Question

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

Answer 1

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...

Answer 2

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`