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

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Answer 1 · 2016-09-04T04:28:09

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#.

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.