# Thread: A very easy math problem I can't do

1. ## A very easy math problem I can't do

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 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

2. that's easy?!! I feel stupid. of course I haven't done math in about three years.

YOUR head hurts?! I haven't __________ in a week! *bows, falls over*

5. I'd show you if I could speak...but I'm not about to write it out.

6. 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 :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 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).

7. Shite.

I actually used to be able to do this.

Sigh.

8. 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.

9. Originally Posted by ygolo
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)].
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.

10. Originally Posted by Jennifer
Shite.

I actually used to be able to do this.