Solving for Delta Squared: Rules & Examples

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SUMMARY

This discussion addresses the mathematical treatment of delta squared in the context of volume calculations involving diameter and length. The formula presented is V = (d²πL)/4, where d represents diameter and L represents length. Participants clarify that squaring a delta, such as (∆d)², is simply the square of the change in diameter, and emphasize that there are no special rules for handling deltas in this manner. Additionally, the term "del" is cautioned against as it can refer to the nabla operator in vector calculus.

PREREQUISITES
  • Understanding of calculus, specifically differentials and volume calculations.
  • Familiarity with the formula for the volume of a cylinder, V = πr²L.
  • Knowledge of the notation for changes in variables, such as ∆d and ∆L.
  • Awareness of the nabla operator and its implications in vector calculus.
NEXT STEPS
  • Study the application of differentials in volume calculations using the formula dV = 2πrLdr + πr²dL.
  • Research the implications of using the nabla operator in vector calculus to avoid confusion with delta notation.
  • Explore advanced calculus topics related to partial derivatives and their applications in physics.
  • Review examples of volume calculations involving multiple variables and their respective changes.
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Students and professionals in mathematics, engineering, and physics who are dealing with volume calculations and the application of calculus in real-world problems.

Zerius
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How does one deal with a delta squared ? For example, volume is a function of del Diameter and del Length.

If i let v = d^2(pi)L/4, and then call the d and L del. d and del L, when i separate for d i get del d^2

So I'm wondering if there are any special rules for squaring a "del"
 
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Hi Zerius! :smile:

(have a del: ∆ and try using the X2 tag just above the Reply box :wink:)
Zerius said:
… So I'm wondering if there are any special rules for squaring a "del"

You mean as in π(∆d)2(∆L)/4 ?

(∆d)2 is just the square of ∆d.

What is the context? Is this part of some larger problem? :confused:
 
Zerius said:
How does one deal with a delta squared ? For example, volume is a function of del Diameter and del Length.

If i let v = d^2(pi)L/4, and then call the d and L del. d and del L, when i separate for d i get del d^2

So I'm wondering if there are any special rules for squaring a "del"
All you have done here is call ordinary numbers "delta d" and "delta L". There is nothing at all different about that.

If you are talking about differentials, then with [itex]V= \pi r^2 L[/itex], [itex]dV= 2\pi r L dr+ \pi r^2 dL[/itex]. There is only one "dr".

By the way "del" is a common way of referring to the [itex]\nabla[/itex] operator so it is not a good idea to shorten "delta" to "del".
 

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