Find the time derivative of an integral of a function over a moving curve, involving partial derivatives and curve geometry | Step-by-Step Solution

What We're Solving:

We need to find how fast an integral changes when both the function being integrated AND the curve we're integrating along are changing with time. This is like asking: "If I'm measuring the total temperature along a rope that's moving through space, and the temperature field itself is changing, how fast is my total measurement changing?"

The Approach:

This is a beautiful application of the transport theorem (also called the Reynolds transport theorem). We'll use the fundamental theorem of calculus in a clever way, treating the moving curve as having endpoints that change with time. The key insight is that the rate of change comes from two sources:

• 1. The function f itself changing at each point (even if the curve stayed still)
• 2. The curve moving to new locations where f has different values

Step-by-Step Solution:

Step 1: Set up the integral we want to differentiate Let's say we have: $$I(t) = \int_0^L f(\mathbf{r}(l,t), t) \, dl$$

where $\mathbf{r}(l,t)$ describes our moving curve parameterized by arc length $l$, and $f(\mathbf{x},t)$ is our time-dependent function.

Step 2: Apply the fundamental theorem of calculus Since we're differentiating an integral with respect to time, we can bring the derivative inside: $$\frac{dI}{dt} = \int_0^L \frac{d}{dt}[f(\mathbf{r}(l,t), t)] \, dl$$

Step 3: Use the chain rule on the integrand The function $f(\mathbf{r}(l,t), t)$ depends on time in TWO ways:

• Explicitly through its second argument
• Implicitly because $\mathbf{r}(l,t)$ changes with time

So: $$\frac{d}{dt}[f(\mathbf{r}(l,t), t)] = \frac{\partial f}{\partial t} + \nabla f \cdot \frac{\partial \mathbf{r}}{\partial t}$$

Step 4: Recognize the velocity vector The term $\frac{\partial \mathbf{r}}{\partial t}$ is exactly our velocity vector $\mathbf{V}(l,t)$! This represents how fast each point on the curve is moving.

Step 5: Substitute back into our integral $$\frac{dI}{dt} = \int_0^L \left[\frac{\partial f}{\partial t} + \nabla f \cdot \mathbf{V}(l,t)\right] dl$$

Step 6: Interpret each term

• $\frac{\partial f}{\partial t}$: How fast $f$ is changing at each fixed location
• $\nabla f \cdot \mathbf{V}$: How much $f$ changes as we move with the curve (directional derivative in the direction of motion)

The Answer:

The time derivative of an integral along a moving curve is: $$\boxed{\frac{d}{dt}\int_{\text{curve}} f(\mathbf{x},t) \, dl = \int_{\text{curve}} \left[\frac{\partial f}{\partial t} + \nabla f \cdot \mathbf{V}\right] dl}$$

This elegant result shows that the rate of change equals the integral of:

• 1. The local time rate of change of $f$, plus
• 2. The convective rate of change due to the curve's motion

Memory Tip:

Think of it like this: You're walking along a moving sidewalk while carrying a thermometer. The temperature reading changes for two reasons - the weather itself is changing (∂f/∂t), AND you're moving to different locations (∇f·V). The total rate of change is the sum of both effects!

This is the same principle behind many physics equations, like the material derivative in fluid mechanics. Once you see this pattern, you'll recognize it everywhere in advanced mathematics and physics!