Quote:
Originally Posted by FDG
I ask for help on this math problem that was among my exercises and that I can't seem to get done properly  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 z-z(0) = f'(x) (x-x(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 |
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 y-1=2(x-1). Clearly this line does not go through (0,0) because 0-1<>2(0-1).
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 y-f(x*)=f'(x*)(x-x*). Now the only way every such line could go through (0,0) is if 0-f(x*)=f'(x*)(0-x*), 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*=>0-f(x*)=f'(x*)(0-x*)=>all tangent lines go through (0,0).