Question

the slope of the line normal to the curve f(x） 2cos4x at x= pi/12 is​

Answer 1

Step-by-step explanation:

We have,

$y = f(x) = 2 \cos(4x)$

$\implies \frac{dy}{dx} = - 8 \sin(4x)$

$\implies \frac{dy}{dx} _{x = \frac{\pi}{12} } = - 8 \sin( \frac{\pi}{3} ) = - 4 \sqrt{3}$

At

$x= \frac{\pi}{12}, y=1$

Equation of normal:

$(y - 1). (\frac{dy}{dx} )_{x = 1} + (x - \frac{\pi}{12} ) = 0$

$(y - 1).( - 4 \sqrt{3} )+ (x - \frac{\pi}{12} ) = 0$

$\implies - 4 \sqrt{3} y + 4 \sqrt{3} + x - \frac{\pi}{12} = 0$

$\implies - 4 \sqrt{3} y + x+ 4 \sqrt{3} - \frac{\pi}{12} = 0$