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# Vector Calculus – Definition, Summary, and Vector Analysis

**Vector calculus**opens the door to different types of functions and analyses we can use in different fields. Learning about the core components and the theorems behind vector calculus allows us to describe and study quantities and relationships defined by vector-valued functions.

**This is why we’re writing this article – to prepared and give you an idea of what to expect and the topics you’ll encounter in vector calculus. In short, our discussion will simply give you a glance at this extensive topic!**

*In vector calculus, we study the differentiation and integration of vector functions. This field is closely related to multivariable calculus.***What is vector calculus?**

**Vector calculus is simply the study of a vector field’s differentiation and integration**

**. It’s a core branch in calculus that covers all key concepts to master differentiating and integrating all kinds of vector functions. In vector calculus, we’ll explore the following:**

- Mastering the fundamentals of vector quantities.
- Understanding how vector-valued functions behave in 2D and 3D coordinate systems.
- Evaluate single and even multiple integrals of vector-valued functions.
- Learning how to integrate vector fields (an important technique in higher physics).

**Summary of Vector Calculus**

As we have mentioned, vector calculus and multivariable calculus are closely related. That’s because we’ll learn how to integrate, differentiate, and apply multiple integral operations on vector and multivariable functions at the same time.\begin{aligned}f(x + y) &= 4x^3 – e^y\\ f(t) &= (t^2, 2\ln t)\end{aligned}These are examples of functions that we’ll be dealing with in vector calculus. To give you a better idea on what to expect from this branch of calculus, we’ve summarized some key formulas and concepts you’ll learn in vector calculus:Name | Common Notation | Derivative | Integral |

Curve | \begin{aligned}\textbf{r}(t) \end{aligned} | \begin{aligned}\dfrac{d}{dt} \textbf{r}(t) &=\textbf{r}^{\prime}(t) \end{aligned} | We can apply line integral. |

Scalar Field | \begin{aligned}f(x)\end{aligned} | \begin{aligned}\text{grad} f &= \overrightarrow{\bigtriangledown} f\end{aligned} | Evaluate integrals using:· Double or triple integral· Surface integrals |

Parametrized Surface | \begin{aligned}\textbf{D}(x,y) \end{aligned} | \begin{aligned}\textbf{D}_x &=\dfrac{\partial D}{\partial x}\\\textbf{D}_y &=\dfrac{\partial D}{\partial y} \end{aligned} | Evaluate integrals using surface integrals. |

Vector Field | \begin{aligned}\textbf{F}(\text) \ebf{x}nd{aligned} | Divergence:\begin{aligned}\overrightarrow{\bigtriangledown} \cdot \textbf{F}\end{aligned}Curl:\begin{aligned}\overrightarrow{\bigtriangledown} \times \textbf{F}\end{aligned} | Evaluate integrals using line integrals and surface integrals of vector fields. |

**How to prepare for vector calculus through vector analysis?**

Vector analysis covers all the fundamental operations we can perform on scalars and vectors. It’s crucial that we know vector analysis by heart before expanding our knowledge base.In the past, we’ve learned how to perform basic operations on vectors such as: vector addition, subtraction, and scalar multiplication. Here are some examples of how we perform these operations:\begin{aligned}\textbf{u} &= <-3, 2, 1>\\ \textbf{v} &= <4, 6, 8>\\\\ \textbf{u} + \textbf{v} &= <-3 + 4, 2 + 6, 1 + 8>\\&= <1, 8, 9>\\ 2\textbf{v} &= <2 \cdot 4, 2\cdot 6, 2\cdot 8>\\&= <8, 12, 16> \end{aligned}In fact, we’ve also defined the magnitude of vectors as $|\textbf{v}| = \sqrt{v_1^2 + v_2^ + v_3^2}. We’ve also covered dot and cross products of vectors as part of our vector analysis tool kit.\begin{aligned}\textbf{u} &= <u_1, u_2, u_3>\\ \textbf{v} &= <v_1, v_2, v_3>\\\\ \textbf{u} \cdot \textbf{v} &= u_1v_1 + u_2v_2 + u_3v3 \\\textbf{u} \times \textbf{v} &= \end{aligned}In fact, through the cross and dot products, we’ve learned how to also relate the two vectors and the angle formed between them.\begin{aligned}\textbf{u} \cdot \textbf{v} &= |\textbf{u}||\textbf{v}|\cos \theta\\ \cos \theta &= \dfrac{\textbf{u} \cdot \textbf{v}}{|\textbf{u}||\textbf{v}|}\\\\\textbf{u} \times \textbf{v} &= |\textbf{u}||\textbf{v}|\sin\theta\\ \sin\theta &= \dfrac{\textbf{u} \times \textbf{v}}{|\textbf{u}||\textbf{v}|}\end{aligned}We’ve also defined normal and orthogonal vectors in the past and knowing these properties will be a great help when simplifying vectors.Orthogonal Vectors | Normal Vectors |

\begin{aligned}\textbf{u} \cdot \textbf{v} &= 0\end{aligned} | \begin{aligned}\textbf{u} \times \textbf{v} &= |\textbf{u}||\textbf{v}|\end{aligned} |

**Given that $\textbf{u} = 4\textbf{i} – 2\textbf{j} + 3\textbf{k}$ and $\textbf{u} = 2\textbf{i} + 3\textbf{j} – \textbf{k}$, calculate the following:a. $4\textbf{u} – 3\textbf{v}$ b. $\textbf{u} \cdot \textbf{v}$ c. $\textbf{u} \times \textbf{v}$**

*Example 1*__Solution__To find $4\textbf{u} – 3\textbf{v}$, simply distribute the scalar factors to $\textbf{u}$ and $\textbf{v}$ then subtract the result.\begin{aligned}4\textbf{u} &= 16\textbf{i} – 8\textbf{j} + 12\textbf{k} \\3\textbf{v} &= 4\textbf{i} + 6\textbf{j} – 2\textbf{k}\\\\ 4\textbf{u} – 3\textbf{v}&= (16 – 4)\textbf{i} + (-8- 6)\textbf{j} + (12 – -2)\textbf{k}\\&= 12\textbf{i} – 14\textbf{j} + 14\textbf{k}\end{aligned}Apply the appropriate formulas to find the two vectors’ dot and cross products.\begin{aligned}\textbf{u} &= <u_1, u_2, u_3>\\ \textbf{v} &= <v_1, v_2, v_3>\\\\ \textbf{u} \cdot \textbf{v} &= u_1v_1 + u_2v_2 + u_3v3 \\\textbf{u} \times \textbf{v} &= \end{aligned}Hence, we have the following:

\begin{aligned}\textbf{u} \cdot \textbf{v} \end{aligned} | \begin{aligned}\textbf{u} \cdot \textbf{v} &= 4(2) + (-2)(3) + 3(-1)\\&= 8 – 6 -3\\&=1\end{aligned} |

\begin{aligned}\textbf{u} \times \textbf{v} \end{aligned} | \begin{aligned}\textbf{u} \times \textbf{v} &= (-2\cdot -1 – 3\cdot 3)\textbf{i} + (3\cdot 2 – 4\cdot -1)\textbf{j} + (4\cdot 3- -2\cdot 2)\textbf{k}\\&= (-7)\textbf{i} + (10)\textbf{j} + (16)\textbf{k} \\&=-7\textbf{i} + 10\textbf{j} + 16\textbf{k} \end{aligned} |

**What is the angle between $\textbf{u} = <2, 3, -1>$ and $\textbf{u} = <1, 2, 1>$?**

*Example 2*__Solution__First, calculate their magnitudes as shown below:\begin{aligned}|\textbf{u}| &= \sqrt{2^2 + 3^2 + (-1)^2}\\&= \sqrt{14}\\|\textbf{v}| &= \sqrt{1^2 + 2^2 + 1^2}\\&= \sqrt{6}\end{aligned}Now, take the dot product of the two vectors first. Hence, we have\begin{aligned}\textbf{u} \cdot \textbf{v} &= 2(1) + (3)(2) + -1(1)\\&= 2 + 6 -1\\&= 7\end{aligned}Use the fact that $\cos \theta = \dfrac{\textbf{u} \cdot \textbf{v}}{|\textbf{u}||\textbf{v}|}$.\begin{aligned}\cos \theta &= \dfrac{7}{ \sqrt{14} \cdot sqrt{6}}\\ \theta &= \cos^{-1} \left(\dfrac{7}{\sqrt{84}}\right) &\approx 40.20^{\circ}\end{aligned}This means that the angle formed between $\textbf{u} = <2, 3, -1>$ and $\textbf{u} = <1, 2, 1>$ is equal to $40.20^{\circ}$.