What is the equation of the line normal to $f(x)= x^2sin^2(2x)$ at $x=pi/6$ ?

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Answer 1 · 2016-09-29T18:45:14

$y - pi^2/48 = -36/(sqrt3pi^2 + 9pi)(x - pi/6)$

Let's begin by finding the y coordinate corresponding to the given x coordinate:

$y = ((pi)/6)^2sin^2(2(pi)/6)$

$y = (pi^2/36)(3/4) = pi^2/48$

The slope, n, of the normal line is:

$n = -1/m$ where m is the slope of the tangent line.

To obtain the slope of the tangent line, we must compute the first derivative:

$f'(x) = 2xSin(2 x) (2xCos(2 x) + Sin(2 x))$

Evaluate at $x = pi/6$ :

$m = f'(pi/6) = 2(pi/6)Sin(2 pi/6) (2(pi/6)Cos(2 pi/6) + Sin(2 pi/6))$

$m = 2(pi/6)(sqrt3/2) ((pi/6)+ sqrt3/2)$

$m = sqrt3(pi/6) ((pi/6)+ sqrt3/2)$

$m = sqrt3(pi/6)^2+ pi/4$

$m = (sqrt3pi^2 + 9pi)/36$

$n = -36/(sqrt3pi^2 + 9pi)$

Using the point-slope form of the equation of a line:

$y - y_1 = n(x - x_1)$

$y - pi^2/48 = -36/(sqrt3pi^2 + 9pi)(x - pi/6)$

What is the equation of the line normal to f(x)= x^2sin^2(2x) f(x)= x^2sin^2(2x) at x=pi/6x=pi/6?