Define a function z = f(x,y) by f(0,0) = 0 and otherwise?

[matthewturner]

please help me with this. At least some thing to start solving...

Define a function z = f(x,y) by f(0,0) = 0 and otherwise.

f(x,y) =(x^2 y) / (x^2+y^2 )

a. Show that in polar coordinates this function may be expressed (for r≠ 0) as z = r 〖cos〗^2 (θ)sin(θ)

b. Show that if θ is fixed then the graph is given by z = mr, a line of slope
m= 〖cos〗^2 (θ)sin(θ).
(Note that this says that the surface z = f(x,y) is what is called a ruled surface.)

c. Compute the directional derivatives of z in the θ direction. Does Df exist at the point (0,0)? Explain.

 

[tiny-tim]

welcome to pf!

hi matthew! welcome to pf!

(try using the X² button just above the Reply box )

show us what you've tried, and where you're stuck, and then we'll know how to help!

start with "a."

 

[DonAntonio]

please help me with this. At least some thing to start solving...

Define a function z = f(x,y) by f(0,0) = 0 and otherwise.

f(x,y) =(x^2 y) / (x^2+y^2 )

a. Show that in polar coordinates this function may be expressed (for r≠ 0) as z = r 〖cos〗^2 (θ)sin(θ)

b. Show that if θ is fixed then the graph is given by z = mr, a line of slope
m= 〖cos〗^2 (θ)sin(θ).
(Note that this says that the surface z = f(x,y) is what is called a ruled surface.)

c. Compute the directional derivatives of z in the θ direction. Does Df exist at the point (0,0)? Explain.

Start for noting that in polar coordinates, $x=r\cos\theta\,\,,\,\,y=r\sin\theta$ , so substitute and get (a) at least . Now show some self effort.

DonAntonio

 

[matthewturner]

Thanks DonAntonio,

thanks for showing me the path. I almost completed the part a and b. have a confusion about the part c. here Df means the derivative of the f. Do you have any idea about that one?

Thanks again for helping me. I appreciate what you did. As you gave me before I just need a hint to proceed, not the whole proof. thanks again.

 

[blue_raver22]

a. To express the function in polar coordinates, we can substitute x = r cos(θ) and y = r sin(θ) into the function. This gives us:

f(r cos(θ), r sin(θ)) = (r^2 cos^2(θ)sin(θ)) / (r^2) = r 〖cos〗^2 (θ)sin(θ)

b. To show that the graph is given by z = mr, we can rewrite the function in terms of r and θ:

z = r 〖cos〗^2 (θ)sin(θ) = (r sin(θ)) 〖cos〗^2 (θ) = mr

This shows that the graph is a line of slope m = 〖cos〗^2 (θ)sin(θ).

c. The directional derivative of z in the θ direction can be computed using the formula:

Df = (∂z/∂θ) = (∂f/∂x)(∂x/∂θ) + (∂f/∂y)(∂y/∂θ)

Substituting in the values for x and y in terms of r and θ, we get:

Df = (r sin(θ) cos(θ))(-r sin(θ)) + (r cos^2(θ)) (r cos(θ)) = -r^2 sin^2(θ) - r^2 cos^2(θ)

Therefore, the directional derivative in the θ direction is dependent on the value of θ.

At the point (0,0), the directional derivative does not exist because the function is not continuous at this point. This is because the function is defined as 0 at (0,0), but as r approaches 0, the function approaches infinity. Therefore, the function is not differentiable at (0,0) and the directional derivative does not exist.