Find the equation of the line tangent to the parametric curve define by

x=cos2t and y=sin3t and t=pi/12
plz i don't no where to start from show working

first, find the slope of the tangent line at any point:

dy/dx = (dy/dt)/(dx/dt)
Yes, these really are fractions!
dy/dx = (3cos3t)/(-2sin2t)
Now, at t=pi/2, the point on the curve is (0,-1), and the slope is 0

So, our tangent is the line

y+1 = 0(x-0)
or
y = -1

See the curve at

http://www.wolframalpha.com/input/?i=parametric+plot+x%3Dcos2t+and+y%3Dsin3t

To find the equation of the line tangent to the parametric curve, we need to find the slope of the tangent line at the given point and then use point-slope form to write the equation.

Step 1: Find the derivative of both x and y with respect to t.
dx/dt = -2sin(2t)
dy/dt = 3cos(3t)

Step 2: Evaluate the derivatives at t = π/12.
dx/dt = -2sin(2(π/12)) = -2sin(π/6) = -1
dy/dt = 3cos(3(π/12)) = 3cos(π/4) = 3/sqrt(2) = 3√2/2

Step 3: Calculate the slope of the tangent line.
slope = dy/dx = (dy/dt)/(dx/dt) = (3√2/2) / (-1) = -3√2/2

Step 4: Use the formula for point-slope form to find the equation of the tangent line.
y - y1 = m(x - x1), where (x1, y1) is the given point on the curve.
The given point is (x, y) = (cos(2(π/12)), sin(3(π/12))) = (cos(π/6), sin(π/4)) = (√3/2, √2/2).

Using this point and the slope, the equation of the tangent line is:
y - (√2/2) = (-3√2/2)(x - √3/2)

Simplifying the equation gives:
y - √2/2 = (-3√2/2)x + 3/2
y = (-3√2/2)x + (√2 + 3)/2

So, the equation of the tangent line to the parametric curve at t = π/12 is y = (-3√2/2)x + (√2 + 3)/2.