Final answer:
To find the unit tangent vector t(t) at the given value of the parameter t, differentiate the position vector r(t) and find its magnitude. Then, divide the derivative vector by its magnitude at the given value of t to get the unit tangent vector.
Step-by-step explanation:
To find the unit tangent vector t(t) at the given value of the parameter t, we first need to find the derivative of the position vector r(t). The position vector is given as r(t) = cos(t)i + 5tj + 3sin(4t)k. So, the derivative is r'(t) = -sin(t)i + 5j + 12cos(4t)k.
Next, we find the magnitude of the derivative vector at the given value of t. The magnitude is given by |r'(t)| = sqrt((-sin(t))^2 + (5)^2 + (12cos(4t))^2). Evaluating this at t = 0, we get a magnitude of |r'(0)| = sqrt(1^2 + 5^2 + 12^2) = sqrt(170).
Finally, we find the unit tangent vector by dividing the derivative vector by its magnitude at the given value of t. So, t(0) = (r'(0))/|r'(0)| = (-sin(0)/sqrt(170))i + (5/sqrt(170))j + (12cos(0)/sqrt(170))k = -1/sqrt(170)i + 5/sqrt(170)j + 0k.