Derivative of the cross and dot product

In summary, to prove the derivatives of functions A(t) and B(t) are as follows: d(A (dot) B) / dt = [A (dot) (dB)/(d(t)] + [d(A)/d(t) (dot) B] and d(A x B) / dt = [A x (dB)/(d(t)] + [d(A)/d(t) x B], differentiate using the product rule and then simplify and show that it is equal to the derivative of the initial expression. It is also helpful to work backward for the cross product. The use of \delta_{ij} and \epsilon_{ijk} is not necessary.
  • #1
bluelagoon
1
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Homework Statement



If you have two functions dependent on t, A(t) and B(t). Prove their derivatives are as follows:

d(A (dot) B) / dt = [A (dot) (dB)/(d(t)] + [d(A)/d(t) (dot) B]

{Where "(dot)" acts as the dot product}




d(A x B) / dt = [A x (dB)/(d(t)] + [d(A)/d(t) x B]

{Where "x" acts as the cross product}




I have little experience with differentiating cross products and dot products and appreciate any help in starting this one!

Thanks,
Janet
 
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  • #2
Well A . B = axbx + ayby + azbz (in 3 dimensions)
Differentiate that using the product rule, then regroup and show that its the expression you need it to be equal to. The cross product is similar.
 
  • #3
I think the easy way around of using [tex] \delta_{ij} [/tex] and [tex] \epsilon_{ijk} [/tex] is a bit too advanced.

For the cross-product I would just say from experience that working backward may be a bit less confusing, e.g. show what two parts are and simplify the output expression, and then showing that it is equal to the derivative of the initial expression
 
  • #4
You've already proven the derivative rule for one kind of product; doesn't the same method work for these products?
 

What is the definition of the derivative of a cross product?

The derivative of a cross product is a vector that represents the instantaneous rate of change of the cross product with respect to some variable. It is obtained by taking the partial derivatives of each component of the cross product.

How is the derivative of a cross product calculated?

The derivative of a cross product is calculated by taking the cross product of the original vectors and then taking the dot product of that result with the derivative of the vector that is being crossed with the other vector. This process is repeated for each component of the cross product.

What is the significance of the derivative of a cross product?

The derivative of a cross product is used to calculate the direction and magnitude of the instantaneous change in a vector quantity. It is commonly used in applications such as physics, engineering, and computer graphics.

What is the derivative of a dot product?

The derivative of a dot product is a scalar that represents the instantaneous rate of change of the dot product with respect to some variable. It is obtained by taking the partial derivatives of each component of the dot product.

What are some real-world applications of the derivative of a cross product?

The derivative of a cross product is used in many real-world applications such as calculating the angular velocity of a rotating object, determining the torque on a rotating body, and finding the direction and speed of a moving object in a three-dimensional space.

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