Exercises

01

Indefinite vector integrals

Compute the indefinite vector integrals:

• (a) $\int \frac{1}{1+t^2}{\mathbf{e}}_{\mathbf{i}}+t{e}^{t^2}{\mathbf{e}}_{\mathbf{j}}+\sqrt{t}{\mathbf{e}}_{\mathbf{k}}dt$
• (b) $\int (t\cos (t^2),t^{-1},{\sec }^2(t))dt$

02

Definite vector integrals

Compute the definite vector integrals:

• (a) ${\int }_a^b(1,t^2,\sin (t))dt$
• (b) ${\int }_0^1\left(\frac{1}{1+s^2},\frac{s}{1+s^2}\right)ds$
• (c) ${\int }_0^{1}t{e}^{-t^2}{\mathbf{e}}_{\mathbf{i}}+t\ln (t^2+1){\mathbf{e}}_{\mathbf{j}}dt$
• (d) ${\int }_0^t(3s,6s^2,9)ds$

03

Differential equations

For the differential equations below, find the general solution, and find the particular solution that satisfies the given initial condition:

• (a) $\frac{d\mathbf{r}}{dt}=(1-3t,4t),\mathbf{r}(0)=(3,1)$
• (b) ${\mathbf{r}}^{\prime }=t^2{\mathbf{e}}_{\mathbf{i}}+5t{\mathbf{e}}_{\mathbf{j}}+{\mathbf{e}}_{\mathbf{k}},\mathbf{r}(1)={\mathbf{e}}_{\mathbf{j}}+2{\mathbf{e}}_{\mathbf{k}}$
• (c) ${\mathbf{r}}^{\prime \prime }(t)=(e^{2t-2},t^2-1,2),\mathbf{r}(0)=(1,0,0),{\mathbf{r}}^{\prime }(0)=(0,1,0)$

04

Prepare to explain the following reasoning in class:

Planet motion stays in some plane: $\mathbf{r}\times {\mathbf{r}}^{\prime }=\mathbf{n}$

Consider the following calculation:

$$\begin{array}{rl}\frac{d}{dt}(\mathbf{r}\times {\mathbf{r}}^{\prime })& ={\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime }+\mathbf{r}\times {\mathbf{r}}^{\prime \prime }\\ & =0+\mathbf{r}\times \left(-\frac{GM}{|\mathbf{r}{|}^3}\mathbf{r}\right)=0.\end{array}$$

By integrating both sides of this vector equation, we find $\mathbf{r}\times {\mathbf{r}}^{\prime }=\mathbf{n}$ for some constant vector $\mathbf{n}$. So $\mathbf{n}$ is perpendicular to $\mathbf{r}(t)$ for all $t$, which means $\mathbf{r}(t)$ lies in the plane through the origin with normal vector $\mathbf{n}$.

05

Prepare to explain the following reasoning in class:

Equation to integrate: $({\mathbf{r}}^{\prime }\times \mathbf{n}{)}^{\prime }=GM{\mathbf{e}}_{\mathbf{r}}^{\prime }$

Next we verify this equation. The product rule gives $({\mathbf{r}}^{\prime }\times \mathbf{n}{)}^{\prime }={\mathbf{r}}^{\prime \prime }\times \mathbf{n}$. Now compute this using the vector differential equation and the definition of $\mathbf{n}$:

$${\mathbf{r}}^{\prime \prime }\times \mathbf{n}=-\frac{GM}{|\mathbf{r}{|}^3}\mathbf{r}\times (\mathbf{r}\times {\mathbf{r}}^{\prime }).$$

Expand the triple product with the Lagrange triple product identity $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})=\mathbf{b}(\mathbf{a}\cdot \mathbf{c})-\mathbf{c}(\mathbf{a}\cdot \mathbf{b})$, and let $r=|\mathbf{r}|$, so $\mathbf{r}=r{\mathbf{e}}_{\mathbf{r}}$, and calculate:

$$\begin{array}{rl}-\frac{GM}{|\mathbf{r}{|}^3}\mathbf{r}\times (\mathbf{r}\times {\mathbf{r}}^{\prime })& =-\frac{GM}{r^3}\left(\mathbf{r}(\mathbf{r}\cdot {\mathbf{r}}^{\prime })-{\mathbf{r}}^{\prime }(\mathbf{r}\cdot \mathbf{r})\right)\\ & =-\frac{GM}{r^3}\left(r{\mathbf{e}}_{\mathbf{r}}(r{\mathbf{e}}_{\mathbf{r}}\cdot (r^{\prime }(t){\mathbf{e}}_{\mathbf{r}}+r{\mathbf{e}}_{\mathbf{r}}^{\prime }))-(r^{\prime }(t){\mathbf{e}}_{\mathbf{r}}+r{\mathbf{e}}_{\mathbf{r}}^{\prime })r^2\right)\\ & =-\frac{GM}{r}\left({\mathbf{e}}_{\mathbf{r}}(r^{\prime }(t)+0)-r^{\prime }(t){\mathbf{e}}_{\mathbf{r}}-r{\mathbf{e}}_{\mathbf{r}}^{\prime }\right)\\ & =GM{\mathbf{e}}_{\mathbf{r}}^{\prime }.\end{array}$$

06

Prepare to explain the following reasoning in class:

Result of integration: ${\mathbf{r}}^{\prime }\times \mathbf{n}=GM{\mathbf{e}}_{\mathbf{r}}+\mathbf{c}$

Take the indefinite vector integral on both sides of $({\mathbf{r}}^{\prime }\times \mathbf{n}{)}^{\prime }=GM{\mathbf{e}}_{\mathbf{r}}^{\prime }$ and obtain ${\mathbf{r}}^{\prime }\times \mathbf{n}=GM{\mathbf{e}}_{\mathbf{r}}+\mathbf{c}$ where $\mathbf{c}$ is some constant vector.

Notice that $\mathbf{c}\perp \mathbf{n}$ by taking $-\cdot \mathbf{n}$ on both sides of the stated equation, using that ${\mathbf{e}}_{\mathbf{r}}\perp \mathbf{n}$. This means $\mathbf{c}$ lies in the same plane of the orbit that $\mathbf{r}(t)$ lies in.

Let $c:=|\mathbf{c}|$ and let $\theta (t)$ be the angle between $\mathbf{r}(t)$ and $\mathbf{c}$ in the plane perpendicular to $\mathbf{n}$. Also let $n:=|\mathbf{n}|$.

07

Prepare to explain the following reasoning in class:

Triple product for scalar equation

We take the dot product $\mathbf{r}\cdot -$ on both sides of the previous stated equation, and rearrange the triple product, in order to find a scalar equation:

$$\begin{array}{rl}\mathbf{r}\cdot ({\mathbf{r}}^{\prime }\times \mathbf{n})& =GM\mathbf{r}\cdot {\mathbf{e}}_{\mathbf{r}}+\mathbf{r}\cdot \mathbf{c}\\ \mathbf{n}\cdot (\mathbf{r}\times {\mathbf{r}}^{\prime })& =GMr+rc\cos (\theta ).\end{array}$$

Since $\mathbf{r}\times {\mathbf{r}}^{\prime }=\mathbf{n}$ and $\mathbf{n}\cdot \mathbf{n}=n^2$, we have:

$$n^2=r(GM+c\cos (\theta )),$$

so:

$$r=\frac{n^2}{GM}\left(\frac{1}{1+e\cos (\theta )}\right),$$

where $e:=\frac{c}{GM}$ is defined to simplify the expression. This, now, is the polar equation of an ellipse with eccentricity $e$.