Unizor - Derivatives - Normal to Parametric Curve

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{x(t);y(t)}, where both functions x(t) and y(t) are given and differentiable.

Our task is to find an equation that describes the normal to this curve at some point {x₀;y₀} that corresponds to a parameter value t=t₀, that is x₀=x(t₀) and y₀=y(t₀).

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 {x₀;y₀} has a point-slope equation
y−y₀ = m·(x−x₀)
So, all we have to determine in the equation for a normal that goes through point {x₀;y₀} is its 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 angleθ with positive direction of the X-axis, and we have determined that tan θ = m (see formula form above), we can determine the angle ν formed by a normal that is perpendicular to a tangential line and positive direction of the X-axis as follows:
ν = θ−90^o

Now we determine the slope n of the normal using the following simple trigonometry:
n = tan(ν) = tan(θ−90^o) =
= −tan(90^o−θ) = −cot(θ) =
= −1/tan(θ) = −1/m =
= −D_t=t0[x(t)] /D_t=t0[y(t)]
since cot(θ)=1/tan(θ) and assuming that derivatives, participating in these calculations, are not equal to zero at point t=t₀.

The equation defining a normal will then look like this:
(y−y₀) = −D_t=t0[x(t)](x−x₀)/D_t=t0[y(t)]