What is the equation of the line that is normal to #f(x)=-sinx +tanx # at # x=-pi/3#?

Question

What is the equation of the line that is normal to #f(x)=-sinx +tanx # at #  x=-pi/3#?

Christopher Allers · Accepted Answer

#y+sqrt3/2=-2/7(x+pi/3)#

First, find the point the normal line will intercept:

#f(-pi/3)=-sin(-pi/3)+tan(-pi/3)=sqrt3/2-sqrt3=ul(-sqrt3/2#

The normal line passes through the point #(-pi/3,-sqrt3/2)#.

To find the slope of the normal line, we must first find the slope of the tangent line. Since the normal line and tangent line are perpendicular, their slopes will be opposite reciprocals.

To find the slope of the tangent line, find the value of the derivative at #x=-pi/3#.

The derivative of the function is

#f'(x)=-cosx+sec^2x#

The slope of the tangent line is

#f'(-pi/3)=-cos(-pi/3)+sec^2(-pi/3)=-1/2+2^2=ul(7/2#

Since the slopes of the tangent and normal lines are opposite reciprocals, the slope of the normal line is #-2/7#.

The point of #(-pi/3,-sqrt3/2)# and slope of #-2/7# can be related as an equation in point-slope form:

#y+sqrt3/2=-2/7(x+pi/3)#

Graphed are the original function and its normal line:

graph{(y+sinx-tanx)(y+sqrt3/2+2/7(x+pi/3))=0 [-4.858, 2.94, -2.89, 1.007]}

Cameron Alexander · Answer

undefined The equation of the line that is normal to f(x)=-sinx +tanx at x=-pi/3 is y = -√3x - 2√3.