Suppose u is a differentiable function of r and r=sqrt(x^2+y^2+z^2). Show that (du/dx)^2+(du/dy)^2+(du/dz)^2=(du/dr)^2.

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

The problem provides the information that the function u is a function of r and `r(x,y,z) = sqrt(x^2+y^2+z^2)` .

Hence `du/dx = ((du)/(dr))*((dr)/(dx)),`  hence, you need to differentiate the function r with respect to x only  such that:

`(dr)/(dx) = 2x/(2sqrt(x^2+y^2+z^2)) =gt (dr)/(dx) = x/(sqrt(x^2+y^2+z^2))`

You need to find `(dr)/(dy)`  and `(dr)/(dz)`  such that:

`(dr)/(dy) = 2y/(2sqrt(x^2+y^2+z^2)) =gt (dr)/(dy) = y/(sqrt(x^2+y^2+z^2))`

`(dr)/(dz) = 2z/(2sqrt(x^2+y^2+z^2)) =gt (dr)/(dz) = z/(sqrt(x^2+y^2+z^2))`

Raising to square all partial derivatives and evaluating the addition yields:

`((du)/(dr))^2*((dr)/(dx))^2 + ((du)/(dr))^2*((dr)/(dy))^2 + ((du)/(dr))^2*((dr)/(dz))^2 = ((du)/(dr))^2*(x^2/(x^2+y^2+z^2) + y^2/(x^2+y^2+z^2) + z^2/(x^2+y^2+z^2))`

`((du)/(dr))^2*((dr)/(dx))^2 + ((du)/(dr))^2*((dr)/(dy))^2 + ((du)/(dr))^2*((dr)/(dz))^2 = ((du)/(dr))^2*((x^2+y^2+z^2)/(x^2+y^2+z^2))`

`((du)/(dr))^2*((dr)/(dx))^2 + ((du)/(dr))^2*((dr)/(dy))^2 + ((du)/(dr))^2*((dr)/(dz))^2 = ((du)/(dr))^2`

Hence, the last line proves the identity `((du)/(dx))^2+((du)/(dy))^2+((du)/(dz))^2=((du)/(dr))^2` .

We’ve answered 317,615 questions. We can answer yours, too.

Ask a question