Let $\Vert\;\Vert$ be the Euclidean norm on $\Bbb{R}^n$. Let $x:[0,a]\to \Bbb{R}^n$ be differentiable. How do I define \begin{align}\frac{d}{dt}\Vert x(t)\Vert^2?\end{align} Please, I need help on this! A detailed answer would suit me. Thanks a lot!
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4It's $$\frac d{dt}(x_1(t)^2+\cdots+x_n(t)^2).$$ – Angina Seng Aug 28 '18 at 06:08
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1Observe that $|x(t)|^2=\langle x(t),x(t)\rangle$ where $\langle \cdot,\cdot\rangle$ denotes the scalar product, which is a bilinear map. – Bebop Aug 28 '18 at 06:08
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If $x(t)=(x_1(t),...,x_n(t))$, then
$f(t)=||x(t)||^2=x_1^2(t)+...+x_n^2(t)$, hence
$f'(t)=2x_1(t)x'_1(t)+...+2x_n(t)x_n'(t)$.
Fred
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In vector form $$\frac{d}{dt}\Vert x(t)\Vert^2=\frac{d}{dt}(x\cdot x)=(\dot{x}\cdot x)+(x\cdot \dot{x})=2\dot{x}\cdot x$$
Nosrati
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Please, why did you take one as a vector as the other as a scalar? – Omojola Micheal Aug 28 '18 at 06:31
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1@Mike They're both vectors. $\dot x$ means $\frac{d}{dt} x$. It's a notation especially common in physics. – Arthur Aug 28 '18 at 06:32
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