Calculate $dF/dt$ when $F(t)=f(x(t))$, $f(x) = \|x\|^2$ , $ x(t) = \cos(t) e_1 + 2 \sin(t) e_2 + t e_3$
Calculate $dF/dt$ when $F(t)=f(x(t))$, $f(x) = \|x\|^2$ , $ x(t) = \cos(t) e_1 + 2 \sin(t) e_2 + t e_3$
So I need to calculate d$F$/dt and I'm not sure if I'm being really silly. But using chain rule I get that it is equal to (d$F$/dx)(dx/dt)
Basically is df/dx = 2||$x$||.dx/dt ?
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$\begingroup$You are probably using the Euclidean norm so $f(x(t))$ for your specified x(t) is, $f(x(t))=cos^2(t)+4sin^2(t)+t^2$. Taking the derivative of this wrt to t gives us $F(t)=f'(x(t))=-2cos(t)sin(t)+8sin(t)cos(t)+2t$, then simplify if you can.
$\endgroup$ $\begingroup$Using chain rule, one has: $$\dot{F}(t)=\mathrm{d}_{x(t)}f\cdot \dot{x}(t)=2\langle x(t),\dot{x}(t)\rangle.$$
Indeed, notice that: $$f(x+h)=f(x)+2\langle x,h\rangle+O(\|h\|^2),$$ with $h\mapsto 2\langle x,h\rangle$ being linear which gives $\mathrm{d}_xf$.
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