Equation of the normal lineWithout using a calculator write an equation of the normal line to the curve `xcosy=1` at `(2,pi/3)`
We'll start by outlining our plan for this problem. When finding a line perpendicular to a curve at a certain point, we are finding a line that is perpendicular to the tangent line of the curve at that point. Thus, we need to first find the derivative of the function at that point. Then, we will find the perpendicular using techniques from algebra.
Let's first find the derivative.
`1 = xcosy`
Here we'll need to use implicit differentiation, where we find a function of `x` and `y` and set it to zero. We can do this with a constant on one sides; however, that may confuse some, so we'll just say "set it to 0." Here it is easy! We will simply subtract 1 from both sides:
`0 = xcosy - 1`
Now we use the derivative operator with respect to `x` on both sides:
`D_x(0) = D_x(xcosy - 1)`
Differentiating using the multiplication rule and chain rule gives us the following result:
`0 = cosy + x*(dy)/(dx) (-siny)`
Now, we simply solve for `dy/dx`:
`-x*(dy)/(dx)(-siny) = cosy`
dividing both sides by `xsiny` gives us our result for the derivative:
`dy/dx = x/tany`
Keep in mind `siny/cosy = tany`, so we have found `1/tany` in the above relation.
Now that we know the derivative, we can calculate the slope of the tangent line at the given point:
`dy/dx = 2/(tan(pi/3))`
Using our values for tangent, we find the following result:
`dy/dx = 2/sqrt3`
Now that we know the slope of the tangent line, we can find the slope of the perpendicular line to the tangent line. To do this, we take the negative reciprocal to give our slope, which we will term `m_p`:
`m_p = -1/(2/sqrt3) = -sqrt3/2`
Now, we need to find the rest of the equation for the perpendicular line. Let's use the following form for an equation of a line:
`y = m_p x + b`
Now, we simply use the given point to calculate `b` in our equation:
`pi/3 = -sqrt3/2*2 + b`
To solve, we simply simplify the multiplied term and subtract the result from both sides:
`pi/3 = -sqrt3 + b`
`pi/3 + sqrt3 = b`
Therefore, our final equation is as follows:
`y = m_px+b`
`y = sqrt3/2x + pi/3 + sqrt3`
Confirming with a graph:
Those are perpendicular, if you zoom correctly, so we have the correct function! Also, note that the blue graph (original function) does not display the entire graph, only the part relevant to our discussion.