Question

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

Answer 1

`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!