在θ=(5pi)/ 8时,极坐标曲线的斜率f(θ)=θ=θ^3θ+θtasin^3θ是多少?

在θ=(5pi)/ 8时,极坐标曲线的斜率f(θ)=θ=θ^3θ+θtasin^3θ是多少?
Anonim

回答:

#DY / DX = -0.54#

说明:

对于极地功能 #F(THETA)#, #DY / DX =(F '(THETA)sintheta + F(THETA)costheta)/(F'(THETA)costheta-F(THETA)sintheta)#

#F(THETA)=θ-秒^ 3theta + thetasin ^ 3theta#

#F'(THETA)= 1-3(秒^的2θ)(d / DX sectheta) - 罪^ 3theta + 3thetasin ^的2θ(d / DX sintheta)#

#F'(THETA)= 1-3sec ^ 3thetatantheta-SIN ^ 3theta + 3thetasin ^ 2thetacostheta#

#F'((5pi)/ 3)= 1-3sec ^ 3((5pi)/ 3)TAN((5pi)/ 3)-sin ^ 3((5pi)/ 3)3((5pi)/ 3 )罪^ 2((5pi)/ 3)cos((5pi)/ 3)-9.98 ~~#

#F((5pi)/ 3)=((5pi)/ 3) - 仲^ 3((5pi)/ 3)+((5pi)/ 3)罪^ 3((5pi)/ 3)~~ -6.16 #

#DY / DX =( - 9.98sin((5pi)/ 3)-6.16cos((5pi)/ 3))/( - 9.98cos((5pi)/ 3)+ 6.16sin((5pi)/ 3)) = -0.54#