Final answer:
The unit vectors parallel to the tangent line at the point (π/6, 4) on the curve y = 8 sin(x) are (1/7, 4√3/7) and (-1/7, -4√3/7), calculated by finding the derivative, evaluating the slope at the point, and normalizing the resulting vectors.
Step-by-step explanation:
To find the unit vectors that are parallel to the tangent line to the curve y = 8 sin(x) at the point (π/6, 4), we first need to calculate the derivative of the curve to determine the slope of the tangent line at the point.
The derivative of y with respect to x is y' = 8 cos(x). Plugging in the value x = π/6 into the derivative yields y'(π/6) = 8 cos(π/6) = 8 * (√3/2) = 4√3, which is the slope of the tangent line at the given point.
The slope of the tangent is equal to the vector's y-component divided by its x-component, implying that for a unit vector with an x-component of 1, the y-component would be the slope. Thus, the tangent vector is (1, 4√3). To make this a unit vector, we need to divide it by its magnitude. The magnitude is √(1² + (4√3)²) = √(1 + 48) = √49 = 7. Thus, the unit vector is (1/7, 4√3/7).
However, we need to consider both directions along the tangent, so the opposite direction is also a unit vector parallel to the tangent. The resulting vectors are (1/7, 4√3/7) and (-1/7, -4√3/7).