What is the slope of the tangent line of #r=5theta+cos(-theta/3-(pi)/2)# at #theta=(-5pi)/6#?

1 Answer
Dec 20, 2015

slope would be #[(-1/2)(5- 1/3 cos ((5pi)/18) )+{(-25pi)/6+ sin ((5pi)/18)} ((-sqrt3)/2)]# / #[{5- 1/3 cos ((5pi)/18)}((-sqrt3)/2) -{(-25pi)/6+sin ((5pi)/18)}(-1/2)]#

Explanation:

write r= #5 theta +cos (theta/3 +pi/2)= 5theta- sin(theta/3)#

For #theta= (-5pi)/6#, #r= (-25pi)/6 + sin ((5pi)/18)#

#(dr)/(d(theta)) = 5- 1/3 cos (theta/3)#

For #theta= (-5pi)/6#

#(dr)/(d(theta)) = 5- 1/3 cos ((5pi)/18)#

#[cos (-theta)= cos theta]#

Formula for slope in polar coordinates is

#dy/dx= ((dr)/(d(theta)) sin theta +rcos theta)/((dr)/(d(theta)) cos theta -rsin theta)#

Also #sin ((-5pi)/6)= -1/2#
#cos ((-5pi)/6)= -sqrt3 /2#

slope would be #[(-1/2)(5- 1/3 cos ((5pi)/18) )+{(-25pi)/6+ sin ((5pi)/18)} ((-sqrt3)/2)]# / #[{5- 1/3 cos ((5pi)/18)}((-sqrt3)/2) -{(-25pi)/6 + sin ((5pi)/18)} (-1/2) ]#