What is the equation of the tangent line of #r=2cos(theta-pi/2) + sin(2theta-pi)# at #theta=pi/4#?

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Answer 1 · 2017-02-16T09:35:32

#rsin(theta-61.325^o)=-(sqrt2-1)sin(16.325^o)#. The Cartesian
form is #0.877x-0.480y-0.116=0#. See Socratic depiction.

Use #rsin(theta-psi)=asin(alpha-psi)#, for the equation to the

tangent at #P(a, alpha)#, where the slope of the tangent

#m=tanpsi=(r'sintheta+rcostheta)/(r'costheta-rsintheta)#.

Here,

#r = 2sintheta-sin2theta#

At #theta = pi/4=alpha, r = a =sqrt2-1#.

So, the point of contact P of the tangent is #(sqrt2-1, 45^o)#.

#r'=2costheta-2cos2theta=sqrt2#, at P.

The slope of the tangent at P is

#m=tanpsi=((sqrt2)(1/sqrt2)+(sqrt2-1)(1/sqrt2))/((sqrt2)(1/sqrt2)-(sqrt2-1)(1/sqrt2))#

#=2sqrt2-1#

#psi=arctan(2sqrt2-1)=61.325^o#.

So, the equation to the tangent is

#rsin(theta-61.325^o)=-(sqrt2-1)sin(16.325^o)#

Expanding and using #(x, y)=r(costheta, sintheta)#, the Cartesian

form is

#0.877x-0.480y-0.116=0#
.

graph{(sqrt(x^2+y^2)(x^2+y^2-2y)+2xy)(0.877x-0.48y-0.116)((x-.293)^2+(y-.293)^2-.009)=0 [-5, 5, -2.5, 2.5]}

The Cartesian form is used for the Socratic graph.