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Comments on Consider the second of these integrals (What's the meaning of second right here?)

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Consider the second of these integrals (What's the meaning of second right here?)

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$$\frac{dJ}{d\alpha}=\int_{x_1}^{x_2}(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}+\frac{\partial f}{\partial \dot{x}}\frac{\partial \dot{x}}{\partial \alpha})\mathrm dx$$ Consider the second of these integrals: $$\int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial \dot{y}}{\partial \alpha}\mathrm dx=\int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial^2 y}{\partial x \partial \alpha}\mathrm dx$$

What did they mean by "second"? There must be a negative on LHS or RHS. What happened to the line? To me, it seems like he had used chain rule for $\partial y$. But, why there's no negative?

picture of book used double blockquote to decrease size of the picture

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What did they mean by "second"?

They've mentally expanded $$\frac{dJ}{d\alpha}=\int_{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}+\frac{\partial f}{\partial \dot{y}}\frac{\partial \dot{y}}{\partial \alpha}\right)\mathrm dx$$ (which I've corrected to be what it says in the image rather than your MathJax) as $$\frac{dJ}{d\alpha}=\int_{x_1}^{x_2}\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha} \mathrm dx + \int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial \dot{y}}{\partial \alpha}\mathrm dx$$

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I wasn't talking about this one (2 comments)
I wasn't talking about this one
deleted user wrote about 3 years ago · edited about 3 years ago

I was saying that $\frac{dJ}{d\alpha}=0$ So, $0=\int_{x_1}^{x_2}\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha} \mathrm dx + \int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial \dot{y}}{\partial \alpha}\mathrm dx$ Hence, $\int_{x_1}^{x_2}\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha} \mathrm dx=-\int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial \dot{y}}{\partial \alpha}\mathrm dx$ that's what I was talking about

Peter Taylor‭ wrote about 3 years ago

deleted user I think (although without more context I can't be certain) that you're misunderstanding the flow of the text. It looks to me as though (2.6) states a condition (which you've misquoted in your comment: that subscript $\alpha = 0$ is there for a reason) and then the following text, from "By the usual methods", begins a subthread to calculate a subexpression in the condition and hence rephrase the condition. Certainly the final equation given in your quoted image does not use $\int_{x_1}^{x_2}\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha} \mathrm dx$ at all: it merely changes one instance of Newton notation in the "second of these integrals" to Liebniz notation.