What is the slope of #r=tantheta-theta# at #theta=pi/8#?

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What is the slope of #r=tantheta-theta# at #theta=pi/8#?

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Answer 1 · 2018-03-19T18:59:33

Derivative with Polar Coordinates is # (\partial x) / (\partial y) = ((\partial r) / (\partial theta) sin(theta)+ rcos(theta))/((\partial r) / (\partial theta) cos(theta)- rsin(theta)#

Explanation:

#(\partial r) / (\partial theta) = 1/cos^2(theta)+1#
#r(theta = pi/8)=0.0 22#
# (\partial x) / (\partial y) = (tan(theta)/cos(theta)+sin(theta)+rcos(theta))/(1/cos(theta)+cos(theta)-rsin(theta))#
# (\partial x) / (\partial y)(theta=pi/8) =1.57 #

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What is the slope of #r=tantheta-theta# at #theta=pi/8#?

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