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Now taking the time derivitive of <math>V</math>
:<math>
\dot{V}=r\dot{r}
</math>
:<math>
\dot{V}=(\dot{e}+\alpha e)(\ddot{e}+\alpha \dot{e})
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And hence the error and error rate, remembering that <math>V=\frac{1}{2}(\dot{e}+\alpha e)^2</math>, exponentially decay to zero.
If you wish to tune a particular response from this, it is necessary to subsititute back into the solution we derived for for <math>V</math> and solve for <math>e</math>. This is left as an exercise for the reader but the first few steps at the solution are shown.
:<math>
r\dot{r}=-\frac{\kappa}{2}r^2
</math>
:<math>
\dot{r}=-\frac{\kappa}{2}r
</math>
:<math>
r=r(0)e^{-\frac{\kappa}{2} t}
</math>
:<math>
\dot{e}+\alpha e= (\dot{e}(0)+\alpha e(0))e^{-\frac{\kappa}{2} t}
</math>
==Notes==
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