# Thread: Hard multivariable calculus problem..

1. ## Hard multivariable calculus problem..

Hi.
I have this one in my textbook.
It is really hard!!!
I can't do anything.

Problem:
For what values of the number r is the function
$\displaystyle f(x,y,z)=\frac{(x+y+z)^r}{x^2+y^2+z^2} \,$ if $\displaystyle \, (x,y,z) \neq 0$
$\displaystyle f(x,y,z) = 0 \,$ if $\displaystyle \, (x,y,z)=0$

continuous on R ?
the problem here is that the question did not specify r, is it natural or real .. etc

2. ok you need to replace the x, y, z cartesian coordinates by their corresponding rho, phi, theta coordinates...

3. Originally Posted by rebghb
ok you need to replace the x, y, z cartesian coordinates by their corresponding rho, phi, theta coordinates...

The bad news is that this problems comes before the spherical coordinates' section in the book.

4. ρ,Φ,θ
f(x(ρ,Φ,θ),y(ρ,Φ,θ),z(ρ,Φ,θ)) = (ρsinΦcosθ + ρsinΦcosθ + ρcosΦ )^r / ρ^2
= [ρ^(r)](sinΦcosθ + sinΦcosθ + cosΦ) / ρ^2
note that the numerator is bounded because its sin terms and cosine terms are all bounded by -1 and 1. Also note that the function is piecewise continuous at (x,y,z) = (0,0,0) or extended by continuity. all we need is to equate the limit of f as ---> ρ <--- approches zero to zero. that is:

ρ^(r) > ρ^2 <=> ρ^(r) / ρ^2 > 1 <=> ρ^(r - 2) > 1 or r - 2 > 0 <=> r > 2

Hope this convinces you ( and is 100% correct)

5. i've been typing this for an hour, sorry, i'll try to cook up another answer while this is supposed to be the only getawat to such problems...

6. Originally Posted by rebghb
i've been typing this for an hour, sorry, i'll try to cook up another answer while this is supposed to be the only getawat to such problems...
Ohhh. sorry for that
Use latex. It saves time .
Thanks anyway.

7. OK, here's another way to look at it (btw i did a minor algebra error, its supposed to be rho^r and not ^1/r)...

(x + y + z)^r / (x2 + y2 + z2) must approach zero thats the numerator is muuuuuuch smaller than the denominator... This is equivelent to saying that (x2 + y2 + z2) / (x + y + z)^r goes to infinity as (x,y,z) tend to (0,0,0)... u know that x2 + y2 + z2 = (x+y+z)2 - 2xy - 2yz - 2xz so:

(x2 + y2 + z2) / (x + y + z)^r = (x+y+z)2 / (x+y+z)r -2(xy + xz +yz)/(x + y + z)^r

the 1st term is muuuuch greater than the second term (x,y and z are all of order 2). So:

(x+y+z)r / (x+y+z)2 ---> +inf so (x+y+z)^2 / (x+y+z)^r goes to zero...
implies that (x+y+z)^r is "greater" than (x+y+z)^2 or simply r>2
QED??!! LETS HOPE