Question

What is the slope of the tangent line of `r=theta-sin((theta)/6+(2pi)/3)` at `theta=(pi)/4`?

Answer 1

The slope is the equation:

`dy/dx =(r'(theta)sin(theta) + r(theta)cos(theta))/(r'(theta)cos(theta)-r(theta)sin(theta))`

evaluated at `theta = pi/4`

Explanation:

From the reference Tangents with polar coordinates , we obtain the equation [1].

`dy/dx =(r'(theta)sin(theta) + r(theta)cos(theta))/(r'(theta)cos(theta)-r(theta)sin(theta))" [1]"`

`r(theta) = theta - sin(theta/6 + (2pi)/3)`

`r'(theta) = 1 - 1/6cos(theta/6 + (2pi)/3)`

`dy/dx =((1 - 1/6cos(theta/6 + (2pi)/3))sin(theta) + (theta - sin(theta/6 + (2pi)/3))cos(theta))/((1 - 1/6cos(theta/6 + (2pi)/3))cos(theta)-(theta - sin(theta/6 + (2pi)/3))sin(theta))`

The slope, `m = dy/dx|_(pi/4)`

`m = 0.776632`