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In the proof of theorem 3.1 they put : $\langle X,\nabla f \rangle =X(f)$ after that they say that: for a curve $c$ on $M$ then $\left\langle\dfrac{\mathrm{d}c}{\mathrm{d}t},\nabla f\right\rangle=\dfrac{\mathrm{d}(f\circ c)}{\mathrm{d}t}$

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i don't understand why $<X,\nabla f > =1$ ?

please

Thank you .

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1 Answer 1

$$\dfrac{dc}{dt}=\left(\dfrac{dc^1}{dt}, \dfrac{dc^2}{dt}, \dots , \dfrac{dc^n}{dt}\right)$$ $$\nabla f=\left(\dfrac{\partial f}{\partial x^1}, \dfrac{\partial f}{\partial x^2}, \dots , \dfrac{\partial f}{\partial x^n}\right)$$ Also, we have: $$\dfrac{d(f\circ c)}{dt}=\dfrac{df(c^1(t),c^2(t),\dots ,c^n(t))}{dt}=\dfrac{\partial f}{\partial x^1}\dfrac{dc^1}{dt}+\dots+\dfrac{\partial f}{\partial x^n}\dfrac{dc^n}{dt}=\left\langle\dfrac{dc}{dt},\nabla f\right\rangle$$

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from the definition $\displaystyle<\frac{dc}{dt},\nabla f>=\frac{dc}{dt}(f)$ why $\displaystyle<\frac{dc}{dt},\nabla f>=\frac{d(f\circ c)}{dt}$ –  Vrouvrou Apr 20 '13 at 6:03
    
How do you define $X$? Above I wrote it like a vector , however it is more than a vector of $\mathbb R ^n$. If you write it in the base : $$X=\chi^1\dfrac{\partial}{\partial x^1}+\dots+\chi^n\dfrac{\partial}{\partial x^n}$$ So $$\dfrac{dc}{dt}=\dfrac{dc^1}{dt}\dfrac{\partial}{\partial x^1}+\dots+\dfrac{dc^n}{dt}\dfrac{\partial}{\partial x^n}$$ Apply $f$ , and you obtain: $$\dfrac{dc}{dt}(f)=\dfrac{dc^1}{dt}\dfrac{\partial f}{\partial x^1}+\dots+\dfrac{dc^n}{dt}\dfrac{\partial f}{\partial x^n}$$ –  Dimitrios Nt Apr 20 '13 at 7:27
    
ok ,thank you i edited my question can you help me ? –  Vrouvrou Apr 20 '13 at 18:15
3  
$\langle X, grad f \rangle= \langle \rho\cdot grad f, grad f\rangle=\rho \langle grad f , grad f \rangle= \dfrac{1}{\langle grad f,grad f \rangle}\langle grad f,grad f \rangle=1 $ –  Dimitrios Nt Apr 20 '13 at 23:16

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