Kinematics Using Curves

Parametric curves are extremely useful in kinematics-style problems. A body moving in 3D or 2D, if well-defined using a parametric curve, can be differentiated to obtain kinematic properties like velocity.

If we have $\vec{c}(t)$ where $\vec{c}:\mathbb{R}\to {\mathbb{R}}^3$ that is, we have:

$$\vec{c}(t)=(x(t),y(t),z(t))x,y,z:\mathbb{R}\to \mathbb{R}$$

we can let this be the function for position along time, in three dimensions.

Velocity §

$$\vec{v}(t)=\frac{\text{d}\vec{c}}{\text{d}t}={\vec{c}}^{{}^{\prime }}(t)=\left(\frac{\text{d}x}{\text{d}t},\frac{\text{d}y}{\text{d}t},\frac{\text{d}z}{\text{d}t}\right)$$

Note that the velocity is always tangential to the the position curve.

Speed §

$$\Vert \vec{v}(t)\Vert =\Vert \frac{\text{d}\vec{c}}{\text{d}t}\Vert =\Vert {\vec{c}}^{{}^{\prime }}(t)\Vert =\Vert \left(\frac{\text{d}x}{\text{d}t},\frac{\text{d}y}{\text{d}t},\frac{\text{d}z}{\text{d}t}\right)\Vert =\sqrt{{\frac{\text{d}x}{\text{d}t}}^2+{\frac{\text{d}y}{\text{d}t}}^2+{\frac{\text{d}z}{\text{d}t}}^2}$$

Acceleration §

$$\vec{a}(t)=\frac{{\text{d}}^2\vec{c}}{\text{d}{t}^2}={\vec{c}}^{{}^{\prime \prime }}(t)=\frac{\text{d}\vec{v}}{\text{d}t}={\vec{v}}^{{}^{\prime }}(t)$$

Tangent Line §

The tangent line at a given value $t=t_0$

$$\ell (t)=\vec{c}(t_0)+(t-t_0)\cdot \vec{v}(t_0)$$

Examples §

1: Kinematics of a 3D curve

Let:

$$\vec{c}(t)=(3t+2,t^2-7,t-t^2)$$

Find:

1. Velocity

2. Speed

3. Acceleration

4. The equation of the tangent line at $t=1$

Solution $\vec{c}:\mathbb{R}\to {\mathbb{R}}^3$, it is a valid parametric curve and we can deduce kinematic properties using differentiation:

Since

$$\vec{v}(t)=\frac{\text{d}\vec{c}}{\text{d}t}=(3,2t,1-2t)$$

$$\Vert \vec{v}(t)\Vert =\Vert (3,2t,1-2t)\Vert =\sqrt{3^2+(2t{)}^2+(1-2t{)}^2}=\sqrt{9+4t^2+1-4t^2+4t^2}=\sqrt{4t^2+10}$$

$$\vec{a}(t)=\frac{\text{d}\vec{v}}{\text{d}t}=(0,2,-2)$$

$$\begin{array}{rl}\ell (t)& =\vec{c}(1)+(t-1)\vec{v}(1)\\ & =(5,-6,0)+(t-1)\times (3,2,-1)\\ & =(2,-8,1)+t\cdot (3,2,-1)\end{array}$$

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