivative is |r'(t)| = √(3^2 + (-3sin(t))^2 + (3cos(t))^2) = 3√(1 + sin^2(t) + cos^2(t)).
The arc length of the curve is then given by the integral of 3√(1 + sin^2(t) + cos^2(t)) with respect to t over the interval [-5, 5]. This integral can be simplified using trigonometric identities to 6√(2)∫dt = 6√(2)(t + C) evaluated from -5 to 5, where C is the constant of integration. Plugging in the limits of integration, we get 6√(2)((5 + C) - (-5 + C)) = 60√(2).