Derivative of Cross Product with Differentiable Functions

In summary, the conversation discusses the process of finding the derivative of the cross product u(f(t)) x v(g(t)), using the chain rule. It is determined that the formula for the derivative is u'(f(t))(f'(t)) x v(g(t)) + u(f(t)) x v'(g(t)(g'(t)).
  • #1
vroomba03
7
0

Homework Statement


Assume that you are given differentiable function f(t) and g(t). Find a formula for the
derivative of the cross product u(f(t)) x v(g(t)).


Homework Equations


d/dt(u(t) x v(t)) = (u'(t) x v(t) + u(t) x v'(t)

The Attempt at a Solution


So in this case I was thinking that you would just substitute f(t) and g(t) where t would be in the regular equation, so it would be U'(f(t)) x v(g(t)) + u(f(t)) x v'(g(t)), for the equation, but I have a feeling that that's not right just because it seems too simple.
 
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  • #2
u'(f(t)) = d u(f(t))/f(t) and similar result for v'(g(t)).
What you want is d/dt ( u(f(t)) × v(g(t)) ), so you will have to use chain rule.
 
  • #3
So then the equation would be U'(du(f(t))/f(t)) x V(g(t)) + U(g(t)) x V'(du(g(t))/g(t)) x U(f(t)) ?
 
  • #4
vroomba03 said:
So then the equation would be U'(du(f(t))/f(t)) x V(g(t)) + U(g(t)) x V'(du(g(t))/g(t)) x U(f(t)) ?
No, start by applying the chain rule to find $$\frac{d}{dt} u(f(t))$$
 
  • #5
Oh okay, so that is u'(f(t))(f'(t)) when using chain rule. So you said u'(f(t)) = d u(f(t))/f(t), so then would I divide what i got by the chain rule and divide it by f(t) and that would be my u'(f(t))?
 
  • #6
vroomba03 said:
Oh okay, so that is u'(f(t))(f'(t)) when using chain rule.
Yes.

So you said u'(f(t)) = d u(f(t))/f(t), so then would I divide what i got by the chain rule and divide it by f(t) and that would be my u'(f(t))?
You need to find $$\frac{d}{dt} \left(u(f(t)) \times v(g(t))\right) = \frac{d}{dt} u(f(t)) \times v(g(t)) + u(f(t)) \times \frac{d}{dt} v(g(t))$$

You correctly found ##\frac{d}{dt} u(f(t))##. Now find ##\frac{d}{dt} v(g(t))## and substitute in.

What I should have wrote is u'(f(t)) ##\equiv## d u(f(t))/f(t), these two expressions denote the derivative of u with respect to f(t).
 
  • #7
So the final equation for the question would be u'(f(t))(f'(t)) x v(g(t)) + u(f(t)) x v'(f(t)(f'(t)) ?
 
  • #8
vroomba03 said:
So the final equation for the question would be u'(f(t))(f'(t)) x v(g(t)) + u(f(t)) x v'(f(t)(f'(t)) ?

Check the last term again. v is a function of g(t), not f(t).
 
  • #9
Oops silly error. u'(f(t))(f'(t)) x v(g(t)) + u(f(t)) x v'(g(t)(g'(t)) correct?
 
  • #10
vroomba03 said:
Oops silly error. u'(f(t))(f'(t)) x v(g(t)) + u(f(t)) x v'(g(t)(g'(t)) correct?
Correct :smile:
 
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  • #11
Thank you a bunch!
 

FAQ: Derivative of Cross Product with Differentiable Functions

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

The derivative of a cross product is a mathematical operation that calculates the rate of change of a cross product with respect to a particular variable. It is commonly denoted as d/dx (A x B), where A and B are vectors and x is the variable of differentiation.

How is the derivative of a cross product calculated?

The derivative of a cross product can be calculated using the product rule of differentiation. This involves taking the derivative of each vector individually and then combining them using the cross product formula. Alternatively, it can also be calculated using the determinant method, where the components of the two vectors are arranged in a matrix and the determinant is taken to find the derivative.

What is the physical interpretation of the derivative of a cross product?

The derivative of a cross product has a physical interpretation as the instantaneous change in the direction of the cross product. It can also be interpreted as the angular velocity of the cross product, or the rate at which the angle between the two vectors changes.

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

The derivative of a cross product has many applications in fields such as physics, engineering, and computer graphics. It is used to calculate the torque in rotational motion, the magnetic field in electromagnetism, and the orientation of objects in 3D computer graphics.

Are there any special properties or rules for the derivative of a cross product?

Yes, there are a few properties and rules that apply to the derivative of a cross product. These include the product rule, chain rule, and quotient rule, as well as the fact that the derivative of a cross product is perpendicular to both of the original vectors. It also follows the right-hand rule, where the direction of the cross product depends on the order of the vectors in the product.

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