Question

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

Answer 1

slope would be `[(-1/2)(5- 1/3 cos ((5pi)/18) )+{(-25pi)/6+ sin ((5pi)/18)} ((-sqrt3)/2)]` / `[{5- 1/3 cos ((5pi)/18)}((-sqrt3)/2) -{(-25pi)/6+sin ((5pi)/18)}(-1/2)]`

Explanation:

write r= `5 theta +cos (theta/3 +pi/2)= 5theta- sin(theta/3)`

For `theta= (-5pi)/6`, `r= (-25pi)/6 + sin ((5pi)/18)`

`(dr)/(d(theta)) = 5- 1/3 cos (theta/3)`

For `theta= (-5pi)/6`

`(dr)/(d(theta)) = 5- 1/3 cos ((5pi)/18)`

`[cos (-theta)= cos theta]`

Formula for slope in polar coordinates is

`dy/dx= ((dr)/(d(theta)) sin theta +rcos theta)/((dr)/(d(theta)) cos theta -rsin theta)`

Also `sin ((-5pi)/6)= -1/2`
`cos ((-5pi)/6)= -sqrt3 /2`

slope would be `[(-1/2)(5- 1/3 cos ((5pi)/18) )+{(-25pi)/6+ sin ((5pi)/18)} ((-sqrt3)/2)]` / `[{5- 1/3 cos ((5pi)/18)}((-sqrt3)/2) -{(-25pi)/6 + sin ((5pi)/18)} (-1/2) ]`