What is the normal of cos(t)+tsin(t),\at t=0 ?

The normal of cos(t)+tsin(t),\at t=0 is t=0

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normal of cos(t)+tsin(t),\at t=0

normal of

cos (t) + t sin (t), at t = 0

t = 0

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 0

Compute the slope of the perpendicular line:

m_{p} = \infty

Find the line with slope m=

\infty

and passing through

(0, 1) : t = 0

t = 0

\frac{d y}{d x} = \frac{x ^{2} + y ^{2} - x y}{x ^{2}}

\int ln (a x) d x

\int \frac{18}{( 81 x ^{2} + 1 ) ^{2}} d x

h (t) = t (1 - t^{2})

x \cdot e^{(- 4 x)}

What is the normal of cos(t)+tsin(t),\at t=0 ?
The normal of cos(t)+tsin(t),\at t=0 is t=0