What is the equation of the normal line of f(x)= xsinx-cos^2x at x = pi/8?

1 Answer
Jun 22, 2018

y =-0.69x-0.43 [To 2D]

Explanation:

f(x) = xsinx-cos^2x

Apply product rule, chain rule and standard derivatives.

f'(x) = xcosx+sinx- 2cosx*(-sinx)

= xcosx+sinx+2sinxcosx

= xcosx+sinx+sin(2x)

The slope of f(x) at x=pi/8 is f'(pi/8)

f'(pi/8) = pi/8*cos(pi/8) +sin(pi/8) + sin(pi/4)

approx 1.542597 (Calculator)

The slope of a tangent (m_1) xx slope of a normal (m_2) at any point on a curve is given by: m_1*m_2 =-1

:. Slope of the normal to f(x) at x=pi/8 approx -1/1.542597

approx -0.688422

The equation of a straight line of slope m passing through the point (x_1,y_1) is given by:

(y-y_1) = m(x-x_1)

In this example, our normal will have the equation:

y-f(pi/8) approx -0.688422 (x-pi/8)

y- (-0.703274) approx -0.688422 (x- 0.392699) (Calculator)

y+0703274 approx -0.688422x+0.270343

y approx -0.688422x-0.432931

y =-0.69x-0.43 [To 2D]

We can see f(x) and the normal at x=pi/8 on the graphic below.

enter image source here