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.

- Give a summary of the ideas you learned in 112, including graphing, derivatives (product, quotient, power, chain, trig, exponential, and logarithm rules), and integration ($u$-sub and integration by parts).
- Compute the differential $dy$ of a function and use it to approximate the change in a function.
- Explain how to perform matrix multiplication and compute determinants of square matrices.
- Illustrate how to solve systems of linear equations, including how to express a solution parametrically (in terms of $t$) when there are infinitely solutions.
- Extend the idea of differentials to approximate functions using parabolas, cubics, and polynomials of any degree.

You'll have a chance to teach your examples to your peers
prior to the exam.

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.

Provide a rough sketch of the following functions, showing
their basic shapes: $$\displaystyle x^2, x^3, x^4, \frac{1}{x},
\sin x, \cos x, \tan x, \sec x, \arctan x, e^x,\ln x.$$ Then
use a computer algebra system, such as
Sage or Wolfram Alpha to
verify your work.

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.

- Power rule {$(x^n)' = nx^{n-1}$}
- Sum and difference rule {$(f\pm g) = f'\pm g'$}
- Product {$(fg)' = f' g + fg'$} and quotient rule {$\ds\left(\frac f g\right)' = \frac{f' g - fg'}{g^2}$}
- Chain rule (arguably the most important) {$(f\circ g)' = f'(g(x))\cdot g'(x)$}

See sections 3.2-3.6 for more practice with derivatives.
The later problems in 3.6 review of most of the entire
differentiation chapter.

Compute the derivative of $e^{\sec
x}\cos(\tan(x)+\ln(x^2+4))$. Show each step in your
computation, making sure to show what rules you used.
If $y(p) = \ds \frac{e^{p^3}\cot(4p+7)}{\tan^{-1}(p^4)}$ find
$dy/dp$. Again, show each step in your computation, making sure
to show what rules you used.

The following problem will help you review some of your trigonometry, inverse functions, as well as implicit differentiation.

See sections 3.7-3.9 for more examples involving inverse
trig functions and implicit differentiation.

Use implicit differentiation to explain why the
derivative of $y=\arcsin x$ is $\ds y'=\frac{1}{\sqrt{1-x^2}}$.
[Rewrite $y=\arcsin x$ as $x=\sin y$, differentiate both sides,
solve for $y'$, and then write the answer in terms of $x$].
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.

For practice with $u$-substitution, see section 5.5 and
5.6. \\ For practice with integration by parts, see section
8.1.

Compute $\ds\int x\sqrt{x^2+4}dx$.
Compute $\ds\int x\sin 2x dx$.

Compute $\ds \int \arctan x dx$.

Compute $\ds \int x^2 e^{3x} dx$.

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 $dx$).”

We call $dx$ the differential of $x$. If $f$ is a function of
$x$, then the differential of $f$ is $df = f'(x) dx$. Since we
often write $y=f(x)$, we'll interchangeably use $dy$ and $df$
to represent the differential of $f$. We will often refer to
the differential notation $dy=f'dx$ as

a change in the output $y$ equals the derivative times a change in the input $x$.

See 3.10:19-38.

If $f(x) = x^2\ln(3x+2)$ and $g(t) = e^{2t}\tan(t^2)$
then compute $df$ and $dg$.
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.

See 3.11:39-44. Also see problems 3.11:1-18. The
linearization of a function is just an equation of the
tangent line where you solve for $y$.

Consider the function $y=f(x) = x^2$. This problem has
multiple steps, but each is fairly short.
- Find the differential of $y$ with respect to $x$.
- Give an equation of the tangent line to $f(x)$ at $x=3$.
- Draw a graph of $f(x)$ and the tangent line on the same axes. Place a dot at the point $(3,9)$ and label it on your graph. Place another dot on the tangent line up and to the right of (3,9). Label the point $(x,y)$, as it will represent any point on the tangent line.
- Using the two points $(3,9)$ and $(x,y)$, compute the slope of the line connecting these two points. Your answer should involve $x$ and $y$. What is the rise (i.e, the change in $y$ called $dy$)? What is the run (i.e, the change in $x$ called $dx$)?
- We already know the slope of the tangent line is the derivative $f'(3)=6$. We also know the slope from the previous part. These two must be equal. Use this fact to give an equation of the tangent line to $f(x)$ at $x=3$.

See 3.11:45-62.

The manufacturer of a spherical
storage tank needs to create a tank with a radius of 3 meters.
Recall that the volume of a sphere is $V(r) = \frac{4}{3}\pi
r^3$. No manufacturing process is perfect, so the resulting
sphere will have a radius of 3 meters, plus or minus some small
amount $dr$. The actual radius will be $3+dr$. Find the
differential $dV$. Then use differentials to estimate the
change in the volume of the sphere if the actual radius is 3.02
meters instead of the planned 3 meters.
A forest ranger needs to estimate the height of a tree. The
ranger stands 50 feet from the base of tree and measures the
angle of elevation to the top of the tree to be about
60$^\circ$. If this angle of 60$^\circ$ is correct, then what
is the height of the tree? If the ranger's angle measurement
could be off by as much as $5^\circ$, then how much could his
estimate of the height be off? Use differentials to give an
answer.

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.

For extra practice, make up two small
matrices and multiply them. Use Sage
or Wolfram
Alpha to see if you are correct (click the links to see how to
do matrix multiplication in each system).

Compute the
following matrix products.
- $\begin{bmatrix}3 & 2& 1\end{bmatrix} \begin{bmatrix} -1 \\ 2\\ 0\end{bmatrix}$
- $\begin{bmatrix}1 &2\\3&4\end{bmatrix}\begin{bmatrix}5&0\\6&1\end{bmatrix}$

Compute the product $\begin{bmatrix}3 & 2& 1\\ 0 &
1& -4\end{bmatrix}
\begin{bmatrix} -1&3 &0 \\ 2&-1 &0\\ 0&1 &2\end{bmatrix}$.

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.

The determinant of a $2\times 2$ matrix is the number
We use vertical bars next to a matrix to state we want the
determinant, so $\det A = |A|$.
\begin{align*}\det \begin{bmatrix}
a&b\\c&d\end{bmatrix} &=
\begin{vmatrix}
a&b\\c&d
\end{vmatrix}= ad-bc.
\end{align*}
The determinant of a $3\times 3$ matrix is the number
Notice the negative sign on the middle term of the $3 \times
3$ determinant. Also, notice that we had to compute three
determinants of 2 by 2 matrices in order to find the
determinant of a 3 by 3 matrix.
\begin{align*}
\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix} &= a\det\begin{vmatrix}e&f\\h&i\end{vmatrix} -b\det\begin{vmatrix}d&f\\g&i\end{vmatrix} +c\det\begin{vmatrix}d&e\\g&h\end{vmatrix}\\
&=a(ei-hf)-b(di-gf)+c(dh-ge).
\end{align*}

For extra practice, create your own square matrix (2 by 2 or
3 by 3) and compute the determinant by hand. Then use Sage or Wolfram
Alpha to check your work. Do this until you feel comfortable
taking determinants.

Compute $\begin{vmatrix}
1&2\\
3&4
\end{vmatrix}
$
and
$\begin{vmatrix}
1&2&0\\
-1&3&4\\
2&-3&1
\end{vmatrix}
$.
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.

\draw[help lines,step=1cm] (0,0) grid (4,2);
\draw[->,>=stealth,red] (0,0) -- (3,0);
\draw[->,>=stealth,red] [shift={(1,2)}](0,0) --
(3,0); \draw[->,>=stealth,blue] (0,0) -- (1,2);
\draw[->,>=stealth,blue] [shift={(3,0)}](0,0) --
(1,2); \draw[->,>=stealth] (0:1cm) node[above
right=1pt,fill=white]{\normalsize $+$} arc (0:64:1cm) ;
\draw[<-,>=stealth] (0:2cm) node[above
right=1pt,fill=white]{\normalsize $-$} arc (0:64:2cm) ;
\node[fill=white] at (2.5, 1.5) {Area $=6$};

\vspace{2pt} $
{3}&{1}\\{0}&{2}

=6$ and $
{1}&{3}\\{2}&0

=-6$
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.

- Use determinants to find the area of the triangle with vertices $(0,0)$, $(-2,5)$, and $(3,4)$.
- What would you change if you wanted to find the area of the triangle with vertices $(-3,1)$, $(-2,5)$, and $(3,4)$? Find this area.

For additional practice, make up your own systems of
equations. Use Wolfram Alpha or Sage to check your work.

Solve the following linear systems of equations.
- $\begin{cases}x+y&=3\\2x-y&=4\end{cases}$
- $\begin{cases}-x + 4y&=8\\3x - 12y&=2\end{cases}$

This
link
will show you how to specify which variable is $t$ when using
Wolfram Alpha.

Find all solutions to the linear system $\begin{cases}x+y+z&=3\\2x-y&=4\end{cases}$.
Since there are more variables than equations, this
suggests there is probably not just one solution, but perhaps
infinitely many. One common way to deal with solving such a
system is to let one variable equal $t$, and then solve for the
other variables in terms of $t$. Do this three different ways.
- If you let $x=t$, what are $y$ and $z$. Write your solution in the form $(x,y,z)$ where you replace $x$, $y$, and $z$ with what they are in terms of $t$.
- If you let $y=t$, what are $x$ and $z$ (in terms of $t$).
- If you let $z=t$, what are $x$ and $y$.

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.

Let $f(x)=e^x$. You should find that your work on each step can
be reused to do the next step.

- Find a first degree polynomial $P_1(x)=a+bx$ so that $P_1(0)=f(0)$ and $P'_1(0)=f'(0)$. In other words, give me a line that passes through the same point and has the same slope as $f(x)=e^x$ does at $x=0$. Set up a system of equations and then find the unknowns $a$ and $b$. The next two are very similar.
- Find a second degree polynomial $P_2(x)=a+bx+cx^2$ so that $P_2(0)=f(0)$, $P'_2(0)=f'(0)$, and $P''_2(0)=f''(0)$. In other words, give me a parabola that passes through the same point, has the same slope, and has the same concavity as $f(x)=e^x$ does at $x=0$.
- Find a third degree polynomial $P_3(x)=a+bx+cx^2+dx^3$ so that $P_3(0)=f(0)$, $P'_3(0)=f'(0)$, $P''_3(0)=f''(0)$, and $P'''_3(0)=f'''(0)$. In other words, give me a cubic that passes through the same point, has the same slope, the same concavity, and the same third derivative as $f(x)=e^x$ does at $x=0$.
- Now compute $e^{0.1}$ with a calculator. Then compute $P_1(.1)$, $P_2(.1)$, and $P_3(.1)$. How accurate are the line, parabola, and cubic in approximating $e^{.1}$?

The polynomial you are
creating is often called a Taylor
polynomial.

Now let $f(x)=\sin x$. Find a 7th degree polynomial so
that the function and the polynomial have the same value and
same first seven derivatives when evaluated at $x=0$. Evaluate
the polynomial at $x=0.3$. How close is this value to your
calculator's estimate of $\sin(0.3)$? You may find it valuable
to use the notation $$P(x) = a_0+a_1x+a_2x^2+a_3x^3 +\cdots+a_7
x^7.$$
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 $x=1$.

Let $f(x)=e^x$.

- Find a second degree polynomial $$T(x)=a+bx+cx^2$$ so that $T(1)=f(1)$, $T'(1)=f'(1)$, and $T''(1)=f''(1)$. In other words, give me a parabola that passes through the same point, has the same slope, and the same concavity as $f(x)=e^x$ does at $x=1$.
- Find a second degree polynomial written in the form Notice that we just replaced $x$ with $x-1$. This centers, or shifts, the approximation to be at $x=1$. The first part will be much simpler now when you let $x=1$. $$S(x)=a+b(x-1)+c(x-1)^2$$ so that $S(1)=f(1)$, $S'(1)=f'(1)$, and $S''(1)=f''(1)$. In other words, find a quadratric that passes through the same point, has the same slope, and the same concavity as $f(x)=e^x$ does at $x=1$.
- Find a third degree polynomial written in the form $$P(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3$$ so that $P(1)=f(1)$, $P'(1)=f'(1)$, $P''(1)=f''(1)$, and $P'''(1)=f'''(1)$. In other words, give me a cubic that passes through the same point, has the same slope, the same concavity, and the same third derivative as $f(x)=e^x$ does at $x=1$.

The problems above introduced the final idea. We will let
$dx=x-c$ (a small change in $x$). Since the polynomial should be
close to the function, a small change in $y$ is $dy=P_n(x)-f(c)$.
We can write the Taylor polynomial notation above in the form
$$P_n(x) -
f(c)=dy=\frac{f'(c)}{1!}dx+\frac{f''(c)}{2!}dx^2+\cdots
+\frac{f^{(n)}(c)}{n!}dx^n. $$ If we want to estimate the
change in $y$ using a first order approximation, this gives us
the differential notation $$dy = f'(c)dx.$$ A second order
approximation is $$dy = f'(c)dx + \frac{f''(c)}{2!}dx^2.$$ A
third order approximation is $$dy = f'(c)dx +
\frac{f''(c)}{2!}dx^2+ \frac{f'''(c)}{3!}dx^3.$$

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.

- A first order approximation yields $P=36\pi+ 36\pi\cdot 0.02 =36.72\pi.$ The volume increased by $0.72\pi$ m$^3$.
- A second order approximation yields $$P=36\pi+ 36\pi\cdot 0.02 +12\pi (0.02)^2 =36.7248\pi.$$
- A third order approximation yields $$P=36\pi+36\pi\cdot 0.02 +12\pi (0.02)^2+\frac{4}{3}\pi(0.02)^3 =36.724810\bar6\pi.$$

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.

Ask me in class to draw a 3D graph which illustrates the
volume added on by each successive approximation. As a
challenge, try to construct this graph yourself first. If you
have it before I put it up in class, let me know and I'll let
you share what you have discovered with the class.

Suppose you are constructing a cube whose side length should be
$s=2$ units. The manufacturing process is not exact, but
instead creates a cube with side lengths $s=2+ds$ units. (You
should assume that all sides are still the same, so any error
on one side is replicated on all. We have to assume this for
now, but before the semester ends we'll be able to do this with
high dimensional calculus.)
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 $V(2.3)-V(2)$.

I would include this problem above in class as a demonstration.
It allows them to see what differentials do. You can easily see
how each new piece adds more to the approximation.