What is the gradient vector problem for a function with dependent variables?

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SUMMARY

The discussion centers on calculating the gradient vector \(\nabla f\) for a function \(z = f(x,y)\) where \(x = r + t\) and \(y = e^{rt}\). The correct approach involves determining the gradient \(\nabla f(r,t)\) as \(\nabla f(r,t) = \), rather than directly substituting \(r\) and \(t\) into the partial derivatives of \(f\). The user confirmed their calculations align with this interpretation, indicating a potential ambiguity in the problem statement.

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evo_vil
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Homework Statement



If [itex]z = f(x,y)[/itex] such that [itex]x = r + t[/itex] and [itex]y = e^{rt}[/itex], then determine [itex]\nabla f(r,t)[/itex]

Homework Equations



[itex]\nabla f(x,y) = <f_x,f_y>[/itex]

The Attempt at a Solution



Now if i follow this the way i think it should be done then i find the partials of f wrt x and y and then simply sub in r and t in the place of x and y respectively...

But if i get del f the normal way i get:

[itex]\nabla f = <f_x+f_y t e^{rt},f_x+f_y r e^{rt}>[/itex]

is this the final/correct answer or am i missing a trick question where i was asked to find del f(r,t) and not del f(x,y)
 
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Welcome to FP, evo_vil! :smile:

Your problem is ambiguous.
If I take it very literal, the answer would be:
[tex]\nabla f(r,t)=<f_x(r,t), f_y(r,t)>[/tex]

However, I can't imagine that this was intended.

I expect that you're supposed to take the gradient from a function f* defined by:
f*(r,t) = f(x(r,t), y(r,t)).
It is not unusual that this function f* is simply called f, although that is ambiguous.

This is what you calculated, and no doubt correct.
 
Thanks!

Ive browsed PF for quite a few years, but never participated, so thanks for the welcome...

I think I am just going to go with what I've calculated and see how it goes...
Maybe see if other people get the same thing.

Thanks for your help
 

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