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\newcommand{\mytitle}[1]{\begin{center} {\Large Intro to Cognitive Science -- 
9.62J Problem Set \psnum \\ Juliet Bourne -- #1} \\ \vspace{-.1in}
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{\Large 9.62J Problem Set \psnum -- Juliet Bourne -- {\bf Page \thepg}}
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\begin{document}

\mytitle{11/29/90}

\vspace*{-.2in}
\begin{center} {\large Motor Control} \end{center}

\prob{1.6}

\beq
\tau & = & I\ddot{\theta} \\
\theta (t) & = & t^2 + 2t + 1 \\
\dot{\theta} (t) & = & 2t + 2 \\
\ddot{\theta} (t) & = & 2 \\
\theta (ct) & = & c^2t^2 + 2ct + 1 \\
\dot{\theta} (ct) & = & 2c^2t + 2c \\
\ddot{\theta} (ct) & = & 2c^2 \\
\tau & = & c^2I\ddot{\theta}
\eeq

\noindent Thus, the torque is scaled by $c^2$ when the trajectory (which
is a 2nd-order polynomial) is a function of $ct$.

\prob{1.7}

\beq
f & = & K_1 (0 - x) - K_2 (x - 0) \\
\mbox{} & = &  - K_1x - K_2x \\
f & = & - (K_1 + K_2) x
\eeq

\noindent Thus, the stiffness of the mass is $K_1 + K_2$.

\prob{1.8}

\beq 
f & = & K_1^{'} (x_1^{'} - x) - K_2^{'} (x - x_2^{'}) \\
\mbox{} & = & K_1^{'}x_1^{'} - K_1^{'}x - K_2^{'}x + K_2^{'}x_2^{'} \\
f & = & - (K_1^{'} + K_2^{'})x + K_1^{'}x_1^{'} + K_2^{'}x_2^{'} \\ 
K_1^{'}x_1^{'} & = & - K_2^{'}x_2^{'}
\eeq

\noindent Thus, we can change the stiffness of the mass while keeping
the equilibrium point the same. This is done by choosing $K_1$ and $K_2$
such that $K_1^{'}x_1^{'} = - K_2^{'}x_2^{'}$.

\end{document}
