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

__Differentiation__

of Parametric Curves

of Parametric Curves

Let's assume that we deal with some curve on the plane that is defined not as a graph of certain function that looks like

**(where**

*y=f(x)***is abscissa and**

*x***- ordinate), but**

*y**parametrically*, where both coordinates are defined as functions of some parameter

**.**

*t*An obvious example is a point moving on a plane, and its position

**{**depends on a time parameter, so it can be described as

*x;y*}**{**, where both functions

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

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

*y(t)*So, one independent parameter

**describes both coordinates through these two function.**

*t*Our task is to find an equation that describes the tangential line 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 tangential line to a sufficiently

*smooth*curve at some point is a limit of a secant line that intersects our curve at point where a tangential line should be and another point close to it, when that other point is getting closer and closer to the point of tangency.

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 tangential line that goes through point

**{**is its

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

**.**

*m*Since our tangential line is a limit of a secant, we can assume that the

*slope*of a tangential line is the limit of a

*slope*of a secant as the other point of secant's intersection with our curve is getting infinitesimally close to point

**{**.

*x*}_{0};y_{0}Consider a point of tangency

**{**and give an increment to parameter

*x*}_{0}=x(t_{0});y_{0}=y(t_{0})**from its value**

*t***to value**

*t*_{0}**Δ**

*t*_{0}+**.**

*t*The new point on a curve that corresponds to an incremented value of parameter

**will be**

*t***{**Δ

*x*_{1}=x(t_{0}+**Δ**

*t);y*_{1}=y(t_{0}+

*t)*}A secant that intersects our curve at points

**{**and

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

*x*}_{1};y_{1}*slope*equal to

*m = (y*_{1}−y_{0})/(x_{1}−x_{0})which can be transformed into

**Δ**

*m =***Δ**

*y/*

*x*where Δ

**Δ**

*y=y(t*_{0}+

*t)−y(t*_{0})and Δ

**Δ**

*x=x(t*_{0}+

*t)−x(t*_{0})Obviously, we want to express the limit of this expression for

*slope*

**as Δ**

*m***in terms of derivatives**

*t→0**D*[

_{t}*] and*

**x(t)***D*[

_{t}*].*

**y(t)**For this we can transform it into

**(Δ**

*m =***Δ**

*y/***)**

*t***(Δ**

*/***Δ**

*x/***)**

*t*from which follows that for Δ

*t→0*

*m→m*_{0}=*D*[

_{t=t0}**]**

*y(t)*

*/**D*[

_{t=t0}**]**

*x(t)*where derivatives are taken at point

**.**

*t=t*_{0}*Example*

A unit circle with a center at the origin of coordinates can be described parametrically with an angle

**from the positive direction of the X-axis to a radius to a point on a circle being a parameter. Let's use the radian measure of this angle.**

*θ*Any point on a circle with coordinates

**{**can be described through functions:

*x;y*}**,**

*x=cos(θ)***.**

*y=sin(θ)*Let's choose a point that corresponds to a parameter value

**and determine the equation of a tangential line at this point.**

*θ=π/4*First of all, determine the coordinates of a point of tangency:

*x=cos(π/4)=√2/2*

*y=sin(π/4)=√2/2*Now the

*slope*should be equal to a ratio of derivatives of functions

**and**

*y(θ)***at point with parameter**

*x(θ)***:**

*θ=π/4**D*[

_{θ=π/4}**]=**

*y(θ)**D*[

_{θ=π/4}**]=**

*sin(θ)*=

*cos(π/4) = √2/2**D*[

_{θ=π/4}**]=**

*x(θ)**D*[

_{θ=π/4}**]=**

*cos(θ)*=

*−sin(π/4) = −√2/2*The

*slope*

**of our tangential line is a ratio of the two values above:**

*m***)**

*m = (√2/2*

*/(−√2/2) = −1*So, the equation of our tangential line at point defined by parameter

**in point-slope form is**

*θ=π/4*

*y−√2/2 = −1·(x−√2/2)*or, in a standard form,

*y = −x + √2*
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