Introduction
In this part we’ll cover the fundamental theorem of line integrals (also called Gradient theorem).
Definition
Recall the fundamental theorem of calculus:
∫ a b f ′ ( x ) d x = f ( b ) − f ( a ) \int_a^b f^\prime(x)\ dx = f(b) - f(a) ∫ a b f ′ ( x ) d x = f ( b ) − f ( a ) For line integrals:
Let f : D ⊂ R 2 → R f: D \subset \mathbb{R}^2 \rarr \mathbb{R} f : D ⊂ R 2 → R
Let C = D C = D C = D r ⃗ ( t ) \vec{r}(t) r ( t ) a ≤ t ≤ b a \leq t \leq b a ≤ t ≤ b
∫ C ∇ f ⋅ d r = f ( r ⃗ ( b ) ) − f ( r ⃗ ( a ) ) \boxed{\int_C \nabla f \cdot dr = f(\vec{r}(b)) - f(\vec{r}(a))} ∫ C ∇ f ⋅ d r = f ( r ( b )) − f ( r ( a )) Notice how this doesn’t depend on the actual curve, but only of the start and endpoint.
Additionally, if we assume that C C C
∫ C ∇ ⋅ d r = 0 \int_C \nabla \cdot dr = 0 ∫ C ∇ ⋅ d r = 0 Theorem
Let D ⊂ R 2 D \subset \mathbb{R}^2 D ⊂ R 2 D D D D D D
Let F : D → R 2 F: D \rarr \mathbb{R}^2 F : D → R 2 F F F f f f ∇ f = F \nabla f = F ∇ f = F
∫ C F ⋅ d r = 0 ∣ for every closed curve C ⊂ D \int_C F \cdot dr = 0 \ | \ \text{for every closed curve } C \subset D ∫ C F ⋅ d r = 0 ∣ for every closed curve C ⊂ D Proposition
NB: This only applies for R 2 \mathbb{R}^2 R 2
Suppose F ( x , y ) = P ( x , y ) i ⃗ + Q ( x , y ) j ⃗ F(x, y) = P(x, y) \vec{i} + Q(x, y) \vec{j} F ( x , y ) = P ( x , y ) i + Q ( x , y ) j
∂ P ∂ y = ∂ Q ∂ x \dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x} ∂ y ∂ P = ∂ x ∂ Q Proof
If F F F f : D ⊂ R 2 → R 2 f: D \subset \mathbb{R}^2 \rarr \mathbb{R}^2 f : D ⊂ R 2 → R 2 ∇ f = F \nabla f = F ∇ f = F f x = P f_x = P f x = P f y = Q f_y = Q f y = Q
P y = ∂ P ∂ y = ∂ ∂ y ( ∂ f ∂ x ) = ∂ 2 f ∂ y ∂ x = ∂ 2 f ∂ x ∂ y = ∂ ∂ x ( ∂ f ∂ y ) = ∂ P ∂ y = Q x \begin{align*}
P_y & = \dfrac{\partial P}{\partial y} \newline
& = \dfrac{\partial}{\partial y} \left(\dfrac{\partial f}{\partial x} \right) \newline
& = \dfrac{\partial^2 f}{\partial y \partial x} \newline
& = \dfrac{\partial^2 f}{\partial x \partial y} \newline
& = \dfrac{\partial}{\partial x} \left(\dfrac{\partial f}{\partial y} \right) \newline
& = \dfrac{\partial P}{\partial y} \newline
& = \boxed{Q_x}
\end{align*} P y = ∂ y ∂ P = ∂ y ∂ ( ∂ x ∂ f ) = ∂ y ∂ x ∂ 2 f = ∂ x ∂ y ∂ 2 f = ∂ x ∂ ( ∂ y ∂ f ) = ∂ y ∂ P = Q x If given a field, F = P i ⃗ + Q j ⃗ F = P \vec{i} + Q\vec{j} F = P i + Q j P y = Q x P_y = Q_x P y = Q x F F F
Definition
D ⊂ R 2 D \subset \mathbb{R}^2 D ⊂ R 2 simply connected if no “holes” exists.
Theorem
Suppose D ⊂ R 2 D \subset \mathbb{R}^2 D ⊂ R 2 F : D → R 2 F: D \rarr \mathbb{R}^2 F : D → R 2 F = P i ⃗ + Q j ⃗ F = P\vec{i} + Q\vec{j} F = P i + Q j P y = Q x P_y = Q_x P y = Q x F F F
Let’s now do a bunch of examples to understand all of this theory.
Example
Determine if the following vector fields are conservative:
F ( x , y ) = ( x − y ) i ⃗ + ( x − 2 ) j ⃗ F ( x , y ) = ( 3 + 2 x y ) i ⃗ + ( x 2 − 3 y 2 ) j ⃗ F(x, y) = (x - y)\vec{i} + (x - 2)\vec{j} \newline
F(x, y) = (3 + 2xy)\vec{i} + (x^2 - 3y^2)\vec{j} \newline F ( x , y ) = ( x − y ) i + ( x − 2 ) j F ( x , y ) = ( 3 + 2 x y ) i + ( x 2 − 3 y 2 ) j As we can see, both of these have the domain D = R 2 D = \mathbb{R}^2 D = R 2
In the first case:
P y = − 1 Q x = 1 P_y = -1
Q_x = 1 P y = − 1 Q x = 1 Which means this is not a conservative vector field.
In the second case:
P y = 2 x Q x = 2 x P_y = 2x
Q_x = 2x P y = 2 x Q x = 2 x Which means this is a conservative vector field.
Example
F ( x , y ) = ( 3 + 2 x y ) i ⃗ + ( x 2 − 3 y 2 ) j ⃗ F(x, y) = (3 + 2xy)\vec{i} + (x^2 - 3y^2)\vec{j} F ( x , y ) = ( 3 + 2 x y ) i + ( x 2 − 3 y 2 ) j f f f ∇ f = F \nabla f = F ∇ f = F
Let, f : R 2 → R f: \mathbb{R}^2 \rarr \mathbb{R} f : R 2 → R ∇ f = F \nabla f = F ∇ f = F
Therefore:
f x = 3 + 2 x y f y = x 2 − 3 y 2 f_x = 3 + 2xy \newline
f_y = x^2 - 3y^2 \newline f x = 3 + 2 x y f y = x 2 − 3 y 2 We generally will use this approach (this doesn’t always work but, it’s a good strategy):
Integrate one of these with respect to “their” variable:
f ( x , y ) = ∫ f x ( x , y ) d x = ∫ 3 + 2 x y = 3 x + x 2 y + g ( y ) f(x, y) = \int f_x(x, y)\ dx = \int 3 + 2xy = 3x + x^2y + g(y) f ( x , y ) = ∫ f x ( x , y ) d x = ∫ 3 + 2 x y = 3 x + x 2 y + g ( y ) Notice the g ( y ) g(y) g ( y ) x x x
Let’s now differentiate this with respect to the other variable:
f y ( x , y ) = 0 + x 2 + g ′ ( y ) f_y(x, y) = 0 + x^2 + g^\prime(y) f y ( x , y ) = 0 + x 2 + g ′ ( y ) Now we have two equations for f y f_y f y
x 2 − 3 y 2 = x 2 + g ′ ( y ) g ′ ( y ) = − 3 y 2 g ( y ) = − y 3 + C x^2 - 3y^2 = x^2 + g^\prime(y) \newline
g^\prime(y) = -3y^2 \newline
g(y) = -y^3 + C x 2 − 3 y 2 = x 2 + g ′ ( y ) g ′ ( y ) = − 3 y 2 g ( y ) = − y 3 + C Which means:
f ( x , y ) = 3 x + x 2 y − y 3 + C f(x, y) = 3x + x^2y -y^3 + C f ( x , y ) = 3 x + x 2 y − y 3 + C Let’s do the other way around as well:
f ( x , y ) = ∫ f y ( x , y ) d y = ∫ x 2 − 3 y 2 = x 2 y − y 3 + g ( x ) f(x, y) = \int f_y(x, y)\ dy = \int x^2 - 3y^2 = x^2y - y^3 + g(x) f ( x , y ) = ∫ f y ( x , y ) d y = ∫ x 2 − 3 y 2 = x 2 y − y 3 + g ( x ) f x ( x , y ) = 2 x y + g ′ ( x ) f_x(x, y) = 2xy + g^\prime(x) f x ( x , y ) = 2 x y + g ′ ( x ) 3 + 2 x y = 2 x y + g ′ ( x ) g ′ ( x ) = 3 g ( x ) = 3 x + C 3 + 2xy = 2xy + g^\prime(x) \newline
g^\prime(x) = 3 \newline
g(x) = 3x + C 3 + 2 x y = 2 x y + g ′ ( x ) g ′ ( x ) = 3 g ( x ) = 3 x + C f ( x , y ) = x 2 y − y 3 + 3 x + C ∣ same! f(x, y) = x^2y - y^3 + 3x + C \ | \ \text{same!} f ( x , y ) = x 2 y − y 3 + 3 x + C ∣ same! Example
Evaluate ∫ C F ⋅ d r \int_C F \cdot dr ∫ C F ⋅ d r F ( x , y ) = ( 3 + 2 x y ) i ⃗ + ( x 2 − 3 y 2 ) j ⃗ F(x, y) = (3 + 2xy)\vec{i} + (x^2 - 3y^2)\vec{j} F ( x , y ) = ( 3 + 2 x y ) i + ( x 2 − 3 y 2 ) j C = r ⃗ ( t ) = ( e t s i n ( t ) ) i ⃗ + ( e t c o s ( t ) ) j ⃗ C = \vec{r}(t) = (e^t sin(t))\vec{i} + (e^t cos(t))\vec{j} C = r ( t ) = ( e t s in ( t )) i + ( e t cos ( t )) j
For 0 ≤ t ≤ π 0 \leq t \leq \pi 0 ≤ t ≤ π
Since we already now that F F F ∇ f = F \nabla f = F ∇ f = F R 2 \mathbb{R}^2 R 2
∫ C F ⋅ d r = ∫ C ∇ f ⋅ d r = f ( r ⃗ ( π ) ) − f ( r ⃗ ( 0 ) ) \int_C F \cdot dr = \int_C \nabla f \cdot dr = f(\vec{r}(\pi)) - f(\vec{r}(0)) ∫ C F ⋅ d r = ∫ C ∇ f ⋅ d r = f ( r ( π )) − f ( r ( 0 )) Remember that:
f ( x , y ) = x 2 y − y 3 + 3 x + C f(x, y) = x^2y - y^3 + 3x + C f ( x , y ) = x 2 y − y 3 + 3 x + C Let’s calculate the start and end point:
r ⃗ ( 0 ) = ⟨ 0 , 1 ⟩ r ⃗ ( π ) = ⟨ 0 , − e π ⟩ \vec{r}(0) = \langle 0, 1 \rangle \newline
\vec{r}(\pi) = \langle 0, -e^{\pi} \rangle \newline r ( 0 ) = ⟨ 0 , 1 ⟩ r ( π ) = ⟨ 0 , − e π ⟩ f ( 0 , 1 ) = − 1 + C f ( 0 , − e π ) = − ( − e π ) 3 + C = e 3 π + C f(0, 1) = -1 + C
f(0, -e^{\pi}) = -(-e^{\pi})^3 + C = e^{3\pi} + C f ( 0 , 1 ) = − 1 + C f ( 0 , − e π ) = − ( − e π ) 3 + C = e 3 π + C Which means:
∫ C F ⋅ d r = ( e 3 π + C ) − ( − 1 + C ) = e 3 π + 1 \int_C F \cdot dr = (e^{3\pi} + C) - (-1 + C) = \boxed{e^{3\pi} + 1} ∫ C F ⋅ d r = ( e 3 π + C ) − ( − 1 + C ) = e 3 π + 1 Conservation of energy
(This is just a motivation of the name, nothing really important, but I found it cool so).
Suppose F F F m m m C C C r ⃗ ( t ) \vec{r}(t) r ( t ) a ≤ t ≤ b a \leq t \leq b a ≤ t ≤ b
F = m a F = ma F = ma We can write this as:
F ( r ⃗ ( t ) ) = m r ′ ′ ⃗ ( t ) F(\vec{r}(t)) = m \vec{r^{\prime\prime}}(t) F ( r ( t )) = m r ′′ ( t ) The work done by the force field:
W = ∫ C F ⋅ d r = ∫ a b F ( r ′ ⃗ ( t ) ) ⋅ r ′ ⃗ ( t ) d t W = \int_C F \cdot dr = \int_a^b F(\vec{r^\prime}(t)) \cdot \vec{r^\prime}(t)\ dt W = ∫ C F ⋅ d r = ∫ a b F ( r ′ ( t )) ⋅ r ′ ( t ) d t Using Newtons second law of motion:
∫ a b m r ′ ′ ⃗ ( t ) ⋅ r ′ ⃗ ( t ) d t = m ∫ a b r ′ ′ ⃗ ( t ) ⋅ r ′ ⃗ ( t ) d t \int_a^b m \vec{r^{\prime\prime}}(t) \cdot \vec{r^\prime}(t)\ dt = m \int_a^b \vec{r^{\prime\prime}}(t) \cdot \vec{r^\prime}(t)\ dt ∫ a b m r ′′ ( t ) ⋅ r ′ ( t ) d t = m ∫ a b r ′′ ( t ) ⋅ r ′ ( t ) d t m 2 ∫ a b d d t ∣ r ′ ⃗ ( t ) ∣ 2 d t \dfrac{m}{2} \int_a^b \dfrac{d}{dt} \left| \vec{r^\prime}(t) \right|^2\ dt 2 m ∫ a b d t d r ′ ( t ) 2 d t m 2 ( ∣ r ′ ⃗ ( b ) ∣ 2 − ∣ r ′ ⃗ ( a ) ∣ 2 ) d t \dfrac{m}{2} \left( \left| \vec{r^\prime}(b) \right|^2 - \left| \vec{r^\prime}(a) \right|^2 \right)\ dt 2 m ( r ′ ( b ) 2 − r ′ ( a ) 2 ) d t m 2 ( ∣ v ⃗ ( b ) ∣ 2 − ∣ v ⃗ ( a ) ∣ 2 ) d t \dfrac{m}{2} \left( \left| \vec{v}(b) \right|^2 - \left| \vec{v}(a) \right|^2 \right)\ dt 2 m ( ∣ v ( b ) ∣ 2 − ∣ v ( a ) ∣ 2 ) d t Using that K E = m v 2 2 K_E = \dfrac{mv^2}{2} K E = 2 m v 2
K E ( b ) − K E ( a ) K_E(b) - K_E(a) K E ( b ) − K E ( a ) Suppose now that ∇ f = F \nabla f = F ∇ f = F potential energy of the object is P E ( t ) = − f ( r ⃗ ( t ) ) P_E(t) = -f(\vec{r}(t)) P E ( t ) = − f ( r ( t ))
W = ∫ C F ⋅ d r = ∫ C − ∇ P ⋅ d r = − ( P E ( b ) − P E ( a ) ) W = \int_C F \cdot dr = \int_C - \nabla P \cdot dr = -(P_E(b) - P_E(a)) W = ∫ C F ⋅ d r = ∫ C − ∇ P ⋅ d r = − ( P E ( b ) − P E ( a )) Which means:
K E ( b ) − K E ( a ) = − ( P E ( b ) − P E ( a ) ) K_E(b) - K_E(a) = -(P_E(b) - P_E(a)) K E ( b ) − K E ( a ) = − ( P E ( b ) − P E ( a )) K E ( b ) + P E ( b ) = K E ( a ) + P E ( a ) \boxed{K_E(b) + P_E(b) = K_E(a) + P_E(a)} K E ( b ) + P E ( b ) = K E ( a ) + P E ( a ) Which is the law of conservation of energy!