Temperature Gradient in a Material

Suppose for a material that the temperature is a function of position, so that  
\[T=T(x,y,z)\]
.
Suppose we have a curve  
\[\mathbf{r}\]
  in the material. The curve is parametrized in terms of its distance from sme point, so that  
\[\mathbf{r}= \mathbf{r}(s)\]
.
The change in  
\[T\]
  as it goes from  
\[(x,y,z)\]
  to  
\[(x+ \delta x , y+ \delta y , z+ \delta z )\]
  along the curve is  
\[\ delta T = \frac{\partial T}{\partial x} \delta x + \frac{\partial T}{\partial y} \delta y + \frac{\partial T}{\partial z} \delta z = (\frac{\partial T}{\partial x} , \frac{\partial T}{\partial y} , \frac{\partial T}{\partial z}) \cdot ( \delta x, \delta y \ \delta z = \mathbf{\nabla} \cdot \delta \mathbf{r}\]

Hence  
\[\ \frac{\delta T}{\delta s} = \mathbf{\nabla} \cdot \frac{\delta \mathbf{r}}{\delta s}\]

Let  
\[\delta \mathbf{r} \rightarrow \mathbf{0}\]
  then
\[\ \frac{d T}{ds} = (\mathbf{\nabla} T) \cdot \frac{d \mathbf{r}}{ds}\]