Ex 1.25 Properties of Dt, 3

Structure and Interpretation of Classical Mechanics

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Demonstrate that …

d . \displaystyle{D_t (H \circ G) = (DH \circ G) D_t G}

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\displaystyle{  \begin{aligned}  &D_t (HG) \circ \Gamma[q] (t) \\  \end{aligned}  }

\displaystyle{  \begin{aligned}        &= \partial_0 \left[ (H \circ G) (t, q, v, a, ...) \right] \\     &+ \partial_1 \left[ (H \circ G) (t, q, v, a, ...) \right] v(t) \\     &+ \partial_2 \left[ (H \circ G) (t, q, v, a, ...) \right] a(t) + ... \\       \end{aligned}  }

\displaystyle{  \begin{aligned}        &= \partial_0 \left[ H (G (t, q, v, a, ...)) \right] \\     &+ \partial_1 \left[ H (G (t, q, v, a, ...)) \right] v(t) \\     &+ \partial_2 \left[ H (G (t, q, v, a, ...)) \right] a(t) + ... \\       \end{aligned}  }

\displaystyle{  \begin{aligned}        &= D (H (G (t, q, v, a, ...))) \partial_0 G (t, q, v, a, ...) \\     &+ D (H (G (t, q, v, a, ...))) \partial_1 G (t, q, v, a, ...) v(t) \\     &+ D (H (G (t, q, v, a, ...))) \partial_2 G (t, q, v, a, ...) a(t) + ... \\       \end{aligned}  }

\displaystyle{  \begin{aligned}        &= D (H (G (t, q, v, a, ...))) \left[ \partial_0 G (t, q, ...)   +  \partial_1 G (t, q, ...) v(t)   +  \partial_2 G (t, q, ...) a(t) + ... \right] \\  \end{aligned}  }

\displaystyle{  \begin{aligned}        &= D ( (H \circ G) \circ \Gamma[q] (t) ) D_t G \circ \Gamma[q] (t) \\  \end{aligned}  }

\displaystyle{  \begin{aligned}        &= D_G  (H \circ G)  \circ \Gamma[q] (t) D_t G \circ \Gamma[q] (t) \\  \end{aligned}  }

\displaystyle{ (fg) [q] \equiv f[q] g[q]}

\displaystyle{  \begin{aligned}        &= \{ [D_G  (H \circ G)] D_t G \} \circ \Gamma[q] (t) \\  \end{aligned}  }

— Me@2022-05-23 03:18:43 PM

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2022.05.23 Monday (c) All rights reserved by ACHK