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What is the equation of the normal line of #f(x)=1/sinx+cscx# at #x=pi/6#?

Answer 1

Ezra Adams

#y = 1/(4sqrt(3))x - (pi -96sqrt(3))/(24sqrt(3))#

The function can be written as #f(x) = cscx + cscx = 2cscx#.

Start by finding the y-coordinate at the point of tangency.

#f(pi/6) = 2csc(pi/6) = 2/sin(pi/6) = 2/(1/2) = 4#

Differentiate.

#f(x) = 2/sinx#

#f'(x) = (0 * sinx - 2 * cosx)/(sinx)^2#

#f'(x) = (-2cosx)/sin^2x#

#f'(x) = -2cotxcscx#

You can find the slope of the tangent by inputting the point #x =a# into the derivative.

#f'(pi/6) = -2cot(pi/6)csc(pi/6)#

#f'(pi/6) = -2/(tan(pi/6)sin(pi/6))#

#f'(pi/6) = -2/(1/sqrt(3) * 1/2)#

#f'(pi/6) = -2/(1/(2sqrt(3))#

#f'(pi/6) = -4sqrt(3)#

The normal line is always perpendicular to the tangent. Therefore, its slope is given by the equation #m_"normal" = -1/m_"tangent"#.

#m_"normal" = -1/(-4sqrt(3))#

#m_"normal" = 1/(4sqrt(3))#

The equation of the normal is therefore:

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

#y - 4 = 1/(4sqrt(3))(x - pi/6)#

#y - 4= 1/(4sqrt(3))x - pi/(24sqrt(3))#

#y = 1/(4sqrt(3))x - (pi -96sqrt(3))/(24sqrt(3))#

Hopefully this helps!

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Answer 2

Scarlett Allison

The equation of the normal line of f(x)=1/sinx+cscx at x=pi/6 is y = -√3x + 2√3.

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