What is the equation of the tangent line to the polar curve # f(theta)=2thetasin(theta)+thetacot^2(4theta) # at #theta = pi/3#?

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Answer 1 · 2018-08-07T13:07:07

Equation of tangent is #y= -3.05 x +5.36 #

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

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

#f(pi/3) ~~ 2.16 :.# Point is #(pi/3,2.16)#

#f(theta) = 2 theta sin theta + theta cot^2 (4 theta); #

#f^'(theta) = 2 sin theta +2 theta cos theta+cot^2 (4 theta)-8 cot (4theta)csc^2(4theta) #

#f^'(pi/3) = 2*sin( (pi)/3) +2*pi/3*cos((pi)/3)+cot^2 (4*(pi)/3))-8 cot (4*(pi)/3)csc^2(4* (pi)/3)#

#f^'(pi/3)~~ -3.05 :. # slope at point #(1.05,2.16)# is #-3.05#

Equation of tangent is #y-2.16= -3.05(x-1.05)# or

#y= -3.05 x+3.2+2.16 or y= -3.05 x +5.36 # [Ans]