Find the points on the given curve where the tangent line is horizontal or vertical.
$r = e^{θ}$

Question

Find the points on the given curve where the tangent line is horizontal or vertical.  r=eθ

encolatgehu · Accepted Answer

To find the points where the tangents are horizontal/vertical, we need to find dy/dx. Using the chain rule fore differentiation, we can write

\frac{d y}{d x} = \frac{d y}{d θ} \cdot \frac{d θ}{d x} = \frac{\frac{d y}{d} θ}{\frac{d x}{d} θ}

Substitute

y = r \sin θ = e^{θ} \sin θ

and

x = r \cos θ = e^{θ}

, to get

\frac{d y}{d x} = \frac{d \frac{e^{θ} \sin θ}{d} θ}{d \frac{e^{θ} \cos θ}{d} θ}

Use product rule for differentiation

\frac{d y}{d x} = \frac{d {(e^{θ})}^{'} \sin θ + e^{θ} {(\sin θ)}^{'}}{{(e^{θ})}^{'} \cos θ + e^{θ} {(\cos θ)}^{'}}

\frac{d y}{d x} = \frac{e^{θ} \sin θ + e^{θ} \cos θ}{e^{θ} \cos θ - e^{θ} \sin θ}

Divide both the numerator and the denominator by

e^{θ}

\frac{d y}{d x} = \frac{\sin θ + \cos θ}{\cos θ - \sin θ}

The tangents will be horizontal when the numerator is 0

\sin θ + \cos θ = 0

Subtract

\cos θ

from both sides

\sin θ = - \cos θ

Divide both sides by

\cos θ

\frac{\sin θ}{\cos θ} = - 1

In the right-hand side, we will rewrite -1 as

\tan (- \frac{π}{4})

t a m θ = \tan (- \frac{π}{4})

Remember that: the general solution of the equation

\tan x = \tan α

is

x = α + n π

Therefore, the tangents are horizontal when

θ = - \frac{π}{4} + n π

The tangents will be vertical when the denomiantor is 0

\sin θ = \cos θ

Divides both sides by

\cos θ

\frac{\sin θ}{\cos θ} = 1

In the right-hand side, we will rewrite 1 as

\tan (\frac{π}{4})

\tan θ = \tan (\frac{π}{4})

Therefore, the tangents are vertical when

Find the points on the given curve where the tangent