Question

Equation of the line in x and y which is tangent to polar curve r=2costheta

Answer 1

Explanation:

First of all this simplifies to

r = 2 sin theta +sin 2theta

the tangent vector is the velocity vector so we differentiate wrt time as a parameter

vec r = (2 sin theta +sin 2theta) hat r

vec r' = vec v

by product rule

= d/dt (2 sin theta +sin 2theta) hat r + (2 sin theta +sin2 theta) d/dt hat r=ddt(2sinθ+sin2θ)ˆr+(2sinθ+sin2θ)ddtˆr

where d/dt hat r = d/dt ((cos theta),(sin theta)) = ((-sin theta),(cos theta)) dot theta = dot theta hat theta

implies (2 cos theta + 2 cos 2theta) \ dot theta \ hat r + (2 sin theta +sin 2theta) \ dot theta \ hat theta⇒(2cosθ+2cos2θ) .θ ˆr+(2sinθ+sin2θ) .θ ˆθ

dropping the dot theta scalar...

= (2 cos theta + 2 cos 2theta) ((cos theta),(sin theta))+ (2 sin theta +sin 2theta) ((-sin theta),(cos theta)) =(2cosθ+2cos2θ)(cosθsinθ)+(2sinθ+sin2θ)(−sinθcosθ)

= ((2 cos^2 theta + 2 cos 2 theta cos theta - 2 sin^2 theta - sin theta sin 2 theta),(2 cos theta sin theta + 2 cos 2 theta sin theta + 2 sin theta cos theta + sin 2 theta cos theta))=⎛⎜⎝2cos2θ+2cos2θcosθ−2sin2θ−sinθsin2θ2cosθsinθ+2cos2θsinθ+2sinθcosθ+sin2θcosθ⎞⎟⎠

= ((1 + 0 - 1 - 1/sqrt 2),(1 + 0 + 1+ 1/sqrt 2))

= (( - 1),(2sqrt 2+ 1))

the slope is therefore

m = -(1+ 2sqrt 2)

r (pi/4) = 1 + sqrt 2

x = r cos theta = 1/sqrt 2+ 1

y = r sin theta = 1/sqrt 2+ 1

So we have

(y - 1/sqrt 2 - 1)/(x - 1/sqrt 2 - 1) = -(1+ 2sqrt 2)

Or
y = -(1+ 2sqrt 2)(x - 1/sqrt 2 - 1) + 1 + 1/sqrt 2

that could be further simplified of course