Why isn't the gradient vector of a parametric curve parallel to the tangent vector?

Question

Consider a parametric curve defined by the equation:

$$\mathbf{r}(t) = X(t)\mathbf{\hat{i}} + Y(t)\mathbf{\hat{j}} + Z(t)\mathbf{\hat{k}}$$

Paul's online math notes indicate that the unit tangent vector to such a curve is defined by:

$$\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$$

More importantly, $\mathbf{r}'(t)$ is a tangent vector.

Now let's consider a scenario wherein a thin, insulated wire follows the path $\mathbf{r}(t)$ between two arbitrary end points with some boundary conditions applied. The wire will attain some temperature distribution $T=T(x,y,z)=T(t)$.

My intuition tells me (and I hope everyone will agree) that the temperature gradient at any point along the wire should be a vector tangent to the wire, i.e.:

$$\mathbf{\nabla}T \propto \mathbf{r}'(t)$$

Now I want to confirm my intuition. I'll start with the definition of the gradient vector in Cartesian co-oridinates:

$$ \mathbf{\nabla} = \frac{\partial}{\partial x}\mathbf{\hat{i}} + \frac{\partial}{\partial y}\mathbf{\hat{j}} + \frac{\partial}{\partial z}\mathbf{\hat{k}} $$

Using the chain rule I can re-write each of the derivative operators in terms of $t$:

\begin{align} \mathbf{\nabla} &= \frac{\partial t}{\partial x}\frac{\partial}{\partial t}\mathbf{\hat{i}} + \frac{\partial t}{\partial y}\frac{\partial}{\partial t}\mathbf{\hat{j}} + \frac{\partial t}{\partial z}\frac{\partial}{\partial t}\mathbf{\hat{k}} \\ \mathbf{\nabla} &= \left(\frac{\partial t}{\partial x}\mathbf{\hat{i}} + \frac{\partial t}{\partial y}\mathbf{\hat{j}} + \frac{\partial t}{\partial z}\mathbf{\hat{k}}\right) \frac{\partial}{\partial t} \\ \mathbf{\nabla} &= \left(\frac{1}{X'(t)}\mathbf{\hat{i}} + \frac{1}{Y'(t)}\mathbf{\hat{j}} + \frac{1}{Z'(t)}\mathbf{\hat{k}}\right) \frac{\partial}{\partial t} \end{align}

The vector in the above equation should be parallel to $\mathbf{r}'(t)$. I can test this by checking the cross product of these two vectors (should be $\mathbf{0}$ if they are parallel):

\begin{align} \left(\frac{1}{X'(t)}\mathbf{\hat{i}} + \frac{1}{Y'(t)}\mathbf{\hat{j}} + \frac{1}{Z'(t)}\mathbf{\hat{k}}\right) \times \left( X(t)\mathbf{\hat{i}} + Y(t)\mathbf{\hat{j}} + Z(t)\mathbf{\hat{k}} \right) \\ = \left( \frac{Z'}{Y'} - \frac{Y'}{Z'} \right)\mathbf{\hat{i}} + \left( \frac{X'}{Z'} - \frac{Z'}{X'} \right)\mathbf{\hat{j}} + \left( \frac{Y'}{X'} - \frac{X'}{Y'} \right)\mathbf{\hat{k}} \end{align}

So it appears these vectors are parallel only when $X'^2 = Y'^2 = Z'^2$, which seems like a really unlikely restriction. What's wrong here: my intuition, my math, or both?

Also please keep in mind what I'm really interested in is the correct transformation of the gradient operator. This question is just the most concise way to explain what I'm struggling with.

---

I realize that similar questions on this topic have been asked previously:

• characteristic curves tangent and gradient
• Gradient vector of parametric curve

This may seem like a duplicate of one of those, but from what I understand about those questions neither of them address the case where the shape of the domain is confined to a parametric curve, which is the focus here.

Answer 1

If your function $T$ is only defined on the curve ${\bf r}(t)$, its gradient is not defined.

Answer 2

If you have $T:\mathbb{R}^3\to\mathbb{R}$ and $r:\mathbb{R}\to\mathbb{R}^3$ where $r(t)=(X(t),Y(t),Z(t))$, then by the chain rule $$(T\circ r)'(t)=\mathbf{\nabla}T(r(t))\cdot r'(t)$$

I am not sure if that answers your question.