What is the equation of the tangent line of #r=cos(-3theta-(2pi)/3) # at #theta=(-5pi)/3#?

1 Answer
Feb 8, 2018

#f'((-5pi)/3)=(3sqrt3)/2#

Explanation:

Now, remember the chain rule and the fact that #d/dx(cosx)=-sinx#

The chain rule states that:
If #f(x)=g(h(x))#, then #f'(x)=g'(h(x))*h'(x)#

Since #r=cos(-3theta-(2pi)/3)#, we have:

#f(x)=cos(-3x-(2pi)/3)#

=>#f'(x)=-sin(-3x-(2pi)/3)*d/dx(-3x-2pi/3)#

=>#f'(x)=-sin(-3x-(2pi)/3)*-3#

=>#f'(x)=3sin(-3x-(2pi)/3)#

=>#f'((-5pi)/3)=3sin(-3*(-5pi)/3-(2pi)/3)#

=>#f'((-5pi)/3)=3sin(-1*(-5pi)/1-(2pi)/3)#

=>#f'((-5pi)/3)=3sin(-(-5pi)-(2pi)/3)#

=>#f'((-5pi)/3)=3sin(5pi-(2pi)/3)#

=>#f'((-5pi)/3)=3sin((13pi)/3)#

=>#f'((-5pi)/3)=3sin(4 pi/3)#

Using our special angles, we know that #sin(pi/3)=sqrt3/2#

We have:

=>#f'((-5pi)/3)=3*sqrt3/2#

=>#f'((-5pi)/3)=(3sqrt3)/2#

This is the answer!