What is the normal of y=10cos(x),(pi/3 ,5) ?

The normal of y=10cos(x),(pi/3 ,5) is y= 1/(5sqrt(3))x+5-pi/(15sqrt(3))

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normal of y=10cos(x),(pi/3 ,5)

normal of

y = 10 cos (x), (\frac{π}{3}, 5)

y = \frac{1}{5 3} x + 5 - \frac{π}{15 3}

Solution steps

Find the slope of

y = 10 cos (x) : \frac{d y}{d x} = - 10 sin (x)

Compute the slope of

y = 10 cos (x)

(\frac{π}{3}, 5) : m = - 53

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{5 3}

Find the line with slope m=

\frac{1}{5 3}

and passing through

(\frac{π}{3}, 5) : y = \frac{1}{5 3} x + 5 - \frac{π}{15 3}

y = \frac{1}{5 3} x + 5 - \frac{π}{15 3}

\frac{d y}{d x} = \frac{( 5 x sec ( \frac{y}{x} ) + y )}{x}

\frac{\partial}{\partial x} (x sin (x + y^{2}))

\frac{d}{d x} (\frac{5 x}{x - 1})

\int_{1}^{6} t ln (t) d t

\frac{d}{d x} (e^{-})

What is the normal of y=10cos(x),(pi/3 ,5) ?
The normal of y=10cos(x),(pi/3 ,5) is y= 1/(5sqrt(3))x+5-pi/(15sqrt(3))