What is the slope of the polar curve #r(theta) = theta^2 - sec(theta)+cos(theta)*sin^3(theta) # at #theta = (2pi)/3#?

1 Answer
Dec 30, 2015

#frac{64pi^2 - 192pisqrt{3} - 27sqrt{3} + 1152}{64pi^2sqrt{3} + 192pi - 81} ~~ 0.42824#

Explanation:

First convert the Polar equation to Parametric form.

#y = rsin(theta)#
#= theta^2sin(theta) - tan(theta) + cos(theta)*sin^4(theta)#

#x = rcos(theta)#
#= theta^2cos(theta) - 1 + cos^2(theta)*sin^3(theta)#

To find the slope, we need to find #frac{dy}{dx}#, which we can get by using the Chain Rule.

Since, from the Chain Rule, #frac{dy}{d theta} = frac{dy}{dx}*frac{dx}{d theta}#,

#frac{dy}{dx} = frac{ frac{dy}{d theta} }{ frac{dx}{d theta} }#.

We find #frac{dy}{d theta}# and #frac{dx}[d theta}# separately.

#frac{dy}{d theta} = frac{d}{d theta}(theta^2sin(theta) - tan(theta) + cos(theta)*sin^4(theta))#

#= [theta^2cos(theta) + sin(theta)(2theta)] - sec^2(theta) + [cos(theta)*(4sin^3(theta)cos(theta)) + sin^4(theta)(-sin(theta))]#

#= theta^2cos(theta) + 2thetasin(theta) - sec^2(theta) + 4cos^2(theta)*sin^3(theta) - sin^5(theta)#

#= theta^2cos(theta) + 2thetasin(theta) - sec^2(theta) + 4(1-sin^2(theta))*sin^3(theta) - sin^5(theta)#

#= theta^2cos(theta) + 2thetasin(theta) - sec^2(theta) + 4sin^3(theta) - 4sin^5(theta) - sin^5(theta)#

#= theta^2cos(theta) + 2thetasin(theta) - sec^2(theta) + 4sin^3(theta) - 5sin^5(theta)#

#frac{dy}{d theta}|_{theta = (2pi)/3} = ((2pi)/3)^2cos((2pi)/3) + 2((2pi)/3)sin((2pi)/3) - sec^2((2pi)/3) + 4sin^3((2pi)/3) - 5sin^5((2pi)/3)#

#= - frac{2pi^2}{9} + frac{2pisqrt{3}}{3} + frac{3sqrt{3}}{32} - 4#

#frac{dx}{d theta} = frac{d}{d theta}(theta^2cos(theta) - 1 + cos^2(theta)*sin^3(theta))#

#= [theta^2(-sin(theta)) + cos(theta)(2theta)] - 0 + [cos^2(theta)(3sin^2(theta)cos(theta)) + sin^3(theta)(2cos(theta)(-sin(theta)))]#

#= -theta^2sin(theta) + 2thetacos(theta) + 3sin^2(theta)cos^3(theta) - 2sin^4(theta)cos(theta)#

#frac{dx}{d theta}|_{theta = (2pi)/3} = -((2pi)/3)^2sin((2pi)/3) + 2((2pi)/3)cos((2pi)/3) + 3sin^2((2pi)/3)cos^3((2pi)/3) - 2sin^4((2pi)/3)cos((2pi)/3)#

#= - frac{2pi^2sqrt{3}}{9} - frac{2pi}{3} + 9/32#

Now, to calculate #frac{dy}{dx}#.

#frac{dy}{dx}_{theta = (2pi)/3} = frac{- frac{2pi^2}{9} + frac{2pisqrt{3}}{3} + frac{3sqrt{3}}{32} - 4}{- frac{2pi^2sqrt{3}}{9} - frac{2pi}{3} + 9/32}#

#= frac{64pi^2 - 192pisqrt{3} - 27sqrt{3} + 1152}{64pi^2sqrt{3} + 192pi - 81}#

#~~ 0.42824#