What is the slope of the polar curve #f(theta) = theta + cot^2theta+thetasin^3theta # at #theta = (3pi)/8#?

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Answer 1 · 2016-10-30T20:28:16

#m = 0.2#

From the reference, Tangents with Polar Coordinates, we write the following equation:

#dy/dx = ((dr)/(dθ)sin(θ) + rcos(θ))/((dr)/(dθ)cos(θ) - rsin(θ))#

#r = f((3pi)/4) = (3pi)/4 + cot^2((3pi)/4) + (3pi)/4sin^3((3pi)/4)#

#r = f((3pi)/4) = (3pi)/4 + 1 + (3pi)/4(sqrt(2)/2)^3#

#r = f((3pi)/4) = (3pi)/4 + 1 + (3pisqrt(2))/16 ~~ 4.2#

I used WolframAlpha to compute #(dr)/(dθ)#:

#(dr)/(dθ) = sin^3(θ)+3 θ sin^2(θ) cos(θ)-2 cot(θ) csc^2(θ)+1#

#(dr)/(dθ) ~~ 2.8#

#m = dy/dx]_((3pi)/4) = (2.8sin((3pi)/4) + 4.2cos((3pi)/4))/(2.8cos((3pi)/4) - 4.2sin((3pi)/4))#

#m = 0.2#