Integration (Related to Physics)

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SUMMARY

The discussion centers on the relationship between differentiation and the equation of a straight line, y = mx + c. The user initially confuses the gradient calculation, attempting to derive m using m = y/x, which is incorrect. The correct approach is to recognize that the slope (m) is defined as m = Δy/Δx, and the derivative of the equation y = mx + c with respect to x is simply m. The user acknowledges the mix-up between integration and differentiation in the thread title.

PREREQUISITES
  • Understanding of basic algebraic equations, specifically linear equations.
  • Familiarity with the concepts of differentiation and integration in calculus.
  • Knowledge of the slope formula, m = Δy/Δx.
  • Basic understanding of the laws of indices.
NEXT STEPS
  • Study the principles of differentiation, focusing on linear functions.
  • Explore the concept of integration and its applications in physics.
  • Learn about the relationship between derivatives and slopes in various contexts.
  • Review the laws of indices and their implications in algebraic manipulation.
USEFUL FOR

Students of mathematics and physics, educators teaching calculus concepts, and anyone looking to clarify the differences between differentiation and integration.

_Mayday_
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[SOLVED] Integration (Related to Physics)

This shouldn't take long :smile:

I have been given the general equation for a straight line which is:

[tex]y=mx+c[/tex]

Now I know that to determine the gradient I can use:

[tex]m=\frac{y}{x}[/tex]

Here is my question. Can I differentiate the initial equation given to get to [tex]m=\frac{y}{x}[/tex]

If so, which I am sure you can, then I seem to have come across a problem, though I think it is a problem in my differentiation.

[tex]y=mx=c[/tex]

[tex]\frac{dy}{dx}=mx^{-1}[/tex]

[tex]m=\frac{y}{x^{-1}}[/tex]

or

[tex]m=\frac{x}{y}[/tex]

This does not agree with my initial statement. Either my differentiation is incorrect or I need to touch up on my laws of indices, and if neither of these maybe I am deluded and this can't be done anyway :-p

_Mayday_
 
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(1) The slope is given by [itex]m = \Delta y / \Delta x[/itex], not [itex]m = y/x[/itex].
(2) The derivative (with respect to x) of mx + c is just m.
 
Doc Al said:
(1) The slope is given by [itex]m = \Delta y / \Delta x[/itex], not [itex]m = y/x[/itex].
(2) The derivative (with respect to x) of mx + c is just m.

Just noticed the thread title is integration not differentiation. Thank you for your help.
 

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