Question

What is the slope of the tangent line of `r=5theta+cos(-theta/3-(pi)/2)` at `theta=(-pi)/6`?

Answer 1

Starting function:
`r(\theta) = 5\theta + cos(-\theta/3 - \pi/2)`

Derivative:
`r'(\theta) = (dr(\theta))/(d\theta) = 5 +1/3 sin(-\theta/3-\pi/2)`

Assessing at `\theta = -pi/6`:
`r'(-pi/6) = 5+1/3sin(-\pi/18-\pi/2)`
`r'(-pi/6) = 5+1/3sin(-(5\pi)/(9)) \approx 4.67`