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

evo_vil
Messages
10
Reaction score
0

Homework Statement



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

Homework Equations



\nabla f(x,y) = <f_x,f_y>

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:

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

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)
 
Physics news on Phys.org
Welcome to FP, evo_vil! :smile:

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

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
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top