A Cartesian equation may be obtained by elimination of
\[\]
.Example: Solve the equation
\[\frac{d^2 \mathbf{r}}{dt^2}+2 \mathbf{i}=0\]
with the initial conditions - where \[t=0\]
, \[\mathbf{v}=2 \mathbf{i}+3 \mathbf{j}\]
m/s, \[\mathbf{r}= \mathbf{j}\]
mWith
\[\mathbf{r}= x \mathbf{i} + y \mathbf{j}\]
, the equations of the components are\[\frac{d^2x}{dt^2}+2=0\]
(1_\[\frac{d^2y}{dt^2}=0\]
(2)(1) has solution
\[x=-t^2+At+B\]
.(2) has solution
\[y=Ct+D\]
.Then
\[\mathbf{r} =(-t^2+At+B) \mathbf{i}+ (Ct+D) \mathbf{j}\]
.Now use the initial conditions.
\[\mathbf{v}=(-2t+A) \mathbf{i} + (C) \mathbf{j}\]
\[\mathbf{v}(0)=2 \mathbf{i}+3 \mathbf{j} \rightarrow 2=(-2(0)+A), \; 3 =C \rightarrow A=2, C=3\]
\[\mathbf{r}(0)= \mathbf{j}= (B) \mathbf{i}+ (D) \mathbf{j} \rightarrow B=0, \; D=1\]
.Then
\[\mathbf{r} =(-t^2-t) \mathbf{i}+ ( 3t+1) \mathbf{j}\]
.Then
\[x=-t^2+2t, \; y=3t+1\]
.From the second equation,
\[t=\frac{y-1}{3}\]
so \[x=(\frac{y-1}{3})^2+2(\frac{y-1}{3})\]