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

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 $\frac{\mathrm{dy}}{\mathrm{dx}}$, 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 $\theta = - \frac{5 \pi}{12}$

#### Explanation:

We have a polar equation:

$r = - 5 \cos \left(- \theta - \frac{\pi}{4}\right) + 2 \sin \left(\theta - \frac{\pi}{12}\right)$

When $\theta = - \frac{5 \pi}{12}$ we have:

$r = - 5 \cos \left(\frac{\pi}{6}\right) + 2 \sin \left(- \frac{\pi}{2}\right)$
$\setminus \setminus = - \frac{5 \sqrt{3}}{2} - 2$
$\setminus \setminus \approx - 6$

Hence, The polar equation is invalid when $\theta = - \frac{5 \pi}{12}$