Final answer:
The slope of the tangent line at t = π/6 is -√3, and the equation of the tangent line is y = -√3x + 3√3/2 + 3.
Step-by-step explanation:
To determine the slope of the tangent line, we need to find the derivative of the given parametric equations. Given x = sin(t) and y = 6cos(t), we can find dy/dx as follows:
dy/dt = -6sin(t)
dx/dt = cos(t)
Now, let's find dy/dx by dividing dy/dt by dx/dt:
dy/dx = (-6sin(t))/cos(t)
At t = π/6, cos(π/6) = √3/2 and sin(π/6) = 1/2. Therefore, dy/dx at t = π/6 is (-(6*(1/2)))/(√3/2) = -3/√3 = -√3.
Now that we have the slope of the tangent line at t = π/6, let's find the equation of the tangent line. We can use the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Since the point of tangency is (sin(π/6), 6cos(π/6)) = (√3/2, 3), the equation of the tangent line is:
y - 3 = -√3(x - √3/2)
Simplifying, we get:
y = -√3x + 3√3/2 + 3