What is the slope of the tangent line of #r=theta^3-thetacos(theta-(pi)/3)# at #theta=(-5pi)/3#?

Question

2251 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-15T18:21:54

The slope, #m ~~ 0.007#

#dy/dx = ((dr)/(d theta)sin(theta) + rcos(theta))/((dr)/(d theta)cos(theta) - rsin(theta) #

#r((-5pi)/3) = -(125pi^3)/27 + (5pi)/3 = (5pi)/3 - (125pi^3)/27#

#(dr)/(d theta) = 3 θ^2-sin(θ+π/6)-θ cos(θ+π/6)#

Evaluate at #(-5pi)/3#

#(dr((-5pi)/3))/(d theta) = (25pi^2)/3 - 1#

#sin((-5pi)/3) = sqrt(3)/2#

#cos((-5pi)/3) = 1/2#

#dy/dx = ((dr)/(d theta)sqrt(3) + r)/((dr)/(d theta) - rsqrt(3) #

#dy/dx = ((25pi^2 - 3)sqrt(3) + 3r)/((25pi^2 - 3) - 3rsqrt(3) #

#dy/dx = ((25pi^2 - 3)sqrt(3) + 3((5pi)/3 - (125pi^3)/27))/((25pi^2 - 3) - 3((5pi)/3 - (125pi^3)/27)sqrt(3) #

The slope, #m ~~ 0.007#