How do you differentiate the following parametric equation: # (3t-sin(t/3-pi/4), 2tcos(pi/3-t/4))#?

1 Answer
Nov 27, 2017

#(dy)/(dx)=(3(4cos(pi/3-t/4)-tsin(pi/3-t/4)))/(18-2cos(t/3-pi/4))#

Explanation:

In a parametric equation of the type #f(x(t),y((t))#, #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))#.

Here #x(t)=3t-sin(t/3-pi/4)# and hence #(dx)/(dt)=3-cos(t/3-pi/4)xx1/3#

and as #y(t)=2tcos(pi/3-t/4)# #(dy)/(dt)=2(cos(pi/3-t/4)-tsin(pi/3-t/4)xx(-1/4))#

Hence, #(dy)/(dx)=(2(cos(pi/3-t/4)+t/4sin(pi/3-t/4)))/(3-cos(t/3-pi/4)xx1/3)#

= #(2/4(4cos(pi/3-t/4)+tsin(pi/3-t/4)))/(1/3(9-cos(t/3-pi/4)))#

= #(3(4cos(pi/3-t/4)-tsin(pi/3-t/4)))/(18-2cos(t/3-pi/4))#