- #1
Ishika_96_sparkles
- 57
- 22
- TL;DR
- Confusion about the gradient term in the derivation.
Definition: Let f be a differentiable real-valued function on ##\mathbf{R}^3##, and let ##\mathbf{v}_P## be a tangent vector to it. Then the following number is the derivative of a function w.r.t. the tangent vector
$$\mathbf{v}_p[\mathit{f}]=\frac{d}{dt} \big( \mathit{f}(\mathbf{P}+ t \mathbf{v}) \big)|_{t=0}$$
Then there is this
Lemma: If ##\mathbf{v}_p= (v_1,v_2,v_3)_P## is a tangent vector to ##\mathbf{R}^3##, then
\mathbf{v}_p[\mathit{f}]= \sum_i v_i \frac{d \mathit{f}}{dx_i}(P)
1) How does this gradient term come into the picture? Since, \mathit{f}(\mathbf{P}+ t \mathbf{v}), do we write the argument as \mathbf{x}(t) and then apply the chain-rule in the definition?
2) Could this be thought of as a dot-product ##\mathbf{v}_P \,. \, \nabla \mathit{f}|_p ##?
$$\mathbf{v}_p[\mathit{f}]=\frac{d}{dt} \big( \mathit{f}(\mathbf{P}+ t \mathbf{v}) \big)|_{t=0}$$
Then there is this
Lemma: If ##\mathbf{v}_p= (v_1,v_2,v_3)_P## is a tangent vector to ##\mathbf{R}^3##, then
\mathbf{v}_p[\mathit{f}]= \sum_i v_i \frac{d \mathit{f}}{dx_i}(P)
1) How does this gradient term come into the picture? Since, \mathit{f}(\mathbf{P}+ t \mathbf{v}), do we write the argument as \mathbf{x}(t) and then apply the chain-rule in the definition?
2) Could this be thought of as a dot-product ##\mathbf{v}_P \,. \, \nabla \mathit{f}|_p ##?
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