What is the tangent of r=1+cos(θ) ?

The tangent of r=1+cos(θ) is r=-sin(a_{0})θ+1+cos(a_{0})+a_{0}sin(a_{0})

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tangent of r=1+cos(θ)

tangent of

r = 1 + cos (θ)

r = - sin (a_{0}) θ + 1 + cos (a_{0}) + a_{0} sin (a_{0})

Solution steps

Compute the tangent line to the general point

θ = a_{0}

Find the tangent point:

(a_{0}, 1 + cos (a_{0}))

Find the slope of

r = 1 + cos (θ) : \frac{d r}{d θ} = - sin (θ)

Compute the slope of

r = 1 + cos (θ)

θ = a_{0} : m = - sin (a_{0})

Find the line with slope m=

- sin (a_{0})

and passing through

(a_{0}, 1 + cos (a_{0})) : r = - sin (a_{0}) θ + 1 + cos (a_{0}) + a_{0} sin (a_{0})

r = - sin (a_{0}) θ + 1 + cos (a_{0}) + a_{0} sin (a_{0})

g (x) = (2 x^{2} + x + 9)^{- 3}

\frac{d}{d y} (e^{3 x} (3 x^{2} y + 2 x y + y^{3}))

\int \frac{cos ( 3 t ) + 3}{sin ( 3 t )} d t

\int \frac{2 ln ( x )}{x ^{3}} d x

\int 2 x + 6 d x

What is the tangent of r=1+cos(θ) ?
The tangent of r=1+cos(θ) is r=-sin(a_{0})θ+1+cos(a_{0})+a_{0}sin(a_{0})