Calculus for kinematics

The idea

Kinematics is calculus with physical meaning. Differentiating position gives velocity, and velocity gives acceleration. Integrating goes the other way. Most of the module is choosing which direction to go.

$v=\frac{ds}{dt}a=\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}\text{(a dot means }\frac{d}{dt}\text{: }\dot{s}=v,\ddot{s}=a\text{)}$

Differentiation rules

Rule	Formula	Example
Power	$\frac{d}{dx}{x}^n=n{x}^{n-1}$	$\frac{d}{dt}(4t^3)=12t^2$
Product	$(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$	$\frac{d}{dx}(x^2\cos x)=2x\cos x-x^2\sin x$
Quotient	${\left(\frac{u}{v}\right)}^{\prime }=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$
Chain	$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$	$\frac{d}{dx}{e}^{2x}=2e^{2x}$

Integration

The reverse of differentiation, and the area under a curve.

• Indefinite (no limits): always add $+c$, then find $c$ from a known condition, like the starting velocity.
• Definite (limits): no $c$. ${\int }_0^{2}2xdx=[x^2{]}_0^2=4$.

Separating the differential is how every kinematics formula is built: $v=\frac{ds}{dt}\Rightarrow ds=vdt\Rightarrow {\int }_{s_0}^{s}ds={\int }_0^{t}vdt$

Techniques you’ll meet: substitution ($\int \cos 3xdx=\frac{1}{3}\sin 3x+c$), by parts ($\int udv=uv-\int vdu$) and partial fractions.

Worked example

A robot arm joint has angular acceleration $\alpha =6t$ rad/s², starting at ${\omega }_0=2$ rad/s and ${\theta }_0=0$. Find $\omega$ and $\theta$ at $t=2$ s.

Answer

$\omega ={\omega }_0+{\int }_0^{t}6tdt=2+3t^2$ → $\omega (2)=14$ rad/s $\theta ={\int }_0^t(2+3t^2)dt=2t+t^3$ → $\theta (2)=12$ rad The starting values become the lower limits, so no $+c$ is needed.

Which way to go?

You’re given	Do
$s(t)$ or $\theta (t)$	Differentiate for $v$ and $a$
$a(t)$ or $\alpha (t)$	Integrate, using the starting values
$a$ as a function of position	Use $vdv=ads$ (or $\omega d\omega =\alpha d\theta$)
Constant $a$	Use the constant-acceleration equations (Rectilinear motion)

Check yourself

1. $s=5t^2-2t$. What's $a$?

$v=10t-2$, so $a=10$ m/s², constant.

2. $v=3t^2$ m/s and $s(0)=1$ m. Find $s(2)$.

$s=1+t^3$, so $s(2)=9$ m.