# Vector Functions

## Vector-Valued Functions

A vector function covers a set of multidimensional vectors at the intersection of the domains of [latex]f[/latex], [latex]g[/latex], and [latex]h[/latex].### Learning Objectives

Describe application of the vector valued functions### Key Takeaways

#### Key Points

- A vector valued function can be made up of vectors and/or scalars.
- Each component function in a vector valued function represents the location of the value in a different dimension.
- The domain of the vector value function is the intersection of the component function domains.
- Vector valued functions can behave the same ways as vectors, and be evaluated similarly.

#### Key Terms

**domain**: the set of all possible mathematical entities (points) where a given function is defined**vector**: a directed quantity, one with both magnitude and direction; the signed difference between two points**scalar**: a quantity that has magnitude but not direction; compare vector

### What is a Vector Valued Function?

Also called vector functions, vector valued functions allow you to express the position of a point in multiple dimensions within a single function. These can be expressed in an infinite number of dimensions, but are most often expressed in two or three. The input into a vector valued function can be a vector or a scalar. In this atom we are going to introduce the properties and uses of the vector valued functions.### Properties of Vector Valued Functions

A vector valued function allows you to represent the position of a particle in one or more dimensions. A three-dimensional vector valued function requires three functions, one for each dimension. In Cartesian form with standard unit vectors (i,j,k), a vector valued function can be represented in either of the following ways: [latex]\mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k}\\ \mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle[/latex] where [latex]t[/latex] is being used as the variable. This is a three dimensional vector valued function. The graph shows a visual representation of [latex]\mathbf{r}(t) = \langle 2 \cos(t), 4 \sin(t), t \rangle[/latex]. This can be broken down into three separate functions called component functions: [latex]x(t) = 2 \cos(t)y(t) = 4 \sin(t)z(t) = t[/latex]. If you were to take a cross section, with the cut perpendicular to any of the three axes, you would see the graph of that function. For example, if you were to slice the three-dimensional shape perpendicular to the [latex]z[/latex]-axis, the graph you would see would be of the function [latex]z(t)=t[/latex].The domain of a vector valued function is a domain that satisfies all of the component functions. It can be found by taking the intersection of the individual component function domains. The vector valued functions can be manipulated in the same way as a vector; they can be added, subtracted, and the dot product and the cross product can be found.### Example

For this example, we will use time as our parameter. The following vector valued function represents time, [latex]t[/latex]: [latex-display]\mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k}[/latex-display] This function is representing a position. Therefore, if we take the derivative of this function, we will get the velocity: [latex-display]\displaystyle{\frac{d\mathbf{r}(t)}{dt}= f(t)\mathbf{i}' + g(t)\mathbf{j}' + h(t)\mathbf{k}' \\ \,\qquad = \mathbf{v}(t)\\ }[/latex-display] If we differentiate a second time, we will be left with acceleration: [latex-display]\displaystyle{\frac{d\mathbf{v}(t)}{dt}= \mathbf{a}(t)}[/latex-display]## Arc Length and Speed

Arc length and speed are, respectively, a function of position and its derivative with respect to time.### Learning Objectives

Distinguish the arc length and speed### Key Takeaways

#### Key Points

- Arc length is found by placing a number of points along a curve, connecting them by line segments, and then adding those segment lengths together.
- The beginning of deriving the formula for arc length starts with Pythagoras's theorem.
- When finding the arc length, the integral used needs to be with respect to position, [latex]x[/latex].
- When finding the speed along a curve, the integral used needs to be with respect to time, [latex]t[/latex].

#### Key Terms

**derivative**: a measure of how a function changes as its input changes**velocity**: a vector quantity that denotes the rate of change of position with respect to time, or a speed with the directional component

### Arc Length and Speed

Since length is a magnitude that involves position, it is easy to deduce that the derivative of a length, or position, will give you the velocity —also known as speed—of a function. This is because a derivative gives you a rate of change with respect to a parameter. Velocity is the rate of change of a position with respect to time. Let's start this atom by looking at arc length with calculus.### Arc Length

The arc length is the length you would get if you took a curve, straightened it out, and then measured the length of that line. The arc length can be found using geometry, but for the sake of this atom, we are going to use integration. The arc length is approximated by connecting a finite number of points along and curve, connecting those lines to create a a string of very small straight lines, and adding them together. To find this using integration, we should start out by using the Pythagorean Theorem for length of the different sides of a triangle: [latex]\displaystyle{ds^2=dx^2+dy^2 \frac{ds^2}{dx^2}\\ \, \quad= 1+\frac{dy^2}{dx^2}ds \\ \, \quad=\sqrt{1+{\frac{dy^2}{dx^2}^2}} \cdot dx\\ \, \quad=\int^b_a\sqrt{1+f'(x)^2} \cdot dx}[/latex] where [latex]s[/latex] is the arc length. If [latex]x=X(t)[/latex] and [latex]y=Y(t)[/latex], [latex]\displaystyle{s =\int^b_a\sqrt{1+{f'(x)}^2} \cdots dx\\ \,\,=\int^b_a\sqrt{[X'(t)]^2+[Y'(t)] \cdots dt}\\ \,\,= \int^b_a\sqrt{dx^2+dy^2}\\ \,\,= \int^b_a\sqrt{\frac{dx}{dt}^2+\frac{dy}{dt}^2 \cdots dt}}[/latex] Since this is a function of position and is defined by x, we need to have a derivative that is in respect to x: [latex-display]\int^b_a\sqrt{1+\frac{dy}{dx}^2*dx}[/latex-display] Obviously some cases require polar coordinates instead of Cartesian. In polar coordinates: [latex-display]r=f(\theta)[/latex-display]### Arc Speed

Now that the hard part is over, we can easily find the speed along this curve. Since speed is in relation to time and not position, we need to revert back to the arc length with respect to time: [latex-display]\displaystyle{\int^b_a\sqrt{\frac{dx}{dt}^2+\frac{dy}{dt}^2\cdot dt}}[/latex-display] Then, differentiate with respect to time: [latex-display]\displaystyle{v(t)=s'=\sqrt{[X'(t)]^2+[Y'(t)]^2}}[/latex-display]## Calculus of Vector-Valued Functions

A vector function is a function that can behave as a group of individual vectors and can perform differential and integral operations.### Learning Objectives

Discuss how vector functions are manipulated### Key Takeaways

#### Key Points

- A vector valued function can be made up of vectors and/or scalars.
- Each component function in a vector valued function represents the location of the value in a different dimension.
- Vector valued functions can behave the same ways as vectors, and be evaluated similarly.
- Vector functions are widely used in the study of electromagnetic fields, gravitation fields, and fluid flow.

#### Key Terms

**vector**: a directed quantity, one with both magnitude and direction; the signed difference between two points**scalar**: a quantity that has magnitude but not direction; compare vector

## Arc Length and Curvature

The curvature of an object is the degree to which it deviates from being flat and can be found using arc length.### Learning Objectives

Explain the relationship between the curvature of an object and the arc length### Key Takeaways

#### Key Points

- The arc length is a function of position, so its derivative will be a function of time. This can give you the rate of change of the position, in relation to time, which is called the curvature.
- The curvature can be found by taking the derivative of the velocity vector, which is given: [latex]\Vert\frac{dT}{ds}\Vert[/latex].
- This same magnitude can also be found using the concept of calculus, the limit.

#### Key Terms

**sharpness**: the fineness of the point a pointed object**curvature**: the degree to which an objet deviates from being flat**normal**: a line or vector that is perpendicular to another line, surface, or plane

### Arc Curvature

The curvature of an arc is a value that represents the direction and sharpness of a curve. On any curve, there is a center of curvature, C. This is the intersection point of two infinitely close normals to this curve. The radius, R, is the distance from this intersection point to the center of curvature. In order to find the value of the curvature, we need to take the parameter time, s, and the unit tangent vector, which in this case is the same as the unit velocity vector, T, which is also a function of time.The curvature is a magnitude of the rate of change of the tangent vector, T: [latex-display]\kappa=\Vert\frac{dT}{ds}\Vert[/latex-display] Where [latex]\kappa[/latex] is the curvature and [latex]\frac{dT}{ds}[/latex] is the acceleration vector (the rate of change of the velocity vector over time).### How Does This Relate to Arc Length?

The curvature can also be approximated using limits. Given the points P and Q on the curve, lets call the arc length s(P,Q), and the linear distance from P to Q will be denoted as d(P,Q). The curvature of the arc at point P can be found by obtaining the limit: [latex-display]\kappa(P) = \frac{\lim}{Q\rightarrow P}\sqrt{\frac{24*(s(P,Q)-d(P,Q))}{s(P,Q)^3}}[/latex-display] In order to use this formula, you must first obtain the arc length of the curve from points P to Q and length of the linear segment that connect points P and Q. In a previous atom, we went into more detail on how to find the arc length, but for the sake of this atom we will just restate those formulas:In Cartesian coordinates: [latex-display]\int^b_a\sqrt{1+\frac{dy}{dx}^2*dx}\\ [/latex-display]## Planetary Motion According to Kepler and Newton

Kepler explained that the planets move in an ellipse around the Sun, which is at one of the two foci of the ellipse.### Learning Objectives

Identify three laws of planetary motion formulated by Johannes Kepler### Key Takeaways

#### Key Points

- Kepler's first law of planetary motion describes the motion of its orbit around the Sun.
- Kepler's second law of planetary motion explains the reason why the planet moves faster as it approaches the Sun, and slower as it moves farther away.
- Kepler's third law of planetary motion explains how the period of an orbit is related to the semi-major axis of its orbit.
- Newton takes the information presented by Kepler and uses it to explain that the value of a force on an object is the product of its mass and its orbital acceleration.
- Newton also clarifies that the orbit is an elliptical shape because while each planet is attracted to the Sun, they are also attracted to each other.

#### Key Terms

**gravitational constant**: an empirical physical constant involved in the calculation of the gravitational attraction between objects with mass**eccentricity**: the ratio—constant for any particular conic section—of the distance of a point from the focus to its distance from the directrix**ellipse**: a closed curve; the locus of a point such that the sum of the distances from that point to two other fixed points (called the foci of the ellipse) is constant; equivalently, the conic section that is the intersection of a cone with a plane that does not intersect the base of the cone

### Kepler's First Law

As we already stated, the first law of planetary motion states that the orbit of every planet is an ellipse with the Sun at one focus. In order to discuss this law, and the laws that follow, we should examine the components of an ellipse a bit more closely. The eccentricity of an ellipse tells you how stretched out the ellipse is. The eccentricity can be from 0 to 1. If the eccentricity is equal to zero, that means it is a circle. In Kepler's time, the extremes of planetary eccentricity were Venus, 0.007, and Mercury, 0.2. The eccentricity is what makes an ellipse different from a circle. [latex-display]\displaystyle{r=\frac{p}{1+{\Sigma} \cdot \cos{\theta}}}[/latex-display] where [latex]p[/latex] is the semi-latus rectum, [latex]\Sigma[/latex] is the eccentricity, [latex]r[/latex] is the distance between the planet and the sun, and [latex]\theta[/latex] is the angle between the position of the planet and its most direct route to the Sun. The minimum distance occurs when the angle is 0. The maximum distance occurs when the angle is 180 degrees. These values are important because the equation for eccentricity is: [latex-display]\displaystyle{\Sigma=\frac{r_{\text{max}}-r_{\text{min}}}{r_{\text{max}}+r_{\text{min}}}}[/latex-display] The semi-major axis, [latex]a[/latex], can be found as follows: [latex-display]\displaystyle{a=\frac{p}{1-\Sigma^2}}[/latex-display] The semi-minor axis, [latex]b[/latex], can be found as follows: [latex-display]\displaystyle{b=\frac{p}{\sqrt{1-\Sigma^2}}}[/latex-display] The area of an ellipse is found as follows: [latex-display]A=\pi \cdot a \cdot b[/latex-display]### Kepler's Second Law

The second law of planetary motion states that in an amount of time, [latex]t[/latex], a line from the planet to the Sun will sweep out a triangle having a base of [latex]r[/latex] and a height of [latex]r \cdot d\theta[/latex]. Therefore, the area of this triangle is: [latex-display]\displaystyle{dA=\frac{1}{2} \cdot r \cdot rd\theta}[/latex-display] and the ratio of the area of this triangle to the time elapsed is: [latex-display]\displaystyle{\frac{dA}{dt}=\frac{1}{2} \cdot r^2 \cdot \frac{d\theta}{dt}}[/latex-display] As the planet moves closer to the Sun, it speeds up. This allows the triangle to have an equal area in an equal amount of time regardless of position of the planet. As we learned in the first section, the area of an ellipse is [latex]\pi \cdot a \cdot b[/latex]. Therefore, the period ([latex]P[/latex]) of the ellipse satisfies: [latex]\pi \cdot a \cdot b=P \cdot \frac{1}{2} \cdot r^2 \cdot \theta\\ r^2\theta=n\cdot a\cdot b[/latex] where [latex]\theta[/latex] is the angular velocity with respect to time, and [latex]n[/latex] is the mean motion of the planet around the Sun.### Kepler's Third Law

Kepler's third law describes the relationship between the distance of the planets from the Sun, and their orbits period. [latex]P^2 \propto a^3[/latex], with a constant of proportionality of: [latex-display]\displaystyle{\frac{P^2_{\text{planet}}}{a^3_{\text{planet}}} = \frac{P^2_{\text{earth}}}{a^3_{\text{earth}}} = 1 \frac{\text{year}}{\text{AU}}}[/latex-display] where AU is an astronomical unit. In the case of a circular orbit, the proportionality constant is as follows: [latex-display]\displaystyle{\frac{4\pi^2}{T^2}=\frac{GM}{R^3}}[/latex-display] where [latex]T[/latex] is the period, [latex]G[/latex] is the gravitational constant, and [latex]R[/latex] is the distance between the center of mass of the two bodies.### How does Newton Relate to Kepler?

Newton derived his theory of the acceleration of a planet from Kepler's first and second laws. Newton theorized that the direction of a planet is always towards the Sun. In addition, the magnitude of the acceleration is inversely proportional to the square of its distance from the Sun. From this, Newton defined the force acting on a planet as the product of its mass and acceleration. Therefore, by Newton's law, every planet is attracted to the Sun, and the force acting on a planet is directly proportional to the mass and inversely proportional to the square of its distance from the Sun.## Tangent Vectors and Normal Vectors

A vector is normal to another vector if the intersection of the two form a 90-degree angle at the tangent point.### Learning Objectives

Distinguish tangent vectors and normal vectors### Key Takeaways

#### Key Points

- In order for one vector to be tangent to another vector, the intersection needs to be exactly 90 degrees. On a curve or an uneven object, each point will have a unique normal vector.
- If you want to check whether two vectors are normal to each other, you can find the dot product of the two and make sure it equals zero.
- If you want to find out exactly what the angle between the two vectors is, you can use the following equation, which also employs the dot product: [latex]\mathbf{a} \bullet \mathbf{b} = \left|\mathbf{a}\right| \left|\mathbf{b}\right| \cos \theta[/latex].
- In order to find the tangent vector to another vector or object, just take the derivative of the reference vector.

#### Key Terms

**tangent**: a straight line touching a curve at a single point without crossing it there**perpendicular**: at or forming a right angle (to)

### Normal Vectors

An object is normal to another object if it is perpendicular to the point of reference. That means that the intersection of the two objects forms a right angle. Usually, these vectors are denoted as [latex]\mathbf{n}[/latex]. Not only can vectors be ‘normal' to objects, but planes can also be normal.### Dot Product

As we covered in another atom, one of the manipulations of vectors is called the Dot Product. When you take the dot product of two vectors, your answer is in the form of a single value, not a vector. In order for two vectors to be normal to each other, the dot product has to be zero. [latex]\mathbf{a} \bullet \mathbf{b} = 0\\ \,\,\,\quad = a_1b_1+a_2b_2+a_3b_3\\ \,\,\,\quad = \left|\mathbf{a}\right| \left|\mathbf{b}\right| \cos \theta[/latex]### Tangent Vectors

Tangent vectors are almost exactly like normal vectors, except they are tangent instead of normal to the other vector or object. These vectors can be found by obtaining the derivative of the reference vector, [latex]\mathbf{r}(t)[/latex]: [latex-display]\mathbf{r}(t) = f(t)\mathbf{i}+g(t)\mathbf{j}+h(t)\mathbf{k}[/latex-display]## Licenses & Attributions

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