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Hunus Group TitleBest ResponseYou've already chosen the best response.0
I'm reading a book on classical mechanics and I can't quite catch on to what the author is saying. The scalar product \[\vec r \cdot (d\vec r)= r(d\vec r)_r\]where \[ (d\vec r)_r\] is the projection of \[d\vec r\] on the vector \[\vec r\] This projection is equal to \[d\vec r\] the increment of the magnitude of the vector \[\vec r\] Therefore \[\vec r \cdot (d\vec r)= r*dr\]
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.2
dw:1335174774073:dwfrom the sketch you can see\[\vec r \cdot d\vec r= \left \vec r \right \left d\vec r \right \cos \theta\]and\[ \left d\vec r \right \cos \theta=dr\]hence:\[\vec r \cdot d\vec r= r\:dr\]
 2 years ago

T.P Group TitleBest ResponseYou've already chosen the best response.0
l dont understand the language of the author but this is a vector as you see that it have direction and magnitube. this vector have components Y&X becouse of the angle on the xaxes so l think this language is explaining that
 2 years ago
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