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Exploring Functions in Advanced Mathematics

Parametric Functions

Definition

Definition of Parametric Functions

A function relationship determined by parametric equations {x=x(t)y=y(t)\begin{cases} x = x(t) \\ y = y(t) \end{cases} is called a parametric function.

SymbolTypePronunciation/ExplanationMeaning in This Document
x(t)x(t)Mathematical symbolx of tFunction of parameter t for x
y(t)y(t)Mathematical symboly of tFunction of parameter t for y
ttMathematical symboltParameter variable
DDMathematical symbolDParameter domain, range of parameter t
Mathematical language

The rigorous definition of a parametric function is: for each tt in the parameter domain DD, there exists a uniquely determined (x,y)=(x(t),y(t))(x, y) = (x(t), y(t)), and both x=x(t)x = x(t) and y=y(t)y = y(t) are functions of tt.

Parametric functions describe originally complex curves using two simple functional relationships by introducing parameter t, making them easier to study and compute.

Characteristics

Parametric functions have the following characteristics:

  • Parameter representation: Both independent and dependent variables are expressed through parameter t
  • Parameter elimination possible: Explicit functions can be obtained through parameter elimination
  • Geometric meaning: Geometrically represents curves
  • Directionality: The direction of change of parameter t determines the direction of the curve

Common Parametric Equations

Parametric Equations of a Circle

Parametric equations of a circle with radius r: {x=rcos⁡ty=rsin⁡t,t∈[0,2π]\begin{cases} x = r\cos t \\ y = r\sin t \end{cases}, \quad t \in [0, 2\pi]

Parametric Equations of an Ellipse

Parametric equations of an ellipse with semi-major axis a and semi-minor axis b: {x=acos⁡ty=bsin⁡t,t∈[0,2π]\begin{cases} x = a\cos t \\ y = b\sin t \end{cases}, \quad t \in [0, 2\pi]

Parametric Equations of a Line

Parametric equations of a line passing through point (x0,y0)(x_0, y_0) with direction vector (a,b)(a, b): {x=x0+aty=y0+bt,t∈R\begin{cases} x = x_0 + at \\ y = y_0 + bt \end{cases}, \quad t \in \mathbb{R}

Parametric Equations of a Cycloid

Parametric equations of a cycloid: {x=a(t−sin⁡t)y=a(1−cos⁡t),t∈R\begin{cases} x = a(t - \sin t) \\ y = a(1 - \cos t) \end{cases}, \quad t \in \mathbb{R}

Parameter Elimination Methods

Basic Steps

  1. Solve for parameter t from the parametric equations
  2. Substitute the expression for t into the other equation
  3. Simplify to obtain the explicit functional relationship

Example

For the parametric equations of a circle {x=rcos⁡ty=rsin⁡t\begin{cases} x = r\cos t \\ y = r\sin t \end{cases}:

  1. From the first equation: cos⁡t=xr\cos t = \frac{x}{r}
  2. From the second equation: sin⁡t=yr\sin t = \frac{y}{r}
  3. Using sin⁡2t+cos⁡2t=1\sin^2 t + \cos^2 t = 1: (xr)2+(yr)2=1(\frac{x}{r})^2 + (\frac{y}{r})^2 = 1
  4. Simplify to obtain: x2+y2=r2x^2 + y^2 = r^2

Derivatives of Parametric Functions

Basic Formula

If {x=x(t)y=y(t)\begin{cases} x = x(t) \\ y = y(t) \end{cases}, then:

dydx=dydtdxdt=y′(t)x′(t)\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{y'(t)}{x'(t)}

Second Derivative

d2ydx2=ddx(dydx)=ddt(y′(t)x′(t))⋅dtdx\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dt}(\frac{y'(t)}{x'(t)}) \cdot \frac{dt}{dx}

Exercises

Exercise 1

Find the explicit functional relationship for the parametric equations {x=t2y=t3\begin{cases} x = t^2 \\ y = t^3 \end{cases}.

parametric functionparameter eliminationpower functiondomain

Approach: Obtain the explicit function relationship through parameter elimination.

Steps:

  1. Solve for parameter t from the first equation: t=xt = \sqrt{x} (note x≥0x \geq 0)
  2. Substitute the expression for t into the second equation: y=(x)3=x3/2y = (\sqrt{x})^3 = x^{3/2}
  3. Therefore, the explicit function is: y=x3/2y = x^{3/2}, domain [0,+∞)[0, +\infty)

Answer:

y=x3/2y = x^{3/2}, domain [0,+∞)[0, +\infty).

Exercise 2

Find the derivative of the parametric equations {x=cos⁡ty=sin⁡t\begin{cases} x = \cos t \\ y = \sin t \end{cases}.

parametric functionderivativetrigonometric functionimplicit differentiation

Approach: Use the derivative formula for parametric functions.

Steps:

  1. Calculate x′(t)x'(t) and y′(t)y'(t): x′(t)=−sin⁡tx'(t) = -\sin t y′(t)=cos⁡ty'(t) = \cos t
  2. Use the derivative formula: dydx=y′(t)x′(t)=cos⁡t−sin⁡t=−cot⁡t\frac{dy}{dx} = \frac{y'(t)}{x'(t)} = \frac{\cos t}{-\sin t} = -\cot t
  3. Since x=cos⁡tx = \cos t, we have t=arccos⁡xt = \arccos x, substitute to get: dydx=−cot⁡(arccos⁡x)=−x1−x2\frac{dy}{dx} = -\cot(\arccos x) = -\frac{x}{\sqrt{1-x^2}}

Answer:

dydx=−x1−x2\frac{dy}{dx} = -\frac{x}{\sqrt{1-x^2}}.

Exercise 3

Find the explicit functional relationship for the parametric equations {x=ety=e−t\begin{cases} x = e^t \\ y = e^{-t} \end{cases}.

parametric functionparameter eliminationexponential functionlogarithmic function

Approach: Obtain the explicit function relationship through parameter elimination.

Steps:

  1. Solve for parameter t from the first equation: t=ln⁡xt = \ln x (note x>0x > 0)
  2. Substitute the expression for t into the second equation: y=e−ln⁡x=1xy = e^{-\ln x} = \frac{1}{x}
  3. Therefore, the explicit function is: y=1xy = \frac{1}{x}, domain (0,+∞)(0, +\infty)

Answer:

y=1xy = \frac{1}{x}, domain (0,+∞)(0, +\infty).

Summary

Symbols Used in This Document

SymbolTypePronunciation/ExplanationMeaning in This Document
x(t)x(t)Mathematical symbolx of tFunction of parameter t for x
y(t)y(t)Mathematical symboly of tFunction of parameter t for y
ttMathematical symboltParameter variable
dydx\frac{dy}{dx}Mathematical symboldy by dxDerivative of y with respect to x
d2ydx2\frac{d^2y}{dx^2}Mathematical symbold squared y by dx squaredSecond derivative of y with respect to x
x′(t)x'(t)Mathematical symbolx prime of tDerivative of x with respect to t
y′(t)y'(t)Mathematical symboly prime of tDerivative of y with respect to t
R\mathbb{R}Mathematical symbolBlackboard bold R (Real numbers)Represents the set of real numbers, the set of all real numbers
[0,2π][0, 2\pi]Mathematical symbolClosed intervalClosed interval from 0 to 2π2\pi
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