A very easy math problem I can't do

[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(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

 

[prplchknz]

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

 

[scantilyclad]

my head hurts.

 

[Jack Flak]

my head hurts.

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

 

[Lateralus]

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

 

[ygolo]

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(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).

 

[Totenkindly]

Shite.

I actually used to be able to do this.
About 20 years ago.

Sigh.

 

[Oberon]

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.

 

[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.

 

[Lexlike]

Shite.

I actually used to be able to do this.
About 20 years ago.

Sigh.

Me too. But its actually just one year ago ;p

 

[The_Liquid_Laser]

It's been a while since I've seen this material, but the notation doesn't look quite right to me. (Perhaps I am just not familiar with it.) From the context of the problem f is a function with multiple independent variables, e.g. f(x,y). Therefore f(x/y) doesn't make sense to me. Is this the same as f(x/y, 0) ?

 

[Valiant]

Good LORD! This is why we entj's need the backup of specialists

 

[ygolo]

It's been a while since I've seen this material, but the notation doesn't look quite right to me. (Perhaps I am just not familiar with it.) From the context of the problem f is a function with multiple independent variables, e.g. f(x,y). Therefore f(x/y) doesn't make sense to me. Is this the same as f(x/y, 0) ?

Aha! That interpretation works. (Clever, LL)

Note the "d" are actually denoting partials.

f(x,y)=x(x/y,0)
df(x,y)/dx=f(x/y,0)+x[df(x/y,0)/d(x/y)](d(x/y)/dx)=f(x/y,0)+(x/y)[df(x/y,0)/d(x/y)]
df(x,y)/dy=x[df(x/y,0)/d(x/y)](d(x/y)/dy)=-(x/y)^2[df(x/y,0)/d(x/y)]

Now note for all x,y:
(df(x,y)/dx)x+(df(x,y)/dx)y=xf(x/y,0)+(x^2/y)[df(x/y,0)/d(x/y)]-(x^2/y)[df(x/y,0)/d(x/y)]=xf(x/y,0)=f(x,y).

Note that this is exactly the condition we need.
f(x,y)=(df(x,y)/dx)x+(df(x,y)/dx)y

So, the equation for a tangent plane at point (x0, y0, f(x0,y0)) is:
z-f(x0,y0)=(df(x,y)/dx|x=x0)(x-x0)+(df(x,y)/dx)(y-y0)

The plane goes throught (0,0,0) if and only if:
0-f(x0,y0)=(df(x,y)/dx|x=x0)(0-x0)+(df(x,y)/dx)(0-y0)

Which is the same equation as:
f(x0,y0)=(df(x,y)/dx|x=x0)x0+(df(x,y)/dx)y0

And we know the above equation is simply:
f(x,y)=(df(x,y)/dx)x+(df(x,y)/dx)y with (x,y)=(x0,y0)

So all the tangent planes intercept (0,0,0).

 

[IlyaK1986]

Good LORD! This is why we entj's need the backup of specialists

This is why we ENTJs know enough that we can hold our own with the specialists, but our strength lies in knowing how to best utilize the resources that the specialists present us with.

 

[FDG]

This is why we ENTJs know enough that we can hold our own with the specialists, but our strength lies in knowing how to best utilize the resources that the specialists present us with.

Yeah allright, but my problem with this problem in the end was about the notation of the function, i wasn't sure on how to take the derivative...I've already thnaked ygolo in private for having clarified it to me : )