What is the slope of the polar curve #f(theta) = theta - sec^2theta+costhetasin^3theta # at #theta = (5pi)/6#?

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Answer 1 · 2017-03-23T17:44:39

From the reference Polar Curve Tangents :
#dy/dx=((dr)/(d""theta)sin(theta)+rcos(theta))/((dr)/(d""theta)cos(theta)-rsin(theta))#

Given: #r(theta) = theta - sec^2(theta)+cos(theta)sin^3(theta)#

Evaluate at #5pi/6#

#r(5pi/6) = 5pi/6 - sec^2(5pi/6)+cos(5pi/6)sin^3(5pi/6)#

#r(5pi/6) ~~ 1.18#

Compute #(dr)/(d""theta)#:

#(dr)/(d""theta)=1-2tan(theta)sec^2(theta)+sin^2(theta)(2cos^2(theta)-2sin^2(theta)+1)#

Evaluate at #5pi/6#

#(dr(5pi/6))/(d""theta)~~3.0#

The slope of the tangent line, m, is #dy/dx# evaluated at #5pi/6#:

#m = (3sin(5pi/6)+1.18cos(5pi/6))/(3cos(5pi/6)-1.18sin(5pi/6)#

#m ~~ -0.15#