The plane curve is given by the equation R(t)= ( ln sin t)i + ( ln cos t)j.

find the unit normal vector to the curve t= ¤Ð/6.

assuming you meant t=π/6

The unit tangent vector
T(t) = r'/|r'| =

cot(t) i - tan(t) j
-------------------------
tan^2(t) + cot^2(t)

The unit normal is T'/|T'| =

-(csc^2(t) i + sec^2(t) j)
----------------------------------
csc^4(t) + sec^4(t)

which at t = π/6 is

-(4i + 4/3 j)/(16 + 16/9)
= -4/3 (3i+j) * 9/156
= 1/13 (3i+j)

as always, check my math

To find the unit normal vector to a plane curve, you need to calculate the first and second derivatives of the curve with respect to the parameter t.

Step 1: Find the derivative of the curve R(t):
R'(t) = [(d/dt) (ln sin t)]i + [(d/dt) (ln cos t)]j

To find the derivatives of ln sin t and ln cos t, you can use the chain rule:
(d/dt) (ln sin t) = (1/sin t) * (d/dt) (sin t) = (1/sin t) * cos t = cos t / sin t = cot t
(d/dt) (ln cos t) = (1/cos t) * (d/dt) (cos t) = (1/cos t) * (-sin t) = -sin t / cos t = -tan t

So, R'(t) = cot t * i - tan t * j

Step 2: Find the second derivative of the curve R(t):
R''(t) = [(d/dt) (cot t)]i - [(d/dt) (tan t)]j

To find the derivatives of cot t and tan t, you can again use the chain rule:
(d/dt) (cot t) = (-csc^2 t) * (d/dt) (sin t) = -csc^2 t * cos t = -cos t / sin^2 t = -sec t / sin t
(d/dt) (tan t) = (sec^2 t) * (d/dt) (sin t) = sec^2 t * cos t = cos t / sin^2 t = 1 / sin t = csc t

So, R''(t) = (-sec t / sin t) * i - (csc t) * j

Step 3: Calculate the unit normal vector N(t):
The unit normal vector is given by N(t) = R''(t) / |R''(t)|, where |R''(t)| is the magnitude of R''(t).

First, calculate |R''(t)|:
|R''(t)| = sqrt((-sec t / sin t)^2 + (csc t)^2)
= sqrt(sec^2 t / sin^2 t + csc^2 t)
= sqrt(1 / sin^2 t + 1 / sin^2 t)
= sqrt(2 / sin^2 t)
= sqrt(2) / |sin t|

Now, calculate N(t):
N(t) = R''(t) / |R''(t)|
= [(-sec t / sin t) * i - (csc t) * j] / (sqrt(2) / |sin t|)
= [(-sec t / sin t) * i - (csc t) * j] * (|sin t| / sqrt(2))
= -sec t / sqrt(2) * i - csc t / sqrt(2) * j

Therefore, the unit normal vector to the curve at t = π/6 is approximately:
N(π/6) = -sec(π/6) / sqrt(2) * i - csc(π/6) / sqrt(2) * j
= -2 / sqrt(2) * i - 2 * √3 / sqrt(2) * j
= -√2 * i - √6 * j