Multivariable Calculus Section III Formulas

Position Vector

The position vector is given by $\vec{r}(t)=<x(t),y(t),z(t)>$ where position is given at any moment $t$ by. This can be plotted in 3-space to yield a curve.

Arc Length

Arc length is the integral of the position vector: ${\int }_a^b(|{\vec{r}}^{\prime }(t)|)dt$ where $a\le t\le b$.

Distance

In general $|{\vec{r}}^{\prime }(t)|=\sqrt{x^2+y^2+z^2...n^2}$

Unit Tangent

Unit tangent is given by $\vec{T}(t)=\frac{{\vec{r}}^{\prime }(t)}{|{\vec{r}}^{\prime }t|}$

Unit Normal

$\vec{N}(t)=\frac{{\vec{T}}^{\prime }(t)}{|{\vec{T}}^{\prime }(t)|}$

• Derivative of unit tangent normalized to unit normal

Curvature

Vector function:

​ Curvature is given by $\kappa (t)=\frac{{\vec{T}}^{\prime }(t)}{|{\vec{r}}^{\prime }(t)|}=\frac{|{\vec{r}}^{\prime }(t)\times {\vec{r}}^{\prime \prime }(t)|}{|{\vec{r}}^{\prime }(t){|}^3}$

Generic $f(x)$:

​ $\kappa (x)=\frac{|f^{\prime \prime }(x)|}{(1+(f^{\prime }(x){)}^2{)}^{\frac{3}{2}}}$

• Always positive
• Measures how bendy

Binormal

Vector:

​ $\vec{B}=\vec{T}(t)\times \vec{N}(t)$

• To quickly calculate

  • Need:
    • Unit tangent
    • Principal unit normal
  • Find by:
    • Start with r r(t) gives space curve
    • Find derivative of R
    • Find Unit tangent
    • Find derivative of the unit tangent
    • Divide by magnitude
    • Take cross product

• For some types of curves the curvature is the same and Binormal is constant

Normal Plane

Spanned by Normal and Binormal vectors at $(x_0,y_0,z_0)$

$0={\vec{r}}^{\prime }(t_0)\cdot <x-x_0,y-y_0,z-z_0>$

Osculating Plane

Kisses space curve

Spanned by Unit Tangent and Principal unit normal at $(x_0,y_0,z_0)$

$0=\vec{B}(t_0)\cdot <x-x_0,y-y_0,z-z_0>$

Physics

Transformation

Velocity

Velocity: $\vec{v}(t)={\vec{r}}^{\prime }(t)$

Speed: $|\vec{v}(t)|$

Acceleration: $\vec{a}(t)={\vec{v}}^{\prime }(t)={\vec{r}}^{\prime \prime }(t)$

Newtons Second Law

$\vec{F}(t)=m\vec{a}(t)$ Where $\vec{f}$ is force and $m$ is mass

Projectile Motion

Defined by $\vec{r}(t)$

Begins with an initial velocity at time zero $\vec{v}(0)$

Goes up at an elevation angle $\alpha$

Acceleration is at magnitude = $g\approx 9.8m/s^2$

To map an initial velocity back to a vector function: $\vec{r}(t)=\int <|\vec{v}(0)|\cos (\alpha ),|\vec{v}(0)|\sin (\alpha )-gt>dt$

This can also be done with. This works for position after t given a starting position: $\vec{r}(0)+<|\vec{v}(0)|\cos (\alpha )t,|\vec{v}(0)|\sin (\alpha )t-\frac{1}{2}g{t}^2>$

Tangential and Normal components of Acceleration

$\vec{a}(t)=a_T\vec{T}(t)+a_N\vec{N}(t)$

$a_T=\frac{d}{dt}|\vec{v}(t)|=\frac{{\vec{r}}^{\prime }(t)\cdot {\vec{r}}^{\prime \prime }(t)}{|{\vec{r}}^{\prime }(t)|}$

$a_N=\kappa (t)|\vec{v}(t){|}^2=\frac{|{\vec{r}}^{\prime }(t)\times {\vec{r}}^{\prime \prime }(t)|}{|{\vec{r}}^{\prime }(t)|}$

Intersections/Collisions

Two space curves $\overrightarrow{r_1}(s)$ and $\overrightarrow{r_2}(t)$ intersect when $\overrightarrow{r_1}(t)=\overrightarrow{r_2}(s)$ and $s=t$

Chain Rule

Assume:

​ $z=f(x_1,x_2,...,x_n)$

​ $x_i=g_i(t_1,t_2,...,t_m)$

Then

​ $\frac{\partial z}{\partial t_k}=\frac{\partial f}{\partial x_1}\frac{\partial g_1}{\partial t_k}+\frac{\partial f}{\partial x_2}\frac{\partial g_2}{\partial t_k}+...+\frac{\partial f}{\partial x_n}\frac{\partial g_n}{\partial t_k}$

Directional Derivative

$D_{\frac{\vec{u}}{|\vec{u}|}}f=\nabla f\cdot \frac{\vec{u}}{|\vec{u}|}$

$D_{\vec{u}}f(x,y,z)=f_x(x,y,z)a+f_y(x,y,z)b+f_z(x,y,z)c$

Where $a,b,c$ are the components of the directional vector normalized to a unit vector

$a=\frac{x}{\sqrt{x^2+y^2+z^2}}$

Or more generally

$<a,b,c>=\frac{<x,y,z>}{\sqrt{x^2+y^2+z^2}}$

Vector Formulas from Earlier Sections

2D Dot Product

$\vec{u}\cdot \vec{v}=|\vec{u}||\vec{v}|\cos \theta$

Scalar Projection

Projection of $\vec{v}$ onto $\vec{u}$

$com{p}_{\vec{u}}\vec{v}=\frac{\vec{u}\cdot \vec{v}}{|\vec{u}|}$

Vector Projection

Projection of $\vec{v}$ onto $\vec{u}$

$pro{j}_{\vec{u}}\vec{v}=\frac{\vec{u}\cdot \vec{v}}{|\vec{u}|}\frac{\vec{u}}{|\vec{u}|}$