What is the equation of the tangent line of #r=cos(theta+(5pi)/4) * sin(theta-(17pi)/12)# at #theta=(-5pi)/12#?

1 Answer
Mar 13, 2017

Equation of tangent is #y+(3+sqrt3)/(8sqrt2)=-(38/37+10/37sqrt3)(x-((3-sqrt3)/(8sqrt2)))#

Explanation:

In polar coordinates

#(dy)/(dx)=(sinthetadr+rcosthetad theta)/(costhetadr-rsinthetad theta)#

= #(sintheta(dr)/(d theta)+rcostheta)/(costheta(dr)/(d theta)-rsintheta)#

As #r=cos(theta+(5pi)/4)sin(theta-(17pi)/12)#

#(dr)/(d theta)=-sin(theta+(5pi)/4)sin(theta-(17pi)/12)+cos(theta+(5pi)/4)cos(theta-(17pi)/12)#

and when #theta=(-5pi)/12# and

#r=cos((-5pi)/12+(5pi)/4)sin((-5pi)/12-(17pi)/12)#

= #-cos((5pi)/6)sin(pi/6)=sqrt3/2xx1/2=sqrt3/4#

#(dr)/(d theta)=-sin((-5pi)/12+(5pi)/4)sin((-5pi)/12-(17pi)/12)+cos((-5pi)/12+(5pi)/4)cos((-5pi)/12-(17pi)/12)#

= #sin((10pi)/12)sin((22pi)/12)+cos((10pi)/12)cos((22pi)/12)#

= #cos((22pi)/12-(10pi)/12)=cospi=-1#

and #(dy)/(dx)=(-sin((5pi)/12)+sqrt3/4cos((5pi)/12))/(cos((5pi)/12)+sqrt3/4sin((5pi)/12))#

= #(-(sqrt3+1)/(2sqrt2)+sqrt3/4((sqrt3-1)/(2sqrt2)))/((sqrt3-1)/(2sqrt2)+sqrt3/4((sqrt3+1)/(2sqrt2))#

= #(-4sqrt3-4+3-sqrt3)/(4sqrt3-4+3=sqrt3)#

= #(-5sqrt3-1)/(5sqrt3-1)=-38/37-10/37sqrt3#

and equation of tangent is #y-rsintheta=(dy)/(dx)(x-rcostheta)# or

#y-sqrt3/4sin((-5pi)/12)=-(38/37+10/37sqrt3)(x-sqrt3/4cos((-5pi)/12))#

or #y+sqrt3/4((sqrt3+1)/(2sqrt2))=-(38/37+10/37sqrt3)(x-sqrt3/4((sqrt3-1)/(2sqrt2)))#

or #y+(3+sqrt3)/(8sqrt2)=-(38/37+10/37sqrt3)(x-((3-sqrt3)/(8sqrt2)))#