Friday, December 16, 2016

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 90o.

From the previous lecture we know that the slope of a tangential line can be calculated as
m = Dt=t0[y(t)] /Dt=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−90o

Now we determine the slope of the normal using the following simple trigonometry:
n = tan(ν) = tan(θ−90o) =
= −tan(90o−θ) = −cot(θ) =
= −1/tan(θ) = −1/m =
= −
Dt=t0[x(t)] /Dt=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) = −Dt=t0[x(t)](x−x0)/Dt=t0[y(t)]