What is the normal of cos(t)+tsin(t),\at t=pi ?

The normal of cos(t)+tsin(t),\at t=pi is y= 1/pi t-2

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normal of cos(t)+tsin(t),\at t=pi

normal of

cos (t) + t sin (t), at t = π

y = \frac{1}{π} t - 2

Solution steps

Find the tangent point:

(π, - 1)

Find the slope of

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

Compute the slope of

y = cos (t) + t sin (t)

t = π : m = - π

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{π}

Find the line with slope m=

\frac{1}{π}

and passing through

(π, - 1) : y = \frac{1}{π} t - 2

y = \frac{1}{π} t - 2

y = x, y = \frac{1}{4} x + 1, x = 0

\int x y cos (x y) d x

\frac{d y}{d x} = (y - y^{2})

\int e^{\frac{2}{t}} d t

\int sin^{3} (5 x) cos (5 x) d x

What is the normal of cos(t)+tsin(t),\at t=pi ?
The normal of cos(t)+tsin(t),\at t=pi is y= 1/pi t-2