Gradient Theorem - Fundamental Theorem for Line Integrals

Prerequisites

• Fundamental Theorem of Calculus
• Line integral

Gradient Theorem

The gradient theorem, also known as the fundamental theorem of calculus for line integrals, says that the line integral of the gradient of a scalar field, $\nabla f$ over a curve $\gamma$ with start and end points $a$ and $b$, is equal to evaluating the difference of the original scalar field $f$ at the endpoints $b$ and $a$:

$${\int }_a^b(\nabla f)\cdot d\vec{r}=f(b)-f(a)$$

Where $d\vec{r}$ is the tangent vector to the curve $\gamma$.

Proof

Let $f:{\mathbb{R}}^n\to \mathbb{R}$ be a scalar field, and $\gamma$ be a differentiable curve in ${\mathbb{R}}^n$ starts at $a\in {\mathbb{R}}^n$ and ends at $b\in {\mathbb{R}}^n$. We can parameterize the curve $\gamma$ with a differentiable function $r(t):[0,1]\to {\mathbb{R}}^n$, so that $r(0)=a$ and $r(1)=b$.

Note that $f(r(t))$ represents the scalar field $f$ evaluated along the curve $\gamma$ at time $t$. $f(r(t))$ maps $\mathbb{R}\to {\mathbb{R}}^n\to \mathbb{R}$ by mapping time to a configuration and then to a scalar. If we take the derivative with respect to time, then using the chain rule, we get the Directional derivative for $f$ in the direction of $r^{\prime }(t)=\frac{d}{dt}r(t)$, which is the tangent vector to the curve $\gamma$:

$$\frac{d}{dt}(f(r(t)))=\nabla f(r(t))\cdot r^{\prime }(t)$$

We can now prove the Gradient Theorem by applying the definition of a line integral, using the chain rule from above, and also applying the Fundamental Theorem of Calculus:

$$\begin{array}{rlrl}{\int }_a^b(\nabla f)\cdot d\vec{r}& ={\int }_0^1\nabla f(r(t))\cdot r^{\prime }(t)dt& & \text{(Def. of line integral)}\\ & ={\int }_0^1\frac{d}{dt}(f(r(t)))dt& & \text{(Substitute via chain rule)}\\ & =f(r(1))-f(r(0))& & \text{(Fundamental Theorem of Calculus)}\\ & =f(b)-f(a)& & \text{(Substitute in }r(0)=a,r(1)=b\text{)}\end{array}$$

References

• Wikipedia on Gradient Theorem