For , do what you always do with the absolute value- look at the cases.
If x and y are both positive, this is . It's graph is the straight line through and . Since x and y must be positive, it is specifically the straight line segment between those points. If x is positive, y negative, . That's the line segment between and . If x and y are both negative, , the segment between and . Finally, if x is negative and y positive, , the segment between and . Graphing those you should see a square with vertical and horizontal diagonals or a "diamond" centered on (0,0). The rest should be clear.
You can only find ln of positive numbers so you must have of . That is the set of points (x,y,z) outside the sphere .b) f(x,y,z) = ln(16-(x^2)-(y^2)-(z^2))
This is obviously continuous everywhere except possibly at (0, 0). If it is continuous at (0,0) then the limit as we approach (0,0) along any path must be the same. For a problem like this, I always recommend changing to polar coordinates. That way the distance to (0,0) depends only on r. The limit exists if the limit as r goes to 0 does not depend on [itex]\theta[/itex]. The function is continuous if the limit exists and equal to 0.2.Determine the set of points at which the function is continuous.
f(x,y)= x(sqrtx)(sqrty)/(x^2)+(y^2) if (x,y) not equal to (0,0)
= 0 if (x,y) = (0,0)
Do you know what the derivatve of arctan(x) is? It is any Calculus text. If you don't know it, look it up.3.Find the partial derivatives of ....
a)Find fx and fy of f(x,y) = arctan(x-y)
Use the chain rule. And those are partial derivatives, , and , aren't they?b)Find df/dx and df/dy of f(x,y) = g(x/y)
The fact that you cannot do it, and show no attempt to do it, concerns me. Either you are saying that you do not know basic Calculus and algebra, or you are just not trying.it's my assignment and i can't do it T_T could someone plz help me....