Parametric Equations Calculus

Introduction for parametric equations calculus :

In mathematics, parametric equation is a method of defining a relation using parameters. A simple kinematical examples are  when one use a time parameter to determine the position, velocity, and other information about a body in motion. Abstractly, a Parametric Equations define a relation as a set of equation. It is therefore somewhat more accurates defined as a parametric representation. It is part of regular parametric equation representation.Calculus includes that differential calculus and integral calculus is used.(Source.Wikipedia)



Examples for Parametric Equations Calculus:

Example 1 : Prove that the sum of the intercept on the co-ordinate axes of any tangent to the curve x = d cos^4c,  y = d sin4c,   0 = ? =p /2   is equal to d.

Solution :

Take any point ‘C’ as (d cos^4c, d sin^4c, )

Now  `dx/dd ` = – 4d cos^3c sin c ;

And  `dx/dd` = 4d sin3c cos c

? `dy/dx` = –sin^2c/cos2c

Slope of the tangent at ‘c’ is = –sin^2c/cos2c

Equation of the tangent at ‘c’ is (y - d sin4c) =- sin^2c/cos2c(x - d cos^4c)

or x sin^2 c + y cos2 c = d sin^2 c cos2 c

?`x/d` cos2c+`y/d ` sin^2c= 1

sum of the intercepts = x cos2 c + y sin^2 c = d

Example 2 : Find the equations of the tangent and normal at B =p/2 to the curve x = b (B + sin B), y = b (1 + cos B).
Solution : We have

`dx/dB` = b (1 + cosB) = 2b cos2 B

2`dy/dB` = – b sin B = – 2b sin`B/2 ` cos`B/2`

Then dy/dx =`dy/dA` / `dx/dA` =  – tan`B/2`

Slope m = `dy/dx` [B = p/2] = – tanp/4 = –1

Also for B =p/2 , the point on the curve is{ b p/2 + b, b .}

Hence the equation of the tangent at B =p/2 is

y – b = (–1) [x – a(p/2 + 1)]

x + y =`1/2` b p + 2b or x + y –`1/2` b p – 2b = 0

Equation of the normal at this point is

y – b = (1) x – b(p/2+ 1)

x – y –`1/2` b p = 0

Practice Problem for Calculus Parametric Equations:

Find the equations of the tangents and normal to the ellipse x = c cosS, y = d sin S at the point S =p/4

Answer: (cx – dy) `sqrt(2)` – (c2 – d2) = 0.