How do you differentiate the following parametric equation: # (t-tsin(t/3), -tcos(pi/2-t/3))#?

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Answer 1 · 2017-12-23T09:22:56

#(dy)/(dx)=-cos(t/3)/(-tcos(t/3)-3sin(t/3)+3)#

We know that:
#x=t-tsin(t/3)#
#y=-cos(pi/2-t/3)#

#(dy)/(dx)=(dy)/(dt)-:(dx)/(dt)#

#(dy)/(dt)=d/(dt)[-cos(pi/2-t/3)]=sin(pi/2-t/3)*-1/3=-sin(pi/2-t/3)/3#

#(dx)/(dt)=d/(dt)[t-tsin(t/3)]=d/(dt)[t]+d/(dt)[-tsin(t/3)]=1+d/(dt)[-t]sin(t/3)-td/(dt)[sin(t/3)]=1-sin(t/3)-(tcos(t/3))/3#

#(dy)/(dx)=(-sin(pi/2-t/3)/3)/(1-sin(t/3)-(tcos(t/3))/3)#

#=-cos(t/3)/(3((-tcos(t/3))/3-sin(t/3)+1))#

#=-cos(t/3)/(-tcos(t/3)-3sin(t/3)+3)#