# What is the slope of the tangent line of r=theta-cos(4theta-(3pi)/4) at theta=(-2pi)/3?

Jul 5, 2017

The slope $m \approx = - 4.53$

#### Explanation:

We need the equation

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr}}{d \theta} \sin \left(\theta\right) + r \cos \left(\theta\right)}{\frac{\mathrm{dr}}{d \theta} \cos \left(\theta\right) - r \sin \left(\theta\right)} \text{ [1]}$

from the reference Tangents with Polar Coordinates

I tried graphing r=theta-cos(4theta-(3pi)/4); -2pi< theta <=0 but the Desmos.com/calculator would not graph it.

When I tried to graph r=theta-cos(4theta-(3pi)/4); 0<= theta <2pi the calculator plotted a graph but the point evaluated ${\theta}_{1} = - \frac{2 \pi}{3}$ did not lie on the graph.

The only consistent graph occurred, when I converted ${\theta}_{1} \to \frac{4 \pi}{3}$ (the positive equivalent of $- \frac{2 \pi}{3}$).

Here is the graph:

Compute $\frac{\mathrm{dr}}{d \theta}$:

$\frac{\mathrm{dr}}{d \theta} = 1 + 4 \sin \left(4 \theta - \frac{3 \pi}{4}\right)$

Substitute $r \mathmr{and} \frac{\mathrm{dr}}{d \theta}$ into equation [1]:

dy/dx=((1 +4sin(4theta-(3pi)/4))sin(theta)+(theta-cos(4theta-(3pi)/4))cos(theta))/(1 +4sin(4theta-(3pi)/4))cos(theta)-(theta-cos(4theta-(3pi)/4))sin(theta))" [2]"

The slope, m, is equation [2] evaluated at $\theta = \frac{4 \pi}{3}$:

m=((1 +4sin(4((4pi)/3)-(3pi)/4))sin((4pi)/3)+((4pi)/3-cos(4((4pi)/3)-(3pi)/4))cos((4pi)/3))/(1 +4sin(4((4pi)/3)-(3pi)/4))cos((4pi)/3)-((4pi)/3-cos(4((4pi)/3)-(3pi)/4))sin((4pi)/3))" [3]"

$m \approx = - 4.53$

I offer, as proof, a graph of the tangent line: