What is the normal of sin(t)-tcos(t),\at t=pi ?

The normal of sin(t)-tcos(t),\at t=pi is t=pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

normal of sin(t)-tcos(t),\at t=pi

normal of

sin (t) - t cos (t), at t = π

t = π

Solution steps

Find the tangent point:

(π, π)

Find the slope of

y = sin (t) - t cos (t) : \frac{d y}{d t} = t sin (t)

Compute the slope of

y = sin (t) - t cos (t)

t = π : m = 0

Compute the slope of the perpendicular line:

m_{p} = \infty

Find the line with slope m=

\infty

and passing through

(π, π) : t = π

t = π

\int_{b}^{2 b} x^{9} d x

\int e^{14 x} d x

\frac{d}{d x} ((x^{2} + 3) (x^{2} - 4 x))

x \to - 3 + lim (\frac{1}{6 + 2 x})

3 sin (- x)

What is the normal of sin(t)-tcos(t),\at t=pi ?
The normal of sin(t)-tcos(t),\at t=pi is t=pi