Question

What is the equation of the tangent line of `r=sintheta + costheta` at `theta=pi/2`?

Answer 1

For the inward normal, `theta = pi`. For the outward normal in the opposite direction, it is `theta = 3/2pi`, obtained by adding `pi` to `theta`.

Explanation:

Here,#

`r = sin theta + cos theta`

`=sqrt 2((1/sqrt 2)cos theta+(1/sqrt 2)sin theta)`

`=sqrt 2(cos (pi/4) cos theta+sin(pi/4)sin theta)`

`=sqrt 2 cos(theta - pi/4)`

This represents the circle with center at the pole r = 0 and diameter

sqrt2. The radius is `sqrt 2/2 = 1/sqrt 2`. The initial line is along a

diameter , `theta = pi/4`.

The normal at a point on the circle is along the radius to the point.

Thus, the radial line `theta = pi/2` is outward normal to the point

under reference, `(sqrt 2, pi/2)`.

For the inward normal, the equation is

`theta =3/2pi`, for the opposite direction.