Lab 14, Instructor’s Notes

MAT 229, Spring 2021

Curve representations

Graphs of functions

Some curves can be represented as graphs of functions y=f(x).

y=(x^2) is a parabola.

Plot[x^2,{x,-2,2},PlotRange->{-1,3}]

y=cos(x)+2sin(2x) is an oscillating curve.

Plot[Cos[x]+2Sin[2x],{x,-7,7},PlotRange->{-4,4}]

Implicitly defined curves

Another way to represent a curve is as a general equation in x and y.

(((x-2))^2)+(((y+1))^2)=9 is a circle centered at (2,-1) with radius 3.

ContourPlot[(x-2)^2+(y+1)^2==9,{x,-6,6},{y,-6,6},Axes->True]

(y^2)=(x^3)+3(x^2) is Tschirnhausen’s curve.

ContourPlot[y^2==x^3+3x^2,{x,-6,6},{y,-6,6},Axes->True]

Parametrically defined curves

One more major way of representing curves is with parametric equations. Each coordinate is given as a function of a third variable, the parameter.

x(t)=5cos(t), y(t)=5sin(t), 0≤t≤π is the upper half of a circle of radius 5.

ParametricPlot[{5Cos[t],5Sin[t]},{t,0,Pi},PlotRange->{{-6,6},{-6,6}}]

x(t)=(e^(-t/3))sin(t), y(t)=(e^(-t/3))cos(t), 0≤t≤20 is a spiral.

ParametricPlot[{E^(-t/3)Sin[t],E^(-t/3)Cos[t]},{t,0,20}]

Note

Parametrically defined curves have a direction associated with them, namely the direction as the parameter increases.

Manipulate[ParametricPlot[{5Cos[t],5Sin[t]},{t,0,s},PlotRange->{{-6,6},{-6,6}}],{s,0.1,2Pi}]

Manipulate[ParametricPlot[{5Sin[t],5Cos[t]},{t,0,s},PlotRange->{{-6,6},{-6,6}}],{s,0.1,2Pi}]

Tangents

Slope is always (rise/run). The slope of the tangent line is always ((d y)/(d x)).

Graphs of functions

For a curve given as the graph of a function, y=f(x), the slope of the tangent line at x=a is f'(a).

The slope of the tangent line to y=(x^2) at x=-2 is ((d y)/(d x))(Subscript[|, x=-2])=2(-2)=-4

Implicitly defined curves

Use implicit differentiation to find the slope of a tangent line to an implicitly defined curve at a given point on that curve.

The point (x,y)=(1,2) is on Tschirnhausen’s curve, (y^2)=(x^3)+3(x^2). Using implicit differentiation:
	2y(((d y)/(d x)))=3(x^2)+6x
⟶((d y)/(d x))=((3(x^2)+6x)/(2y))
⟶((d y)/(d x))(Subscript[|, (x,y)=(1,2)])=((3(((1))^2)+6(1))/(2(2)))=(9/4)

Parametrically defined curves

The chain rule implies ((d y)/(d x))=((d y/d t)/(d x/d t)).

The slope of the tangent line to x(t)=5cos(t), y(t)=5sin(t) at t=π/3 is
	((d y)/(d x))=((y'(π/3))/(x'(π/3)))=((5cos(π/3))/(-5sin(π/3)))=((5(1/2))/(-5((Sqrt[3])/2)))=-(1/(Sqrt[3]))

Converting between representations

From graphs

Given y=f(x).

It is automatically an implicitly defined curve since it is an equation in x and y.

Write it parametrically with x as the parameter.
	x=x
y=f(x)

ContourPlot[y==x^2,{x,-6,6},{y,-6,6},Axes->True]

From parametric equations

Given x=f(t), y=f(t).

You may be able to use an identity to eliminate the parameter. 
	x=2cos(t)+1
y=2sin(t)-3
We know that (cos^2)(t)+(sin^2)(t)=1. Since the first equation rewritten is cos(t)=((x-1)/2) and the second is sin(t)=((y+3)/2), we get
	(((((x-1)/2)))^2)+(((((y+3)/2)))^2)=1
or
	(((x-1))^2)+(((y+3))^2)=4

You may be able to solve one of the equations for the parameter and then use that value of the parameter in the other equation.
	x=((5-t)^(1/3))
y=(t/(t+1))
Solve the top equation for t.
	(x^3)=5-t
⟶t=5-(x^3)
Use this value of t in the bottom equation.
	y=(t/(t+1))=((5-(x^3))/((5-(x^3))+1))=((5-(x^3))/(4-(x^3)))