The Mystery of Green-Colored Steps: A Homework Tutorial

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Homework Help Overview

The discussion revolves around a problem involving the manipulation of differentials in a mathematical context, specifically related to the chain rule. The original poster expresses confusion about a particular step highlighted in green and seeks clarification on the technique used.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the manipulation of differentials and the application of the chain rule. The original poster questions the terminology associated with the highlighted step and requests further examples. Some participants reflect on their own difficulties in understanding the step and acknowledge corrections made in their reasoning.

Discussion Status

The discussion is ongoing, with participants exploring the concept of differentials and the chain rule. There is an acknowledgment of confusion and corrections, indicating a productive exchange of ideas, though no consensus has been reached.

Contextual Notes

Participants note the challenge of understanding the specific step in the solution and the reliance on checking solutions for clarity. There may be assumptions about prior knowledge of calculus concepts that are being questioned.

athrun200
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Homework Statement



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Homework Equations





The Attempt at a Solution


In part a, there is an important step, I don't know how to have the step that written in green colour.
Does this technque have any name?
Can you give me more example of it?

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You just manipulate the differentials.
[itex]\frac{d \dot{r}}{dt}=\frac{d \dot{r}}{dt} \frac{dr}{dr}=\frac{d \dot{r}}{dr} \frac{dr}{dt}=\dot{r} \frac{d \dot{r}}{dt}[/itex]

edit. corrected last part
 
Last edited:
bp_psy said:
You just manipulate the differentials.
[itex]\frac{d \dot{r}}{dt}=\frac{d \dot{r}}{dt} \frac{dr}{dr}=\frac{d \dot{r}}{dr} \frac{dr}{dt}=\dot{r} \frac{dr}{dr}[/itex]

It is quick difficult to think of this step.
If I don't check the solution, I wouldn't write out this.
 
athrun200 said:
It is quick difficult to think of this step.
If I don't check the solution, I wouldn't write out this.

It appears that I messed up the last equality. I corrected the last post.
It is clearer if you see that it is the chain rule.
let [itex]\dot{r}=\dot{r}(r(t))[/itex] then [itex]\frac{d \dot{r} }{dt}=\frac{d \dot{r} }{dr} \frac{d{r} }{dt}[/itex]
 

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