Changing V(x) to V(t): Chain Rule Application?

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SUMMARY

The discussion focuses on converting the velocity function V(x) = a*x² + b*x + c, defined in terms of position x, to a function of time t using the chain rule. The key steps involve recognizing that velocity V is defined as dx/dt, leading to the equation dt = dx/(a*x² + b*x + c). By integrating both sides, one can express time as a function of position, subsequently allowing for the derivation of position as a function of time. The integration process is crucial for successfully transitioning from V(x) to V(t).

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  • Understanding of calculus, specifically integration and differentiation.
  • Familiarity with the chain rule in calculus.
  • Knowledge of functions and their transformations.
  • Basic concepts of motion, including velocity and position relationships.
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StephenSF8
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I have a function for velocity, V in terms of position, x. The equation is of the form V(x) = a*x2+b*x+c. Initial conditions are x=0, t=0.

How do I change from V(x) to V(t)? It seems this would be an application of the chain rule, dy/dx = dy/du * du/dx, but I'm struggling to adapt it to this situation. Am I way off base?

Thanks!
 
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Assuming t means time, then V=dx/dt. So dt = dx/(a*x2+b*x+c). Integrate both sides to get t as a function of x. Solve for x as a function of t. Then take the derivative to get V. Good luck!
 
Velocity is the result of differentiating a position equation. To reverse that you have two options, (the same pretty much): Integrate or find the Antiderivative.

Enjoy.
 
mathman said:
Assuming t means time

You assume correct.

mathman said:
Integrate both sides to get t as a function of x. Solve for x as a function of t.

This is what I was missing. I was able to perform the integration and solve for x and all is well. Thanks for the help!
 

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