Math 1C with Jeff Anderson
Part 1: Multidimensional Differential Calculus
Lesson 7: Graphs and Level Curves
©Jeffrey A. Anderson

Key Terms List: Briggs Section 12.2 p. 873
□  Functions of several variables (p. 873)
□  Definition: (Two-variable) Function (p. 874)
□  Definition: Domain (of a two-variable  function) (p. 874)
□  Definition: Range (of a two-variable function) (p. 874)
□  Graph of a two-variable function (p. 874)
□  Viewpoints to study multivariable functions (Jeff’s Lesson Notes)
	□  Verbal description
	□  Numerical Description (table of values)
	□  Algebraic Description (explicit formula)
	□  Graphical Description
□  Contour Map (p. 877)
□  Level Curves (or contour lines) (p. 877)
□  Definition: Function (with n-independent variables) (p. 880)
□  Definition: Domain (of a function with n-independent variables) (p. 880)
□  Definition: Range (of a function with n-independent variables) (p. 880)
□  Level Surfaces (p. 881)

Example 12.2.1 p. 874: Half Sphere

Let’s visualize our function using Mathematica’s Plot3D routine.

F[x_,y_]:=(Sqrt[9 -(x^2 + y^2)]);
Manipulate[
	GraphicsRow[{
	Show[
		Plot3D[
			F[x,y],
			{x,-4,4},
			{y,-4,4},
			PlotRange-> {-4,4},
			AxesLabel->{x,y,z},
			BoxRatios->{1,1, 1},
			ImageSize->{300,300},
			PlotStyle->Directive[Blue,Opacity[0.4]]
		],
		ContourPlot3D[
			z == h,
			{x,-4,4},
			{y,-4,4},
			{z, -4,4},
			ContourStyle->Directive[Red,Opacity[0.4]]
		],
		ParametricPlot3D[
			{(Sqrt[(9-(h^2))])*Sin[t], (Sqrt[(9-(h^2))])*Cos[t], h},
			{t, 0, 2π},
			PlotStyle->Directive[Red,Thickness[0.01],Opacity[0.4]]
		]
	],
	Show[
	ContourPlot[
		F[x,y],
		{x,-4,4},
		{y,-4,4},
		Contours->Table[i,{i, 0, 3, 0.3}],
		ContourStyle->Directive[Black,Opacity[0.4]]
	],
	ContourPlot[
		F[x,y]==h,
		{x,-4,4},
		{y,-4,4},
		PlotLegends->Automatic,
		ContourStyle->Red
	]	
	]
}],
{{h,2},0,3,0.3}
]

Similar to Exercise 12.2.17 p. 882: Find Domain & Range

Suppose that we are working with a function:

	g(x,y) = ln(9 - 9 (x^2) - (y^2))

and we want to find the domain and range of this function. We’ll start with the domain of this function. We recall from Math 1A that 

	z = ln(w) ⟺ w = (e^z)
	
We know by the behavior of the exponential function, that w > 0 must hold true for the inverse function to exist. In this case, we want to find x and y values such that 

	9 - 9 (x^2) - (y^2) > 0

We can visualize which values of x and y satisfy this inequality using Mathematica’s RegionPlot:

RegionPlot[
	9x^2+y^2 < 9,
	{x,-2,2},
	{y,-4,4},
	BoundaryStyle->Dashed,
	PlotLegends->"Expressions"
]

F[x_,y_]:=Log[9-9x^2-y^2];
Plot3D[
	F[x,y],
	{x,-5,5},
	{y,-5,5},
	PlotRange-> {-5,4},
	AxesLabel->{x,y,z}
]

Exercise 12.2.41 p. 883: Electric Potential Function

ϕ[x_,y_]:= (2/(Sqrt[(x^2)+(((y-1))^2)]))+(1/(Sqrt[(x^2)+(((y+1))^2)]));
Row[
	{
		Show[
		Plot3D[
			ϕ[x,y],
			{x,-5,5},
			{y,-5,5},
			PlotRange-> {0,10},
			AxesLabel->{x,y,z},
			BoxRatios->{1,1, 1},
			ImageSize->{400,400},
			PlotStyle->Directive[Yellow,Opacity[0.4]]
			],
			Graphics3D[{PointSize[.02],Point[{{0,-1,0}, {0,1,0}}]},Axes->True],
			Graphics3D[{Thickness[0.008],Line[{{0,-5, 0},{0,5, 0}}]}],
			Graphics3D[{Thickness[0.008],Line[{{5,0,0},{-5, 0,0}}]}]
		],
		Show[
		ContourPlot[
		{
		ϕ[x,y]==0.8, 
		ϕ[x,y]==1, 
		ϕ[x,y]==1.2,
		ϕ[x,y]==1.4, 
		ϕ[x,y]==1.6, 
		ϕ[x,y]==1.8,
		ϕ[x,y]==2, 
		ϕ[x,y]==2.2,
		ϕ[x,y]==2.4, 
		ϕ[x,y]==2.6, 
		ϕ[x,y]==2.8,
		ϕ[x,y]==3, 
		ϕ[x,y]==3.2,
		ϕ[x,y]==3.4, 
		ϕ[x,y]==3.6, 
		ϕ[x,y]==3.8
		},
		{x,-5,5},
		{y,-5,5},
		ImageSize->{300,300},
		ContourStyle->RGBColor[0.1,.5,1]
		],
		Graphics[Line[{{0,-5},{0,5}}]],
		Graphics[Line[{{5,0},{-5, 0}}]],
		Graphics[{PointSize[.02],Point[{{0,-1}, {0,1}}]},Axes->True]
]}
]

}