Differentiating Vector Products

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SUMMARY

The discussion focuses on proving the equation d/dt[r.(vxa)] = r.(vxda/dt) using vector calculus. Participants clarify that the derivative of the velocity vector v with respect to time is the acceleration vector a, which simplifies the expression significantly. The key insight is recognizing that the cross product of a vector with itself results in zero, allowing for the reduction of the equation to r.(vxda/dt). The use of the product rule for differentiation and properties of vector products is essential in this proof.

PREREQUISITES
  • Understanding of vector calculus, specifically dot and cross products.
  • Familiarity with the product rule for differentiation.
  • Knowledge of the properties of vector triple products.
  • Basic concepts of kinematics involving position, velocity, and acceleration vectors.
NEXT STEPS
  • Study the properties of vector triple products in depth.
  • Learn advanced techniques in vector calculus, including differentiation of vector functions.
  • Explore applications of dot and cross products in physics problems.
  • Review kinematic equations and their derivations involving vector quantities.
USEFUL FOR

This discussion is beneficial for students and professionals in physics, particularly those studying mechanics, as well as mathematicians and engineers dealing with vector calculus and its applications in motion analysis.

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Homework Statement



Prove that d/dt[r.(vxa)] = r.(vxda/dt)

Homework Equations



r, v, a are position, velocity and acceleration vectors.
..r.(v.. is the dot product.
..vxa.. is the cross product


The Attempt at a Solution



I expand the equation using the product rule for dot and cross products to get:

dr/dt.(vxa)+r.(dv/dt x a+v x da/dt)

I've expanded this further on paper using the x,y,z components of each vector but i can't manipulate it to get the desired result? Have I missed a step or overlooked something here?
 
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well its way more simple than you are assuming. derivative of v with respect to time is a, so your second term is a cross product between a and a, so that becomes zero. In the first term, the time derivative of r is v. So it becomes
[itex]\mathbf{v}\cdot (\mathbf{v}\times \mathbf{a})[/itex] Use the property of vector triple product to make this zero. So you are left with the last term
 
Last edited:
Thanks very much.
 

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