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Multivariable calculus, also known as calculus of several variables, is a branch of calculus that deals with the study of functions of several variables. In contrast to single-variable calculus, where functions typically have only one independent variable, multivariable calculus involves functions that have more than one independent variable.

The fundamental concepts in multivariable calculus include partial derivatives, gradients, vector fields, line integrals, surface integrals, and volume integrals. These concepts are used to study a wide range of phenomena in mathematics, physics, engineering, and other scientific fields.

In multivariable calculus, one of the key ideas is that the behavior of a function of several variables can be analyzed by considering its behavior along various lines, curves, or surfaces. This leads to the concept of partial derivatives, which describe how a function changes when one of its variables is varied while holding all the other variables constant.

The gradient of a function is a vector that points in the direction of the steepest increase of the function and its magnitude gives the rate of increase in that direction. Vector fields, which assign a vector to each point in space, are used to model phenomena such as fluid flow, electric and magnetic fields, and gravitational fields.

Line integrals involve integrating a function along a curve, while surface integrals involve integrating a function over a surface. These concepts are used to calculate quantities such as work, flux, and circulation in a variety of physical systems.

Volume integrals involve integrating a function over a three-dimensional region and are used to calculate quantities such as mass, center of mass, and moments of inertia in physical systems.

Multivariable calculus is an essential tool in many areas of mathematics, physics, and engineering, and has applications in fields such as computer graphics, economics, and biology.

A parking lot has 66 vehicles (cars, trucks, motorcycles and bicycles) in it. There are four times as many cars as trucks. The total number of tires ( 4 per car or truck, 2 per motorcycle or bicycle) is 252 . How many cars are there? How many bicycles?

Let $c, t, m, b$ denote the number of cars, trucks, motorcycles, and bicycles. Then the statements from the problem yield the equations:

$$

\begin{aligned}

c+t+m+b & =66 \

c-4 t & =0 \

4 c+4 t+2 m+2 b & =252

\end{aligned}

$$

We form the augmented matrix for this system and row-reduce

$$

\left[\begin{array}{ccccc}

1 & 1 & 1 & 1 & 66 \

1 & -4 & 0 & 0 & 0 \

4 & 4 & 2 & 2 & 252

\end{array}\right] \stackrel{\text { RREF }}{\longrightarrow}\left[\begin{array}{ccccc}

1 & 0 & 0 & 0 & 48 \

0 & 1 & 0 & 0 & 12 \

0 & 0 & 1 & 1 & 6

\end{array}\right]

$$

The first row of the matrix represents the equation $c=48$, so there are 48 cars. The second row of the matrix represents the equation $t=12$, so there are 12 trucks. The third row of the matrix represents the equation $m+b=6$ so there are anywhere from 0 to 6 bicycles. We can also say that $b$ is a free variable, but the context of the problem limits it to 7 integer values since you cannot have a negative number of motorcycles.

A wholesaler supplies products to 10 retail stores, each of which will independently make an order on a given day with chance $0.35$. What is the probability of getting exactly 2 orders? Find the most probable number of orders per day and the probability of this number of orders. Find the expected number of orders per day.

Using the independence of orders the chance that only the first two stores place orders is $0.35^2 \cdot 0.65^8$. As there are $10 \times 9 / 2=45$ distinct pairs of stores that could order we have

$$

P(X=2)=450.35^2 0.65^8=0.1757

$$

A similar argument works for any number of orders. We say that the number of orders placed has the $\operatorname{Bin}(10,0.35)$ distribution. The formula for $x$ orders is

$$

P(X=x)=\left(\begin{array}{c}

10 \

x

\end{array}\right) 0.35^x 0.65^{10-x}

$$

The most probable number of orders is 3 (either calculate $P(X=x)$ for a few different $x$ values or look at binomial tables in a textbook) and $P(X=3)=120(0.35)^3(0.65)^7 \approx 0.2522$. The expected number of orders is

$$

\sum_0^{10} x \cdot P(X=x)=1 \cdot 0.0725+2 \cdot 0.1757+3 \cdot 0.2522+4 \cdot 0.2377+\cdots

$$

which (barring numericals errors) will give the same answer as the formula $E(X)=n p=10 \times 0.35=$ $3.5$.

(The problem of which number is most likely for general $n$ and $p$ was not set but is not all that hard – show that $P(X=x+1)(n+1) p$ and think about what that means $)$

Consider the system of linear equations $\mathcal{L S}(A, \mathbf{b})$, and suppose that every element of the vector of constants $\mathbf{b}$ is a common multiple of the corresponding element of a certain column of $A$. More precisely, there is a complex number $\alpha$, and a column index $j$, such that $[\mathbf{b}]i=\alpha[A]{i j}$ for all $i$. Prove that the system is consistent.

The condition about the multiple of the column of constants will allow you to show that the following values form a solution of the system $\mathcal{C S}(A, \mathbf{b})$,

$$

x_1=0 \quad x_2=0 \quad \ldots \quad x_{j-1}=0 \quad x_j=\alpha \quad x_{j+1}=0 \quad \ldots \quad x_{n-1}=0 \quad x_n=0

$$

With one solution of the system known, we can say the system is consistent (Definition CS).

A more involved proof can be built using Theorem RCLS. Begin by proving that each of the three row operations (Definition RO) will convert the augmented matrix of the system into another matrix where column $j$ is $\alpha$ times the entry of the same row in the last column. In other words, the “column multiple property” is preserved under row operations. These proofs will get successively more involved as you work through the three operations.