What is the equation of the tangent line of #r=-5cos(-theta-(pi)/4) + 2sin(theta-(pi)/12)# at #theta=(-5pi)/12#?

2 Answers
Oct 6, 2017

See explanation

Explanation:

Recall that in polar coordinates, you have #x = r*cosθ, y=r*sinθ#. The equation for the slope of tangent line to the curve at any given point is still #dy/dx#, despite being in polar coordinates. Since y and x are functions of r and θ, this means...

#dy/dx = (dy/(dθ))/(dx/(dθ)) = ((dr)/(dθ)sinθ + rcosθ)/((dr)/(dθ)cosθ - rsinθ)#.

Since #r(θ) = -5cos(-θ-pi/4) + 2sin(θ-pi/12), (dr)/(dθ) = -5sin(-θ-pi/4) + 2 cos(θ-pi/12)#...

Then...

Oct 6, 2017

The polar equation is invalid when #theta = -(5pi)/12#

Explanation:

We have a polar equation:

# r = -5cos(-theta-pi/4) + 2sin(theta-pi/12) #

When #theta = -(5pi)/12# we have:

# r = -5cos(pi/6) + 2sin(-pi/2) #
# \ \ = -(5sqrt(3))/2-2 #
# \ \ ~~-6 #

Hence, The polar equation is invalid when #theta = -(5pi)/12#