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

Question

Ezekiel Alexander · Accepted Answer

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

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!

Luke Alvarez · Answer

undefined To find the equation of the tangent line of $ r = \cos(-3\theta - \frac{2\pi}{3}) $ at $ \theta = -\frac{5\pi}{3} $, we first need to find the polar coordinates $(r, \theta)$ at the given point. Then, we can use the polar coordinate form of the equation of a tangent line, which is $ r = x\cos(\theta) + y\sin(\theta) $, where $ (x, y) $ is a point on the tangent line.

By substituting $ \theta = -\frac{5\pi}{3} $ into $ r = \cos(-3\theta - \frac{2\pi}{3}) $, we find the value of $ r $ at the given point. Then, we can convert the polar coordinates to Cartesian coordinates to get $ (x, y) $. Finally, we substitute these values into the equation of the tangent line to obtain the equation.