The normal line of a point on a function is the line perpendicular to the tangent line at that point. This involves three basic steps: finding the slope of the tangent line, taking the opposite reciprocal of that slope (giving us the slope of the normal line), and finding the y-intercept.
Step 1: Find the Slope of the Tangent Line
All we do here is find the derivative of sinx and evaulate it at x=pi/6 - because the slope of the tangent line at some point is the derivative at that point.
f(x)=sinx
f'(x)=cosx->the derivative of sine is cosine
Step 2: Opposite Reciprocal
Taking the opposite reciprocal of the tangent line slopes gives us the normal line slope. The reciprocal of sqrt(3)/2 is 2/sqrt(3). We'll have to rationalize this denominator - having a square root down there is a no-no:
2/sqrt(3)*sqrt(3)/sqrt(3)=(2sqrt(3))/3
To find the opposite of this result, we simply make it negative:
(2sqrt(3))/3*-1=-(2sqrt(3))/3
Step 3: Finding the Equation of the Line
We have the slope, but we need a y-intercept. Remember, a normal line has the form y=mx+b, where x and y are points on the line, m is the slope, and b is the y-intercept. We know the normal line passes through the tangent line - they're perpendicular, meaning they intersect at a 90^o angle. And since we're finding the normal line at x=pi/6, we use that for our point:
f(x)=sinx
f(pi/6)=sin(pi/6)=1/2
We have now identified the point (pi/6,1/2). We will now use this point to find the equation of the line:
y=mx+b
1/2=-(2sqrt(3))/3*pi/6+b
1/2=-(pisqrt(3))/9+b
1/2+(pisqrt(3))/9=b->b=(9+2pisqrt(3))/18
Bingo! The y-intercept is (9+2pisqrt(3))/18. That means the equation of the normal line is y=-(2sqrt(3))/3x+(9+2pisqrt(3))/18. We could combine this into one whole fraction, but we don't have to, and I like avoiding unnecessary work :). If you prefer hard numbers, we can approximate this to y=-1.155x+1.105.
