- #1
ibkev
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I'm working through the discussion of calculus of variations in Taylor's Classical Mechanics today. There's a step where partial differentiation is involved that I don't understand.
Given:
$$S(\alpha)=\int_{x_1}^{x_2} f(y+\alpha\eta, y'+\alpha\eta', x)\,dx$$
The goal is to determine ##y(x)## when the derivative of ##S(\alpha)=0## and ensure that ##\alpha=0## at this point.
So this means:
$$\frac {dS(\alpha)} {d\alpha}=\int_{x_1}^{x_2} \frac {\partial f} {\partial \alpha}\,dx=0$$
and now here is the step that I don't understand. It's apparently an application of the chain rule but I'm just not seeing it...
$$\frac {dS(\alpha)} {d\alpha}=\int_{x_1}^{x_2} \left( \eta\frac {\partial f} {\partial y}+\eta'\frac {\partial f} {\partial y'} \right)\,dx=0$$
Suggestions are appreciated! :)
Given:
$$S(\alpha)=\int_{x_1}^{x_2} f(y+\alpha\eta, y'+\alpha\eta', x)\,dx$$
The goal is to determine ##y(x)## when the derivative of ##S(\alpha)=0## and ensure that ##\alpha=0## at this point.
So this means:
$$\frac {dS(\alpha)} {d\alpha}=\int_{x_1}^{x_2} \frac {\partial f} {\partial \alpha}\,dx=0$$
and now here is the step that I don't understand. It's apparently an application of the chain rule but I'm just not seeing it...
$$\frac {dS(\alpha)} {d\alpha}=\int_{x_1}^{x_2} \left( \eta\frac {\partial f} {\partial y}+\eta'\frac {\partial f} {\partial y'} \right)\,dx=0$$
Suggestions are appreciated! :)
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