Objectives
In this section you will...
identify the domain and range of a multivariable or vector-valued function
express curves and surfaces with parametric equations
Section6.4Functions of Several Variables: \(f\colon {\mathbb{R}}^n\to {\mathbb{R}}\)
¶In this section we'll focus on functions of the form \(f\colon \mathbb{R}^2\to\mathbb{R}^1\) and \(f\colon \mathbb{R}^3\to\mathbb{R}^1\text{;}\) we'll keep the output as a real number. In the next problem, you should notice that the input is a vector \((x,y)\) and the output is a number \(z\text{.}\) There are two ways to graph functions of this type. The next two problems show you how.
Subsection6.4.13-D Surface Plots
Exercise6.4.1
A computer chip has been disconnected from electricity and sitting in cold storage for quite some time. The chip is connected to power, and a few moments later the temperature (in Celsius) at various points \((x,y)\) on the chip is measured. From these measurements, statistics is used to create a temperature function \(T=z=f(x,y)\) to model the temperature at any point on the chip. Suppose that this chip's temperature function is given by the equation \(T=z=f(x,y)=9-x^2-y^2\text{.}\) Eventually we'll be creating a 3D model of this function in this problem, so you'll need to place all your graphs on the same \(x,y,z\) axes. However, to begin with you may find it easier to start with several 2-D plots.
(a)
What is the temperature at \((0,0)\text{,}\) \((1,2)\text{,}\) and \((-4,3)\text{?}\)
(b)
Let \(y=0\text{,}\) Substitute this value into the temperature function \(T=z=f(x,y)=9-x^2-y^2\text{.}\) Plot the resulting function in the \(xz\)-plane.
Hint
Treat \(z\) as “y”. It should be a (upside-down) parabola
(c)
Now let \(x=0\text{.}\) Substitute this value into the temperature function and simplify. Plot the result. This plot should appear in the \(yz\) plane and be another parabola.
(d)
Now let \(z=0\text{.}\) Draw the resulting curve in the \(xy\) plane. If you drew the above curves in 2-D it is time to combine them. You should end up with 3 lines or a basic wire frame. To make a more complete wireframe we need to add more curves.
(e)
State the simplified equation, then plot and label the curves for the values below.
Note: You could have done this with \(y\) also.
(i)
\(x=-2\)
(ii)
\(x=-1\)
(iii)
\(x=1\)
(iv)
\(x=2\)
(v)
Describe the shape. Add any extra features to your graph to convey the 3D image you are constructing.
See this Wolfram Alpha link or use the SageMathCell below to verify your answer.
Subsection6.4.22-D Contour Plots
Exercise6.4.2
We'll be using the same function \(T=z=f(x,y)=9-x^2-y^2\) as the previous problem. However, this time we'll construct a graph of the function by only studying places where the temperature is constant. We'll do this by creating a graph in 2D of the surface (similar to a topographical map).
(a)
How can we find all the points which have zero temperature?
Hint
Think about what a temperature of zero means in the function
(b)
Use your process to find the curve corresponding to a temperature of zero. Plot this curve in the \(xy\)-plane. Be sure to label it with \(T=0\) This curve is called a level curve. As long as you stay on this curve, your temperature will remain level, it will not increase nor decrease.
(c)
Which points in the plane have temperature \(z=5\text{?}\) Add this level curve to your 2D plot and write \(z=5\) next to it.
(d)
Repeat the above for \(z=8\text{,}\) \(z=9\text{,}\) and \(z=1\text{.}\)
(e)
What's wrong with letting \(z=10\text{?}\)
(f)
Using your 2D plot, construct a 3D image of the function by lifting each level curve to its corresponding height.
See this Wolfram Alpha link or use the SageMathCell below to verify your answer.
Definition6.4.1
A level curve of a function \(z=f(x,y)\) is a curve in the \(xy\)-plane found by setting the output \(z\) equal to a constant. Symbolically, a level curve of \(f(x,y)\) is the curve \(c=f(x,y)\) for some constant \(c\text{.}\) A 2D plot consisting of several level curves is called a contour plot of \(z=f(x,y)\text{.}\)
Subsection6.4.3Parametric Surfaces Continued
Exercise6.4.3
Consider the function \(f(x,y)=x-y^2\text{.}\)
(a)
Use the same process as Exercise 1 to construct a 3D surface plot of \(f\text{.}\) [So just graph in 3D the curves given by \(x=0\) and \(y=0\) and then try setting \(x\) or \(y\) equal to some other constants, like \(x=1\text{,}\) \(x=2\text{,}\) \(y=1\text{,}\) \(y=2\text{,}\) etc.]
(b)
Use the same process as Exercise 2 to construct a contour plot of \(f\text{.}\) [So just graph in 2D the curves given by setting \(z\) equal to a few constants, like \(z=0\text{,}\) \(z=1\text{,}\) \(z=-4\text{,}\) etc.]
(c)
Which level curve passes through the point \((2,2)\text{?}\) Draw this level curve on your contour plot.
See this Wolfram Alpha link or use the SageMathCell below to verify your answer.
Notice that when we graphed the previous two functions (of the form \(z=f(x,y)\)) we could either construct a 3D surface plot, or we could reduce the dimension by 1 and construct a 2D contour plot by letting the output \(z\) equal various constants.
The next function is of the form \(w=f(x,y,z)\text{,}\) so it has 3 inputs and 1 output. We could write \(f\colon \mathbb{R}^3\to\mathbb{R}^1\text{.}\) We would need 4 dimensions to graph this function, but graphing in 4D is not an easy task. Instead, we'll reduce the dimension and create plots in 3D to describe the level surfaces of the function.
Exercise6.4.4
Suppose that an explosion occurs at the origin \((0,0,0)\text{.}\) Heat from the explosion starts to radiate outwards. Now suppose that a few moments after the explosion, the temperature at any point in space is given by \(w=T(x,y,z)=100-x^2-y^2-z^2.\)
(a)
Which points in space have a temperature of 99? Use algebra to simplify this to \(x^2+y^2+z^2=1\text{.}\)
Hint
What does the 99 replace in your function?
(b)
Draw the resulting object/function.
(c)
Which points in space have a temperature of 96? of 84? Draw the surfaces.
(d)
What is your temperature at \((3,0,-4)\text{?}\) Draw the level surface that passes through \((3,0,-4)\text{.}\)
(e)
The 4 surfaces you drew above are called level surfaces. If you walk along a level surface, what happens to your temperature?
(f)
As you move outwards, away from the origin, what happens to your temperature?
See Wolfram Alpha currently does not support drawing level surfaces. You could also use Mathematica or Wolfram Demonstrations.
Exercise6.4.5
Consider the function \(w=f(x,y,z)=x^2+z^2\text{.}\) This function has an input \(y\text{,}\) but notice that changing the input \(y\) does not change the output of the function. This means that when \(y=0\) you can draw one curve. When \(y=1\text{,}\) you should draw the same curve. When \(y=2\text{,}\) again you draw the same curve, etc.
(a)
Draw a graph of the level surface \(w=4\text{.}\)
(b)
Graph the surface \(9=x^2+z^2\) (so the level surface \(w=9\)).
(c)
Graph the surface \(16=x^2+z^2\text{.}\)
This kind of graph is called a cylinder, and is important in manufacturing where you extrude an object through a hole.
Most of our examples of function of the form \(w=f(x,y,z)\) can be drawn by using our knowledge about conic sections. We can graph ellipses and hyperbolas if there are only two variables. So the key idea is to set one of the variables equal to a constant and then graph the resulting curve. Repeat this with a few variables and a few constants, and you'll know what the surface is. Sometimes when you set a specific variable equal to a constant, you'll get an ellipse. If this occurs, try setting that variable equal to other constants, as ellipses are generally the easiest curves to draw.
If you feel a little lost and skipped the review unit on conic sections, now would be a good time to go back and go through it.
Exercise6.4.6
Go back to the table you started in Exercise 6.2.3.
(a)
Add in the new problems you've completed.
(b)
What do you observe about the domain (\(n\)), co-domain (\(m\)) and the dimensions required for plotting?
Exercise6.4.7More Surfaces
Consider the function \(w=f(x,y,z)=x^2-y^2+z^2\text{.}\)
We'll probably have a few people present this problem.
(a)
Draw a graph of the level surface \(w=1\text{.}\) [You need to graph \(1=x^2-y^2+z^2\text{.}\) Let \(x=0\) and draw the resulting curve. Then let \(y=0\) and draw the resulting curve. Let either \(x\) or \(y\) equal some more constants (whichever gave you an ellipse), and then draw the resulting ellipses.]
(b)
Graph the level surface \(w=4\text{.}\) [Divide both sides by \(4\) (to get a 1 on the left) and the repeat the previous part.]
(c)
Graph the level surface \(w=-1\text{.}\) [Try dividing both sides by a number to get a 1 on the left. If \(y=0\) doesn't help, try \(y=1\) or \(y=2\text{.}\)]
(d)
Graph the level surface that passes through the point \((3,5,4)\text{.}\)
Hint
what is \(f(3,5,4)\text{?}\)
Your friendly SageMathCell for plotting..
Subsection6.4.4Vector Fields and Transformations:
We will finish this section by considering vector fields and transformations.
\(\vec T(u,v)=(x,y)\) or \(f\colon \mathbb{R}^2\to\mathbb{R}^2\) (2D transformation)
\(\vec T(u,v,w)=(x,y,z)\) or \(f\colon \mathbb{R}^3\to\mathbb{R}^3\) (3D transformation)
\(\vec F(x,y)=(M,N)\) or \(f\colon \mathbb{R}^2\to\mathbb{R}^2\) (vector fields in the plane)
\(\vec F(x,y,z)=(M,N,P)\) or \(f\colon \mathbb{R}^3\to\mathbb{R}^3\) (vector fields in space)
Notice that in all cases, the dimension of the input and output are the same. The difference between vector fields and transformations has to do with the application.
Subsubsection6.4.4.1Transformations
In this section you will...
identify the domain and range of multivariable or vector-valued functions
We've already seen examples of transformations with polar, cylindrical, and spherical coordinates.
Exercise6.4.8Review of Coordinate Transformations
Consider the coordinate transformation
\begin{equation*} \vec T(r,\theta) = (r\cos\theta,r\sin\theta). \end{equation*}Let \(r=3\) and then graph \(\vec T(3,\theta)=(3\cos\theta,3\sin\theta)\) for \(\theta\in[0,2\pi]\text{.}\)
Let \(\theta=\frac{\pi}{4}\) and then, on the same axes as above, add the graph of \(\vec T\left(r,\frac{\pi}{4}\right)=\left(r\frac{\sqrt 2}{2},r \frac{\sqrt 2}{2}\right)\) for \(r\in[0,5]\text{.}\)
Exercise6.4.9
Consider the spherical coordinates transformation as defined in
\begin{equation*} \vec T(\rho,\theta,\phi)=(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi), \end{equation*}which could also be written as
\begin{align*} x\amp =\rho\sin\phi\cos\theta\\ y\amp =\rho\sin\phi\sin\theta\\ z\amp =\rho\cos\phi \end{align*}Graphing this transformation requires \(3+3=6\) dimensions. In this problem we'll construct parts of this graph by graphing various surfaces. We did something similar for the polar coordinate transformation in Exercise 5.1.5.
Recall that \(\phi\) (“phi”) is the angle down from the \(z\) axis, \(\theta\) (“theta”) is the angle counterclockwise from the \(x\)-axis in the \(xy\)-plane, and \(\rho\) (“rho”) is the distance from the origin. Review Exercise 5.2.3 if you need a refresher.
For all of the tasks below, do your best to figure out the shape BEFORE using the sage plotting. You should be able to explain (logically) why the graphs generated match the sage plots.
(a)
See this Wolfram Alpha link.
Let \(\rho=2\) and graph the resulting surface. What do you get if \(\rho = 3\text{?}\)
(b)
Let \(\phi=\pi/4\) and graph the resulting surface. What do you get if \(\phi=\pi/2\text{?}\)
See Wolfram Alpha.
(c)
Let \(\theta=\pi/4\) and graph the resulting surface. What do you get if \(\theta=\pi/2\text{?}\)
If you want to see this plotted, you can just modify the previous sage cell.
Subsubsection6.4.4.2Vector Fields
In this section you will...
identify the domain and range of multivariable or vector-valued functions
define and sketch two- or three- dimensional vector fields
We now explore a vector field example.
Exercise6.4.10
Consider the vector field \(\vec F(x,y)=(2x+y,x+2y)\text{.}\) In this problem, you will construct a graph of this vector field by hand.
(a)
Compute \(\vec F(1,0)\text{.}\) Then draw the vector \(F(1,0)\) with its base at \((1,0)\text{.}\)
(b)
Compute \(\vec F(1,1)\text{.}\) Then draw the vector \(F(1,1)\) with its base at \((1,1)\text{.}\)
(c)
Repeat the above process for the points \((0,1)\text{,}\) \((-1,1)\text{,}\) \((-1,0)\text{,}\) \((-1,-1)\text{,}\) \((0,-1),\) and \((1,-1)\text{.}\) Remember, at each point draw a vector.
See Wolfram Alpha or the cell below. The computer will shrink the largest vector down in size so it does not overlap any of the others, and then reduce the size of all the vectors accordingly.
Exercise6.4.11The Spin field
Use the links above to see the computer plot this.
Consider the vector field \(\vec F(x,y)=(-y,x)\text{.}\) Construct a graph of this vector field. Remember, the key to plotting a vector field is “at the point \((x,y)\text{,}\) draw the vector \(\vec F(x,y)\) with its base at \((x,y)\text{.}\)” Plot at least 8 vectors (a few in each quadrant), so we can see what this field is doing.
Sage can also help us visualize 3d vector fields, like \(\vec F(x,y,z)=(y,z,x)\text{.}\)
Subsection6.4.5Computational Practice
These are provided to help you achieve better skills in basic computational answers.