*Notes to a video lecture on http://www.unizor.com*

__Normal to Parametric Curves__

We continue dealing with a curve on the plane that is parametrically defined by two functions - its coordinates that depend on some parameter

**{**, where both functions

*x(t);y(t)*}**and**

*x(t)***are given and differentiable.**

*y(t)*Our task is to find an equation that describes the normal to this curve at some point

**{**that corresponds to a parameter value

*x*}_{0};y_{0}**, that is**

*t=t*_{0}**and**

*x*_{0}=x(t_{0})**.**

*y*_{0}=y(t_{0})Geometrically speaking, a normal to a curve at some point is defined as a line perpendicular to a tangential line at that same point.

Here is how it looks.

Generally, a straight line that goes through point

**{**has a point-slope equation

*x*}_{0};y_{0}

*y−y*_{0}= m·(x−x_{0})So, all we have to determine in the equation for a normal that goes through point

**{**is its

*x*}_{0};y_{0}*slope*

**.**

*m*Since a normal to a curve at some point is, by definition, perpendicular to a tangential line at the same point, we can find the slope of a tangential line first and then turn the line by

**.**

*90*^{o}From the previous lecture we know that the slope of a tangential line can be calculated as

*[*

**m =**D_{t=t0}**]**

*y(t)*

*/**D*[

_{t=t0}**]**

*x(t)*If a tangential line forms angle

**with positive direction of the X-axis, and we have determined that**

*θ***(see formula for**

*tan θ = m***above), we can determine the angle**

*m***formed by a normal that is perpendicular to a tangential line and positive direction of the X-axis as follows:**

*ν*

*ν = θ−90*^{o}Now we determine the slope

**of the normal using the following simple trigonometry:**

*n*

*n = tan(ν) = tan(θ−90*

= −tan(90

= −1/tan(θ) = −1/m =

= −^{o}) == −tan(90

^{o}−θ) = −cot(θ) == −1/tan(θ) = −1/m =

= −

*D*[

_{t=t0}**]**

*x(t)*

*/**D*[

_{t=t0}**]**

*y(t)*since

**and assuming that derivatives, participating in these calculations, are not equal to zero at point**

*cot(θ)=1/tan(θ)***.**

*t=t*_{0}The equation defining a normal will then look like this:

*(y−y*_{0}) = −*D*[

_{t=t0}**]**

*x(t)*

*(x−x*_{0})/*D*[

_{t=t0}**]**

*y(t)*
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