How does these two equations relate?

  • Context: Undergrad 
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Discussion Overview

The discussion revolves around understanding the relationship between the coefficient of volume expansion and the coefficient of linear expansion, specifically through the manipulation of equations involving volume and linear dimensions. The scope includes mathematical reasoning and conceptual clarification related to calculus and physical principles.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion about the derivation of the formula relating volume expansion to linear expansion, particularly how dV/dL leads to the expression 3L^2.
  • Another participant suggests expressing volume as a function of length, proposing that volume equals L^3 for a cube.
  • There is a discussion on the differentiation of the function y = x^3, leading to the conclusion that dy/dx = 3x^2, which is then applied to the context of volume and length.
  • Participants explore the manipulation of differentials, questioning how to treat dy and dx as quantities in calculus.
  • One participant notes that while these manipulations are common in engineering and physics, they may be viewed differently by pure mathematicians.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the clarity of the derivation process, with some expressing understanding while others remain confused about specific steps in the calculus involved.

Contextual Notes

The discussion highlights potential gaps in understanding calculus concepts such as differentiation and the manipulation of differentials, which may depend on participants' varying levels of familiarity with these topics.

Zerius
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Hi, I'm in first year engineering and I have trouble understanding where this formula comes from in determining the relationship between the coefficient of volume expansion and the coeff of linear expansion.

dV=(dV/dL)*dL = 3L^2*dL

now i know that they changed the dV in the brackets to dL^3, but then I don't understand how dL^3/dL became 3L^2? Is there a part of calculus I am missing? does it have to do with a rate of change divided by a rate of change?

Thanks!
 
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Can you express V as a function of L?
 
Volume = L^3, for a cube? i think that's the relationship there
 
Zerius said:
Volume = L^3, for a cube? i think that's the relationship there

What is dV/dL?
 
If y = x^3, what is dy/dx?
 
then it is 3x^2, that's would be dy/dx. :S how about dX^3/dX? is it the same? OH or is it like dy/dx is the same as d/dX so doesn't matter what it is at the top?
 
Right, dy/dx = 3x^2.

Now treat dy and dx as symbols small quantities that can be manipulated like other quantities.

Multiplying both sides of dy/dx = 3x^2 by dx gives?
 
umm dy = (3x^2)dx?
 
  • #10
Zerius said:
umm dy = (3x^2)dx?

Right. Now change y to V and x to L throughout.

Even though engineers and physicists routinely do these types of manipulations, they make pure mathematicians cringe.
 
  • #11
>.< ok that makes more sense now. I will try to hatch this in my brain. Thanks George Jones!
 

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