Solving Calculus Equation: a=dv/dt =>adt=dv

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SUMMARY

The discussion centers on the calculus equation a = dv/dt and its manipulation to derive displacement equations. The transformation from a = dv/dt to adt = dv is clarified as an accepted notation in calculus, despite dt not being a number. The conversation highlights the integration of the equation to yield v = at + C and subsequently s = (1/2)at² + Ct + D, representing the displacement of an object under constant acceleration. This demonstrates the application of differentials in solving basic differential equations.

PREREQUISITES
  • Understanding of basic calculus concepts, specifically differentiation and integration.
  • Familiarity with differentials and their notation in calculus.
  • Knowledge of the relationship between acceleration, velocity, and displacement.
  • Basic understanding of differential equations and their applications.
NEXT STEPS
  • Study the concept of differentials in calculus, focusing on their mathematical properties.
  • Learn how to solve basic differential equations, particularly those involving constant acceleration.
  • Explore the integration techniques used in calculus, including definite and indefinite integrals.
  • Investigate the physical interpretations of calculus in motion, such as kinematics equations.
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Students of calculus, educators teaching differential equations, and anyone interested in the mathematical foundations of motion and acceleration.

Calculus 142
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Hi,
I am new to calculus, and in some books I read
a= dv/dt
=>adt=dv.

If dt means with respect to t, how is it possible multiply both sides of the equation by dt?
Is there a theorem stating this?

Thankyou.
 
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adt=dv means nothing but a= dv/dt; just a prevalent abuse of notation. Also, dt is not a number which can be multiplied.
 


dv and dt are differentials. For more information, see http://en.wikipedia.org/wiki/Differential_(infinitesimal ).

Differentials come up in the study of differential equations, a simple example of which is a = dv/dt. If a is a constant, we can separate this equation to dv = a dt, and integrate both sides with respect to t, to get v = at + C, where C is an arbitrary constant.

If we realize that v = ds/dt, the time rate of change of position, then we have ds/dt = at + C, which implies that ds = (at + C)dt. Integrating again with respect to t, we get s = (1/2)at^2 + Ct + D, which gives us the displacement of an object moving with a constant acceleration as a function of t.
 
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Thank you for the response.
 

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