# Finding the gradient

1. Mar 4, 2006

### UrbanXrisis

Suppose distances are measured in lightyears and that the temperature T of a gaseous nebula is inversely proportional to the distance from a fixed point, which is the origin. The temperature 1 lightyear from the origin is 100 degrees celsius. Find the gradient of T at (x,y,z).

here's what I have:

$$d=\sqrt{x^2+y^2+z^2}=1$$
$$d=x^2+y^2+z^2=1$$
$$T=\frac{1}{x^2+y^2+z^2}$$
so the gradient is:
$$T_x=-\frac{2x}{(x^2+y^2+z^2)^2}$$
$$T_y=-\frac{2y}{(x^2+y^2+z^2)^2}$$
$$T_z=-\frac{2z}{(x^2+y^2+z^2)^2}$$

but this is not right, i where did I go wrong?

2. Mar 4, 2006

### arildno

First of all, read your info properly!
You are told that T(x,y,z) is inversely proportional to the distance from the origin. That means there exist a constant K, so that we have:
$$T(x,y,z)=\frac{K}{\sqrt{x^{2}+y^{2}+z^{2}}}$$
Furthermore, you have the condition:
$$T(x,y,z)=100, \sqrt{x^{2}+y^{2}+z^{2}}=1$$
Can you determine K from this?

3. Mar 4, 2006

### 0rthodontist

The first equation you have for distance is correct but I don't see why you are setting it to 1. Then you squared it but didn't write d^2...

The distance from the origin is sqrt(x^2 + y^2 + z^2). T is inversely proportional to that distance so T(x, y, z) = k/sqrt(x^2 + y^2 + z^2) where k is some constant. You know that T(x, y, z) = 100 when sqrt(x^2 + y^2 + z^2) = 1, so what is k? Then you can find the gradient.

4. Mar 4, 2006

### arildno

Do you need a confirmation on that question?

5. Mar 4, 2006

### UrbanXrisis

nope, i got it! thanks!