What is the slope of the polar curve #f(theta) = theta^2 - thetasintheta # at #theta = (11pi)/8#?

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Answer 1 · 2017-02-21T15:35:43

11.2206#

#f(theta) = theta^2 -theta sin theta#

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

at #theta = (11 pi)/8#,

#f'(theta) = 2 *(11 pi)/8 -[(11 pi)/8 cos ((11 pi)/8)+sin( (11 pi)/8)]#

#f'(theta) = (11 pi)/4 -[(11 pi)/8(-0.3827)+(-0.9239)]#

# f'(theta)= 8.6429 - (-1.6538 - 0.9239)#

#f'(theta) =8.6429 - ( -2.5777)#

#f'(theta)=11.2206#