What is the slope of the polar curve #f(theta) = -4theta - thetacos^2theta # at #theta = (15pi)/8#?

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Answer 1 · 2018-07-05T12:59:28

Slope of the polar curve at # theta = ((15 pi)/8 )# is # -9.02#

#f(theta) = -4 theta - theta cos^2 theta#

# theta = (15 pi)/8 =337.5^0#

#f^'(theta) = -4 - (-theta* 2 cos theta sin theta+cos^2 theta) #

#f^'(theta) = -4 +theta* 2 cos theta sin theta- cos^2 theta #

#f^'((15 pi)/8)= -4 +theta* sin 2 theta- cos^2 theta #

#f^'((15 pi)/8)= -4 +(15 pi)/8* sin ((15 pi)/4)- cos^2 ((15 pi)/8) #

#f^'((15 pi)/8)= -9.02(2 dp)#

Slope of the polar curve at # theta = ((15 pi)/8 )# is # -9.02# [Ans]