Theory

Vector integrals

A parametric space curve $\mathbf{r}(t)=(x(t),y(t),z(t))$ has a definite vector integral that is defined componentwise:

$${\int }_a^b\mathbf{r}(t)dt=\left({\int }_a^{b}x(t)dt,{\int }_a^{b}y(t)dt,{\int }_a^{b}z(t)dt\right).$$

The vector integral relates directly to the componentwise derivative ${\mathbf{r}}^{\prime }(t)$ by a vector form of the Fundamental Theorem of Calculus:

$${\int }_a^b\mathbf{r}(t)dt=\mathbf{R}(b)-\mathbf{R}(a)$$

where $\mathbf{R}(t)$ is a vector antiderivative of $\mathbf{r}(t)$, meaning that ${\mathbf{R}}^{\prime }(t)=\mathbf{r}(t)$.

Using the standard calculus argument applied to each component separately, one has the fact that:

$${\mathbf{R}}_1^{\prime }(t)={\mathbf{R}}_1^{\prime }(t)⟹{\mathbf{R}}_1(t)={\mathbf{R}}_1(t)+\mathbf{c},$$

where $\mathbf{c}=(c_1,c_2,c_3)$ is some constant vector. This means that if ${\mathbf{R}}^{\prime }(t)=\mathbf{r}(t)$, so $\mathbf{R}(t)$ is a particular antiderivative of $\mathbf{r}(t)$, then the general antiderivative is given by $\mathbf{R}(t)+\mathbf{c}$ for a constant vector $\mathbf{c}$. So we can write the indefinite vector integral:

$$\int \mathbf{r}(t)dt=\mathbf{R}(t)+\mathbf{c}.$$

The vector integral ${\int }_a^{b}v(t)dt$ may be interpreted as the total motion of a particle whose velocity is given by $v(t)$.

Integration for components, not coordinates

If a parametric curve is given using spherical coordinate functions, for example $(r(t),\theta (t),\phi (t))$, then the componentwise derivatives and integrals do not have vector interpretations as tangent vector and total motion vector. The meaning of vector derivative and integral depends on the componentwise addition and scaling operations applicable to vectors.

Position, velocity, acceleration

If the position of a moving particle is identified using the vector-valued parametric function $\mathbf{r}(t)$, then its velocity is given by ${\mathbf{r}}^{\prime }(t)$ and its vector acceleration is ${\mathbf{r}}^{\prime \prime }(t)$.

The definite vector integral reverses this progression. We have ${\int }_a^b{\mathbf{r}}^{\prime \prime }(t)dt={\mathbf{r}}^{\prime }(b)-{\mathbf{r}}^{\prime }(a)$, the total change in velocity over the time interval $a\le t\le b$. Similarly ${\int }_a^b{\mathbf{r}}^{\prime }(t)dt=\mathbf{r}(b)-\mathbf{r}(a)$, the total change in position over the same interval.

Kepler’s Law of Ellipses

Isaac Newton’s laws of motion were used to derive and explain Johannes Kepler’s observed law that planets orbit the sun with elliptical orbits. We can use vector integration to solve a differential equation coming from Newton’s laws to give a modern derivation of this law of elliptical orbits.

Set the sun at the origin $(0,0,0)$, and let $\mathbf{r}(t)$ describe the orbit of some planet. Let $G$ be the gravitational constant, $M$ the mass of the sun, and $m$ the mass of the planet. Let $\mathbf{F}(t)$ be the total force vector on the planet. By Newton’s law of gravitation:

$$\mathbf{F}=-\frac{GMm}{|\mathbf{r}{|}^2}{\mathbf{e}}_{\mathbf{r}}.$$

By Newton’s law of motion, $\mathbf{F}=m\mathbf{a}$ where the acceleration is $\mathbf{a}(t)={\mathbf{r}}^{\prime \prime }(t)$. By identifying the force in each equation, we obtain the vector differential equation:

$${\mathbf{r}}^{\prime \prime }(t)=-\frac{GM}{|\mathbf{r}(t){|}^2}{\mathbf{e}}_{\mathbf{r}}(t)=-\frac{GM}{|\mathbf{r}(t){|}^3}\mathbf{r}(t).$$

Kepler’s Law of Ellipses (homework setup)

By Kepler’s First Law, a planetary orbit is an ellipse, which lies in a plane. Assume this plane is the $xy$-plane, so $\mathbf{r}(t)=(x(t),y(t),0)$. Therefore $\mathbf{n}=\mathbf{r}\times {\mathbf{r}}^{\prime }$ may be written $\mathbf{n}=(0,0,n)$. Now rotate the $xy$-coordinates so that $\mathbf{c}=(c,0,0)$ and therefore the angle $\theta$ between $\mathbf{r}$ and $\mathbf{c}$ is also the polar angle of $\mathbf{r}$, so $\mathbf{r}(t)=(r(t)\cos (\theta (t)),r(t)\sin (\theta (t)),0)$ where $(r(t),\theta (t))$ describes the trajectory in polar coordinates in the $xy$-plane.