This unit covers the following ideas. In preparation for the quiz and exam, make sure you have a lesson plan containing examples that explain and illustrate the following concepts.
We'll need to know how to graph by hand some basic functions. If you have not spent much time graphing functions by hand before this class, then you should spend some time graphing the following functions by hand. When we start drawing functions in 3D, we'll have to piece together infinitely many 2D graphs. Knowing the basic shape of graphs will help us do this.
In first semester calculus, one of the things you focused on was learning to compute derivatives. You'll need to know the derivatives of basic functions (found on the end cover of almost every calculus textbook). Computing derivatives accurately and rapidly will make learning calculus in high dimensions easier. The following rules are crucial.
The following problem will help you review some of your trigonometry, inverse functions, as well as implicit differentiation.
Each derivative rule from the front cover of your calculus text is also an integration rule. In addition to these basic rules, we'll need to know three integration techniques. They are (1) $u$-substitution, (2) integration-by-parts, and (3) integration by using software. There are many other integration techniques, but we will not focus on them. If you are trying to compute an integral to get a number while on the job, then software will almost always be the tool you use. As we develop new ideas in this and future classes (in engineering, physics, statistics, math), you'll find that $u$-substitution and integrations-by-parts show up so frequently that knowing when and how to apply them becomes crucial.
The derivative of a function gives us the slope of a tangent line to that function. We can use this tangent line to estimate how much the output ($y$ values) will change if we change the input ($x$-value). If we rewrite the notation $\ds\frac{dy}{dx}=f'$ in the form $dy=f' dx$, then we can read this as “A small change in $y$ (called $dy$) equals the derivative ($f'$) times a small change in $x$ (called $dx$).”
a change in the output $y$ equals the derivative times a change in the input $x$.
Most of higher dimensional calculus can quickly be developed from differential notation. Once we have the language of vectors and matrices at our command, we will develop calculus in higher dimensions by writing $d\vec y = Df(\vec x) d\vec x$. Variables will become vectors, and the derivative will become a matrix. This problem will help you see how the notion of differentials is used to develop equations of tangent lines. We'll use this same idea to develop tangent planes to surfaces in 3D and more.
We will soon discover that matrices represent derivatives in high dimensions. When you use matrices to represent derivatives, the chain rule is precisely matrix multiplication. For now, we just need to become comfortable with matrix multiplication. We perform matrix multiplication “row by column”. Wikipedia has an excellent visual illustration of how to do this. See Click on the links to go to the websites. Wikipedia for an explanation. See texample.net for a visualization of the idea.
Associated with every square matrix is a number, called the determinant, which is related to length, area, and volume. We'll use the determinant to generalize the concept of volume to higher dimensions as well. Determinants will appear as we study cross products and when we get to the high dimensional version of $u$-substitution. Determinants are only defined for square matrices.
What good is the determinant? The determinant was discovered as a result of trying to find the area of a parallelogram and the volume of the three dimensional version of a parallelogram (called a parallelepiped) in space. If we had a full semester to spend on linear algebra, we could eventually prove the following facts that I will just present here with a few examples.
Consider the 2 by 2 matrix $\begin{bmatrix}3&1\\0&2\end{bmatrix}$ whose determinant is $3\cdot 2-0\cdot 1=6$. Draw the column vectors $\begin{bmatrix}3\\0\end{bmatrix}$ and $\begin{bmatrix}1\\2\end{bmatrix}$ with their base at the origin (see figure \ref{detfig}). These two vectors give the edges of a parallelogram whose area is the determinant $6$. If I swap the order of the two vectors in the matrix, then the determinant of $\begin{bmatrix}1&3\\2&0\end{bmatrix}$ is $-6$. The reason for the difference is that the determinant not only keeps track of area, but also order. Starting at the first vector, if you can turn counterclockwise through an angle smaller than 180$^\circ$ to obtain the second vector, then the determinant is positive. If you have to turn clockwise instead, then the determinant is negative. This is often termed “the right-hand rule” since rotating the fingers of your right hand from the first vector to the second vector will cause your thumb to point up precisely when the determinant is positive.
For a 3 by 3 matrix, the columns give the edges of a three dimensional parallelepiped and the determinant produces the volume of this object. The sign of the determinant is related to orientation. If you can use your right hand and place your index finger on the first vector, middle finger on the second vector, and thumb on the third vector, then the determinant is positive. For example, consider the matrix $A = \begin{bmatrix}\cl{1\\0\\0}&\cl{0\\2\\0}&\cl{0\\0\\3}\end{bmatrix}$. Starting from the origin, each column represents an edge of the rectangular box $0\leq x\leq 1$, $0\leq y\leq 2$, $0\leq z\leq 3$ with volume (and determinant) $V=lwh=(1)(2)(3)=6$. The sign of the determinant is positive because if you place your index finger pointing in the direction (1,0,0) and your middle finger in the direction (0,2,0), then your thumb points upwards in the direction (0,0,3). If you interchange two of the columns, for example $B = \begin{bmatrix} \cl{0\\2\\0}&\cl{1\\0\\0}&\cl{0\\0\\3}\end{bmatrix}$, then the volume doesn't change since the shape is still the same. However, the sign of the determinant is negative because if you point your index finger in the direction (0,2,0) and your middle finger in the direction (1,0,0), then your thumb points down in the direction (0,0,-3). If you repeat this with your left hand instead of right hand, then your thumb points up.
When you ask a calculator to tell you what $e^{0.1}$ means, your calculator uses an extension of differentials to give you an approximation. The calculator only uses polynomials (multiplication and addition) to give you an answer. This same process is used to evaluate any function that is not a polynomial (so trig functions, square roots, inverse trig functions, logarithms, etc.) The key idea needed to approximate functions is illustrated by the next problem.
The previous two problems involved finding polynomial approximations to the function at $x=0$. The next problem shows how to move this to any other point, such as $x=1$.
This example refers back to this problem above. We wanted a spherical tank of radius 3m, but due to manufacturing error the radius was slightly off. Let's now illustrate how we can use polynomials to give a first, second, and third order approximation of the volume if the radius is 3.02m instead of 3m. We start with $V=\frac{4}{3} \pi r^3$ and then compute the derivatives $$V'=4\pi r^2, V''=8\pi r, \text{ and } V'''=8\pi.$$ Because we are approximating the increase in volume from $r=3$ to something new, we'll create our polynomial approximations centered at $r=3$. We'll consider the polynomial $$P(r)=a_0+a_1(r-3)+a_2(r-3)^2+a_3(r-3)^3,$$ whose derivatives are $$P'=a_1+2a_2(r-3)+3a_3(r-3)^2, P''=2a_2+6a_3(r-3), P'''=6a_3.$$ So that the derivatives of the volume function match the derivatives of the polynomial (at $r=3$), we need to satisfy the equations in the table below.
| $k$ | Value of $V$ at the $k$th derivative | Value of $P$ at the $k$th derivative | Equation |
|---|---|---|---|
| $0$ | $V(3) = \frac{4}{3}\pi (3)^3 = 36\pi$ | $P(3) = a_0$ | $a_0=36\pi$ |
| $1$ | $V'(3) = 4\pi (3)^2=36\pi$ | $P'(3) = a_1$ | $a_1=36\pi$ |
| $2$ | $V''(3) = 8\pi (3)=24\pi$ | $P''(3) = 2a_2$ | $2a_2=24\pi$ |
| $3$ | $V'''(3) = 8\pi$ | $P'''(3) = 6a_3$ | $6a_3=8\pi$ |
This tells us that the third order polynomial is $$P(r)=a_0+a_1(r-3)+a_2(r-3)^2+a_3(r-3)^3 =36\pi+36\pi(r-3)+12\pi(r-3)^2+\frac{4}{3}\pi(r-3)^3 .$$ We wanted to approximate the volume if $r=3.2$, so our change in $r$ is $dr=3.2-3=0.2$. We can rewrite our polynomial as $$P(r)=36\pi+36\pi(dr)+12\pi(dr)^2+\frac{4}{3}\pi(dr)^3.$$ We are now prepared to approximate the volume using a first, second, and third order approximation.
With each approximation, we add on a little more volume to get closer to the actual volume of a sphere with radius $r=3.02$. The actual volume of a sphere involves a cubic function, so when we approximate the volume with a cubic, we should get an exact approximation (and $ V(3.02) = \frac 43 \pi (3.02)^3 =(36.724810\bar6)\pi$.)
We'll end this section with a problem to practice the example above.
Suppose that the machine creates a cube with side length $2.3$ units instead of 2 units. Note that the volume of the cube is $V=s^3$. Use a first, second, and third order approximation to estimate the increase in volume caused by the .3 increase in side length. Then compute the actual increase in volume $V(2.3)-V(2)$.