- #1

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I have a problem with the derivation above I dont get how

Can someone derive this and illustrate this visually for example by using Figure 2 or using another drawing?

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- Thread starter fisher garry
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- #1

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I have a problem with the derivation above I dont get how

Can someone derive this and illustrate this visually for example by using Figure 2 or using another drawing?

- #2

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It looks like the book has two wrongs making a right.

PS with the above I get ##dx = \frac{r d\theta}{\cos \theta}##

- #3

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and the equations afterwards. Could you derive how to get to:

starting from

- #4

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well it is not from a textbook it is a document I recieved. Unfortunately I am a bit lost from

View attachment 254268

You're lost because that is wrong. Try what I posted.

- #5

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I have tried to illustrate my problem in the drawing above. Since ##r d\theta## is normal to the radius r and approximately linear since it is a short part of the bowlength the angle between dx and ##r d\theta## should be the same as the angle between r and x that makes cosinus. But what if the fraction ##\frac{r d\theta}{dx}## and the ##\frac{r }{x}## is not the same? We don't know the length of ##r d\theta## measured up to r and the length of dx measured up to x?

- #6

256bits

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Angles and infinitesimals - ugh.View attachment 254289

I have tried to illustrate my problem in the drawing above. Since ##r d\theta## is normal to the radius r and approximately linear since it is a short part of the bowlength the angle between dx and ##r d\theta## should be the same as the angle between r and x that makes cosinus. But what if the fraction ##\frac{r d\theta}{dx}## and the ##\frac{r }{x}## is not the same? We don't know the length of ##r d\theta## measured up to r and the length of dx measured up to x?

Maybe this explanation helps.

##r d\theta## is the arc length, call that da,which for a radius r perpendicular to the axis would have dx = da.

A we deviate from the vertical, r increases in length, the arc length da is no longer parallel to the axis. We have to find the dx portion of da.

- #7

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View attachment 254289

I have tried to illustrate my problem in the drawing above. Since ##r d\theta## is normal to the radius r and approximately linear since it is a short part of the bowlength the angle between dx and ##r d\theta## should be the same as the angle between r and x that makes cosinus. But what if the fraction ##\frac{r d\theta}{dx}## and the ##\frac{r }{x}## is not the same? We don't know the length of ##r d\theta## measured up to r and the length of dx measured up to x?

As I said above, I'll take ##\theta## going clockwise from the vertical, so that ##\theta## and ##x## have the same sign.

##x = R\tan \theta, \ \ R = r\cos \theta, \ \ x = r\sin \theta##

##dx = R \sec^2 \theta d\theta = \frac{r d\theta}{\cos \theta}##

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