Determine the unit tangent vector

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SUMMARY

The discussion centers on the justification of the equation |dr/dt| = ds/dt, which relates the derivative of the position vector to the differential arc length in vector calculus. The participants confirm that ds is equivalent to the modulus of the differential position vector, |d𝑟|. This relationship is foundational in understanding unit tangent vectors in calculus and physics. The equation is crucial for deriving the unit tangent vector from a parametric curve.

PREREQUISITES
  • Understanding of vector differentiation
  • Familiarity with parametric equations
  • Knowledge of differential calculus
  • Concept of arc length in calculus
NEXT STEPS
  • Study the derivation of the unit tangent vector from parametric curves
  • Explore the relationship between arc length and vector functions
  • Learn about the applications of unit tangent vectors in physics
  • Review vector calculus concepts related to differentiation
USEFUL FOR

Students studying calculus, particularly those focusing on vector calculus and physics, as well as educators seeking to clarify concepts related to unit tangent vectors and their derivations.

chwala
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Homework Statement
see attached
Relevant Equations
vector differentiation
1644910436015.png


I need a justification that ##|\dfrac {dr}{dt}|##=##\dfrac {ds}{dt}## cheers guys... all the other steps are easy and clear to me...
 
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chwala said:
Homework Statement:: see attached
Relevant Equations:: vector differentiation

View attachment 297138

I need a justification that [Modulus ##\big | \dfrac {dr}{dt} \big |##=##\dfrac {ds}{dt}##] cheers guys... all the other steps are easy and clear to me...
We have ##ds \equiv |d\vec r|##.
 
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Aaaaah Perok cheers!
PeroK said:
We have ##ds \equiv |d\vec r|##.
 

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