Just got back from my Calc 3 final. There was one problem that nobody I talked to had any idea how to do. I'll throw it out here and see if anyone here knows how to handle this.
The equation: G(x,y) = F( F(x,y), x+y) )
You're given the following partial derivatives: dG/dx = 5, dF/dy = 1
F(x,y) = 0
Find all possible values of dF/dx (partial derivative of F w/ respect to x)
Normally, you'd use the chain rule, for instance, if G(x,y) = F(u,v) and u and v are functions of x + y. But w/ the chain rule, you'd be taking something like dF/dF, so I didn't know where to start.
Anyone here have an idea?
Blah, that notation is confusing the hell out of me, mainly the usage of two capital F's, along with the x+y bit and being told that F(x,y)=0. Are you sure this is written correctly? (I also notice mismatched delimiters).
If by any chance the problem is accurately written like this:
G(x,y)=F(u(x,y),v(x,y)), with additional conditions ∂G/∂x=5, ∂F/∂y=1, and u(x,y)=0 (not sure on last one), then that might make more sense.
Given that ∂G/∂x=5, that means G(x,y) = ∫5dx+?y = (5x+C
1+?y) where C
1 is a constant and ?y represents any function of y (eg 5y, cosy, e
y, etc).
For F you need to use the Chain Rule, which says that ∂F/∂y=(∂F/∂u)(du/dy)+(∂F/∂v)(dv/dy). Being told that ∂F/∂y=1 as well as v(x,y)=0, that leaves (∂F/∂v)(dv/dy)=1.
Thus our F function equals (0,?x+1y+C
2), which also equals our G function. In short,
(5x+C
1+?y)=(?x+C
2+y).
The problem asks for all possible values of ∂F/∂x that satisfy this.
...but then that still doesn't make sense, since it could be anything... :|
:massive shrug:
