Scalar Product of a diffrential.

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SUMMARY

The discussion centers on the scalar product of differential expressions involving vector functions, specifically the equations r·r = 1/2 · r² and r·r = 1/2 · ṙ². The user seeks clarification on these expressions, which utilize the chain rule and product rule for differentiation. The correct application of these rules leads to the conclusion that (r·r)' = 2r·r', demonstrating the relationship between the derivatives of scalar and vector functions.

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  • Understanding of vector calculus
  • Familiarity with differentiation rules (chain rule and product rule)
  • Knowledge of scalar and vector products
  • Basic concepts of differential equations
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  • Study vector calculus, focusing on scalar and vector products
  • Learn about differentiation techniques, specifically the chain rule and product rule
  • Explore applications of differential equations in physics
  • Review examples of scalar products in vector functions
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Students of mathematics and physics, particularly those studying vector calculus and differential equations, as well as educators seeking to clarify these concepts for their students.

raisin_raisin
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Hey, in my textbook they keep doing this and I can't follow
for example r.[tex]\ddot{}r[/tex] = 1/2 [tex]\ddot{}r^{}^2{}[/tex]
and r.[tex]\dot{}r[/tex] = 1/2 [tex]\dot{}r^2{}[/tex].

Can anyone explain this to me? I know I should probably know it.

P.S Can't quite get the dot product to look right apologies.
 
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(r²)' = 2rr' by chain rule. Or if the r's are supposed to be vector functions:

(r.r)'
= r'.r + r.r' by product rule for dot product
= r.r' + r.r' by commutativity of dot product
= 2r.r'
 

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