Function with 2 variables

[hyper]

This is the problem, we have:

f(0,0)=0 for (x,y) not(0,0) f(x,y)= $$\frac{2*x^3*y}{x^6+y^2}$$

a) What is lim f(x,y) (x,y)=->(0,0) alonng any straign line.

b) Show that in every circle centered at the origin there is a point where f(x,y)=1

c) Decide if f(x,y) is continious at the origin.

d)Decide if the partial derivatives at the origin exicsts, and if the do what are they?

my attempt at the sollution:

a) Here I used polarcoordinates and found out that the limit valuse was zero, so I don't need you to try on this.

b) If you choose y=x^3 it is allways one there, so I think I got this, but I want you more to look at the last 2 questions.

c) Now comes the tricky part. Along every straight line the limit valuse is zero, and f(0,0)=0 so it might excist. But if we try the let y have the value y=k*x^3, f(x,y) is allways $$\frac{2k}{1+k^2}$$ and it should also have this value when it goes to the origin, so so if we choose k not 0 we don't get the limit zero and it isn't contious. Do you guys have any ideas here?



d) Here I tried partial differensiation with the expression and evaluating in (0,0), but you can't do this because the you would divide with zero, and that is not ok to do. So maybe I should combine f(0,0)=0 and f(x,y)= $$\frac{2*x^3*y}{x^6+y^2}$$ ? Maybe the partial derivative do not excist there continious?

Help would be greatly appreciated.:)

 

[mighty2000]

a) To find the limit along any straight line, we can substitute y=mx into the expression and take the limit as x approaches 0. This gives us:

lim f(x,mx) = lim \frac{2*x^3*mx}{x^6+m^2*x^2} = lim \frac{2mx^4}{x^2(x^4+m^2)} = 0

Therefore, the limit along any straight line is 0.

b) To show that there is a point where f(x,y)=1 for every circle centered at the origin, we can use the fact that every point on a circle can be represented in polar coordinates as (rcosθ, rsinθ), where r is the radius and θ is the angle. Substituting this into the expression, we get:

f(rcosθ, rsinθ) = \frac{2*r^3*cos^3θ*sinθ}{r^6*cos^6θ+r^2*sin^2θ}

Simplifying, we get:

f(rcosθ, rsinθ) = \frac{2*r*cos^3θ*sinθ}{r^4*cos^6θ+sin^2θ}

Since r is a constant, we can factor it out of the denominator:

f(rcosθ, rsinθ) = \frac{2*r*cos^3θ*sinθ}{r^4*(cos^6θ+sin^2θ)}

Since cos^6θ+sin^2θ is always greater than or equal to 1, the denominator is always greater than or equal to r^4. This means that the expression is always less than or equal to \frac{2}{r^3}. Therefore, as r approaches 0, the expression approaches 0. This means that there is a point on every circle where f(x,y)=1.

c) To determine if f(x,y) is continuous at the origin, we need to check if the limit of f(x,y) as (x,y) approaches (0,0) exists and is equal to f(0,0). We have already shown in part a) that the limit along any straight line is 0, and in part b) that there is a point on every circle where f(x,y)=1. Since these two limits do not agree, the limit does not exist and therefore f(x,y) is not continuous at