What is the equation of the tangent line of #r=5theta + 2sin(4theta+(2pi)/3) # at #theta=(2pi)/3#?

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Answer 1 · 2017-01-27T10:43:43

#rsin(theta^o-20.3^o)=8.74sin(99.7^o)#

I recall that, I have given here the formula, for the polar equation io

the tangent at #P( a, alpha )#, with slope

#m =tanpsi=(r'sin theta + r cos theta)/(r'costheta-rsintheta)#,

evaluated at #(a, alpha)#, as

#rsin(theta-psi)=asin(alpha-psi)#

Here,

#r =5theta+2sin(4theta+2/3pi)#

#alpha=2/3pi=2.0944 radian =120^o#

#a = 10/3pi -sqrt3=8.74#

#r' =5=8cos(4theta+2/3pi) =1,# at P

#m=0.3400 and psi=20.3^o#.

Now, the equation to the tangent at #P(8.74, 120^o)# is

#rsin(theta^o-20.3^o)=8.74sin(120^o-20.3^o)=8.74sin(99.7^o)#