Question

Find the slope of the tangent line to the polar curve
r = sin(6theta) at theta = pi / 12
slope = ?

Answer 1

\[r = 6 \sin \]

\[r' = 6\cos(6\theta)\]

Answer 2

next step is \[(6\cos(6\theta)\sin \theta + \sin(6\theta)\cos \theta) \div (6\cos(6\theta)\cos \theta - \sin(6\theta)\sin(\theta))\]

then plug in
\[\pi / 12 \to \theta\]

Answer 3

after plugging in I came up with
\[-(\cos(\pi/12) \over \sin(\pi/12))\]

I am stuck here

Answer 4

Solve the for the values to get constants and that is your slope of the tangent line

Answer 5

I solved for them and my answer isnt correct\[\sqrt{2 + \sqrt{3}} /2 \over \sqrt{2-\sqrt{3}/2}\]

Answer 6

What is the correct answer?

Answer 7

not sure we wont get the correct one on our hw till tomorrow after its closed it only accepts the answer if its correct

Answer 8

so we know r = sin(6theta) evaluate at r = pi/12 for the y co-ordinate. r =sin(6(pi/12)) = 1 we have co-ordinates (pi/12 , 1)

r' = 6cos(6theta)

Evaluate theta at pi/12

r'(pi/12) = 6cos(6(pi/12)) = 6cos(pi/2) = 6(0) = 0 so the slope is 0.

y - yo = m( x - xo )

y - 1 = 0(1-pi/12)

y = 1 Is that right or completely wrong as well???