Derivative of Cross Product with Differentiable Functions

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Homework Help Overview

The discussion revolves around finding a formula for the derivative of the cross product of two differentiable functions, u(f(t)) and v(g(t)). Participants are exploring the application of the chain rule in this context.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss substituting f(t) and g(t) into the derivative formula for the cross product. There are attempts to clarify the application of the chain rule and the correct differentiation of the functions involved.

Discussion Status

The discussion has progressed with participants providing guidance on applying the chain rule correctly. There is an ongoing exploration of the correct formulation of the derivative, with some participants correcting earlier misunderstandings about the functions involved.

Contextual Notes

Some participants express uncertainty about the proper application of derivatives and the relationships between the functions u and v with respect to their arguments f(t) and g(t). There are indications of confusion regarding the notation and the differentiation process.

vroomba03
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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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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.
 
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)) ?
 
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 $$\frac{d}{dt} u(f(t))$$
 
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))?
 
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).
 
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)) ?
 
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).
 
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!
 

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