I ask for help on this math problem that was among my exercises and that I can't seem to get done properly :steam: even if it's tagged as easy
given the function f = x f( x/y )
prove that every tangent plane passes through the point (0,0,0)
now I know that I just have to use the equation of the tangent plane, which in 2 variables is zz(0) = f'(x) (xx(0)) + f'(y) (y  y(0))
and then substiute 0 0 0 for z0 x0 y0
yet I don't seem to get the derivative right, probably, because I can't prove the equality right
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09062008, 08:38 AM #1
A very easy math problem I can't do
ENTj 738 sx/sp

09062008, 10:22 AM #2
that's easy?!! I feel stupid. of course I haven't done math in about three years.
In no likes experiment.
that is all
i dunno what else to say so

09062008, 10:24 AM #3
my head hurts.

09062008, 10:25 AM #4
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09062008, 10:36 AM #5
I'd show you if I could speak...but I'm not about to write it out.

09082008, 02:25 PM #6
I just saw this. Unfortunately, you probably had this due today or something.
Notation is rather important, and we would really need to know what convention the book is following. The initial equation is something I am having trouble interpreting.
(note: <> means "does not equal")
If f=xf(x/y)=x^2(f/y), then f is irrelevant, we can remove a dimension, and we have y=x^2, y<>0. At any point on this parabola, the instantaneous slope is 2x. At the point (x,y)=(1,1), the slope is 2. So the equation of the tangent line is y1=2(x1). Clearly this line does not go through (0,0) because 01<>2(01).
So I am guessing f=xf(x/y) means something else. If we were to interpret f(x/y) as "f of x/y," then what are the arguments of f on the left side of the equation? just x? If so, then what does y mean? f(x)?
Then we have f(x)=xf(x/f(x)). So f'(x)=f(x/f(x))+xf'(x/f(x))[(f(x)xf'(x))/f^2(x)].
Again, there is one fewer dimension here than expected. The equation for any tangent line through (x*,y*)=(x*,f(x*)) is yf(x*)=f'(x*)(xx*). Now the only way every such line could go through (0,0) is if 0f(x*)=f'(x*)(0x*), IOW, f(x*)=f'(x*)x* for all x*. This is supposed to be easy?
So, I am going to interpret your initial equation to be just the condition stated above. That is:
f=xf'.
Now, the problem is trivial. Because for all x*, f(x*)=x*f'(x*)=>f(x*)=f'(x*)x*=>0f(x*)=f'(x*)(0x*)=>all tangent lines go through (0,0).
Accept the past. Live for the present. Look forward to the future.
Robot Fusion
"As our island of knowledge grows, so does the shore of our ignorance." John Wheeler
"[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
"[P]etabytes of [] data is not the same thing as understanding emergent mechanisms and structures." Jim Crutchfield

09082008, 02:29 PM #7
Shite.
I actually used to be able to do this.
About 20 years ago.
Sigh."Hey Capa  We're only stardust." ~ "Sunshine"
“Pleasure to me is wonder—the unexplored, the unexpected, the thing that is hidden and the changeless thing that lurks behind superficial mutability. To trace the remote in the immediate; the eternal in the ephemeral; the past in the present; the infinite in the finite; these are to me the springs of delight and beauty.” ~ H.P. Lovecraft

09082008, 03:00 PM #8OberonGuest
Pardon me... I can't help but note that that is a calculus problem, particularly dealing with analytical geometry.
What I'm saying is, this ain't exactly "find the area of the rectagle" we're talking about here.

09082008, 03:17 PM #9
Another interesting thing is in this interpretation, if the problem has stated that for all f(x) satisfying, f(x)=xf(x/f(x)) the tangent line for x=0, goes through (0,0), you are essentially done also since f'(x)=f(x/f(x))+xf'(x/f(x))[(f(x)xf'(x))/f^2(x)]=f(x/f(x)) at x=0.
Accept the past. Live for the present. Look forward to the future.
Robot Fusion
"As our island of knowledge grows, so does the shore of our ignorance." John Wheeler
"[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
"[P]etabytes of [] data is not the same thing as understanding emergent mechanisms and structures." Jim Crutchfield

09082008, 03:20 PM #10
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