


Rapid Recall ...or is this actually Jivin' Generation?
- For $\vec F = (xy^2, 3x+4y+5z, 2xyz)$, compute $D\vec F(x,y,z)$.
Solution
The differential of $\vec F$, written as a linear combination of partial derivatives, is $$ d\vec F = \begin{pmatrix}y^2\\3\\2yz\end{pmatrix}dx + \begin{pmatrix}2xy\\4\\2xz\end{pmatrix}dy + \begin{pmatrix}0\\5\\2xy\end{pmatrix}dz .$$ The above, when we write is as the product of a matrix and a vector, gives $$ d\vec F = \begin{bmatrix} \begin{matrix}y^2\\3\\2yz\end{matrix} &\begin{matrix}2xy\\4\\2xz\end{matrix} &\begin{matrix}0\\5\\2xy\end{matrix} \end{bmatrix} \begin{pmatrix}dx\\dy\\dz\end{pmatrix} .$$ The derivative of $\vec F$ is hence the matrix $$ D\vec F(x,y,z) = \begin{bmatrix} \begin{matrix}y^2\\3\\2yz\end{matrix} &\begin{matrix}2xy\\4\\2xz\end{matrix} &\begin{matrix}0\\5\\2xy\end{matrix} \end{bmatrix} . $$
- For $\vec F (x,y) = (2x+3y,3x)$, find a function $f(x,y)$ so that $\vec \nabla f=\vec F$. Such a function $f(x,y)$ we call a potential for $\vec F$.
Solution
We need $\vec F(x,y)=(f_x,f_y)$. This means we need a function $f(x,y)$ such that $f_x = 2x+3y$ and $f_y=3x$. Integrating each partial, with respect to the corresponding variable, gives $$\int f_x dx = \int2x+3ydx = x^2+3xy+C_1(y)\quad\text{and} \quad \int f_y dy = \int 3x dy=3xy+C_2(x).$$ Because we assumed $y$ is constant in the first integral, note that $C_1(y)$ could be any function of $y$. Similarly note that $C_2(x)$ could be any function of $x$. One option for the function $f(x,y)$ we seek is $$f(x,y) = x^2+3xy,$$ as this fits both
- $f(x,y)=x^2+3xy+C_1(y)$ with $C_1(y)=0$, and
- $f(x,y)=3xy+C_2(x)$ with $C_2(x) = x^2$.
- For the vector field $\vec F$ from the previous problem (units in Newtons), compute the work done by $\vec F$ on an object that moves along the curve $C$ that starts at $A=(1,2)$, follows the path $\vec r(t)=(-3,4)t+(1,2)$, and ends at $B=(-2,6)$ (units in meters).
Solution
Since $\vec F = \vec \nabla f$, we know that $\int_C\vec F\cdot d\vec r = \int_C df = f(B)-f(A)$, work done is the difference in potential. We compute
- $f(B) = f(-2,6)=(-2)^2+3(-2)(6) = -32$,
- $f(A) = f(1,2)=(1)^2+3(1)(2) = 7$, and
- $f(B)-f(A) = -32-7 = -39$.
The work done is -39 Nm.
Operators
- What is an operator?
- The derivative $\frac{df}{dx}$
- The integral $\int_a^b f(x) dx$
- The del operator $\vec \nabla = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right)$ or $\vec \nabla = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$.
- Three new quantities (A great book: "Div, Grad, Curl, and all that" by H. M. Schey - Electrostatics)
- Gradient of a function $f$: $\vec \nabla f$.
- Divergence of a vector field $\vec F$: $\vec \nabla \cdot \vec F$.
- Curl of a vector field $\vec F$: $\vec \nabla\times \vec F$.
Practice
- Let $f(x,y,z) = 3xe^{yz^2}$. Compute the gradient $\vec \nabla f$.
- Let $\vec F(x,y,z) = (3x+4y+5z, xy, yz^2)$. Compute the divergence $\vec \nabla \cdot \vec F$.
- Let $\vec F(x,y,z) = (3x+4y+5z, xy, yz^2)$. Compute the curl $\vec \nabla \times \vec F$.
- As a class, let's pick a function $f$ and vector field $\vec F$. Then compute (1) the gradient $\vec \nabla f$, (2) the divergence $\vec \nabla \cdot \vec F$ and (3) the curl $\vec \nabla \times \vec F$.
- Use the function $f$ from #4, and compute $\vec \nabla f$ followed by $\vec \nabla \times \vec \nabla f$. Simplify your answer.
- Let $f$ now represent any function. Symbolically compute $\vec \nabla f$ followed by $\vec \nabla \times \vec \nabla f$. Simplify your answer.
- Use the vector field $\vec F$ from #4. Then compute $\vec \nabla \times \vec F$ followed by $\vec \nabla \cdot ( \vec \nabla \times \vec F)$.
- Let $\vec F=(M,N,P)$ represent any vector field. Symbolically compute $\vec \nabla \times \vec F$ followed by $\vec \nabla \cdot ( \vec \nabla \times \vec F)$. Simplify your answer.
- Pick a specific vector field $\vec F$. Then compute $\vec \nabla \cdot \vec F$ followed by $\vec \nabla ( \vec \nabla \cdot \vec F)$.
- Let $\vec F=(M,N,P)$ represent any vector field. Symbolically compute $\vec \nabla \cdot \vec F$ followed by $\vec \nabla ( \vec \nabla \cdot \vec F)$.
Connecting Gradients and Vector Fields - Potential Functions
- For $f(x,y)$, recall the gradient is the vector field $\vec \nabla f = (f_x,f_y)$ and the differential is $df = f_xdx+f_ydy = (f_x,f_y)\cdot(dx,dy)$.
- The derivative of the gradient is $D(\vec \nabla f) = \begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}$.
- For a vector field $\vec F = (M,N)$, recall the derivative is $D\vec F = \begin{bmatrix}M_{x}&M_{y}\\N_{x}&N_{y}\end{bmatrix}$.
- The differential of $f$ is $df = f_x dx+f_y dy$. The differential of work done by the vector field $\vec F = (M,N)$ is $dW = Mdx+Ndy$.
- Physically, what does the integral $\int_C df$ compute? (Add up ... to get ...)
- Physically, what does the integral $\int_C dW$ compute? (Add up ... to get ...)
- Suppose $f_x dx+f_y dy = Mdx+Ndy$. then how are $f$ and $\vec F$ related?
- When $f_x dx+f_y dy = Mdx+Ndy$, then we have $\int_C dW = \int_C df$. What does this mean in words?
The work done by $\vec F$ along a curve $C$ is equal to ....
Practice
- Let $\vec F=(2x+3y, 3x+4y)$. Let $C$ be the curve parametrized by $\vec r(t) = (2-2t,3t)$ for $0\leq t\leq 1$.
- Draw the curve. At a few points on the curve, draw the vector field.
- From your picture, will the work done by $\vec F$ along $C$ be positive or negative?
- Compute the work done by $\vec F$ along the curve.
- Let $f(x,y) = x^2+3xy+2y^2$.
- Compute $\vec \nabla f$ and $df$.
- State $\vec r(0)$ and the value of $f$ at $t=0$? Then repeat this at $t=1$.
- Compute $\int_C df$. In other words, state the total change in $f$ along the curve $C$. Compare this to the first question.
- Let $f = xy^2+3x$.
- Compute $\vec \nabla f$.
- Compute $D^2f$.
- Compute $\int_{ (2,1) }^{ (-1,3) } df$
- Let $\vec F = (2xy+4, x^2+2y)$.
- Compute $D\vec F$.
- Find a function $f$ so that $\vec F = \vec \nabla f$.
- Find the work done by $f$ to get from $(2,0)$ to $(0,3)$. (Hint, what is $\int_C df$?)
- Let $\vec F = (2x+3y, 4x+5y)$.
- Compute $D\vec F$.
- Explain why is it impossible to find a function $f$ so that $\vec F = \vec \nabla f$.
- Under what conditions will a vector field $\vec F$ have a function $f$ such that $\vec\nabla f = \vec F$. We call such function $f$ a potential for $\vec F$.
- Find a potential for each of the following, or explain why none exists.
- $\vec F = (2x,3y)$
- $\vec F = (2y,3x)$
- $\vec F = (3y,3x)$
- $\vec F = (4x,5y,6z)$
- $\vec F = (4x,5z,6y)$
- $\vec F = (4x,5z,5y)$
- $\vec F = (2x-y,-x+4y)$
- $\vec F = (y^2+2x,2xy)$
- $\vec F = (x+yz,xz+4yz,xy+2y^2)$
- $\vec F = (x+yz,4yz,xy+2y^2)$
- $\vec F = (x+yz,xz+4yz,xy)$
- $\vec F = (yz,xz+4yz,xy+2y^2)$
Boundaries
- Given a line segment on the $x$-axis, what's the boundary? How do we describe the segment mathematically. How do we describe the boundary mathematically? What does $\int_C \frac{df}{dx}dx = f(b)-f(a)$ say in terms of boundaries?
- Given a wire, what's the boundary? How do we describe the wire mathematically. How do we describe the boundary mathematically?
- Given a region $R$ in the plane, what's the boundary? How do we describe the region mathematically? How do we describe the boundary mathematically?
- Given a solid $D$ in space, what's the boundary? How do we describe the solid mathematically? How do we describe the boundary mathematically?
Wires and Surfaces
- What is the difference between $y=f(x)$ and $\vec r(x) = (x,f(x))$?
- What is the difference between $z=f(x,y)$ and $\vec r(x,y) = (x,y,f(x,y))$?
- How do we mathematically describe a wire?
- How do we mathematically describe a surface?
- How do we find the length $s$ of a wire? We add up little bits of length $ds$.
- What does each part of $\int_C ds = \int_C |\frac{d\vec r}{dt}|dt$ mean?
- How do we find the surface area $\sigma$ of a surface? We add up little bits of surface area $d\sigma$.
- What does each part of $\iint_S d\sigma = \iint |\frac{\partial\vec r}{\partial u}\times\frac{\partial\vec r}{\partial v}|dudv$ mean?
Fundamental Theorems
In the work below, note that $\vec T$ is a unit tangent vector to a curve, while $\vec n$ is a unit vector that is normal to a boundary and pointing away from a region.
- $f(b)-f(a) = \int_a^b \frac{df}{dx} dx$ - Fundamental Theorem of Calculus
- $f(B)-f(A) = \int_C \vec \nabla f\cdot \vec T ds$ - Fundamental Theorem of Line Integrals
- $\int_C \vec F\cdot \vec T ds = \iint_R (\vec \nabla \times \vec F)\cdot (0,0,1)dA$ - Green's Theorem
- $\int_C \vec F\cdot \vec T ds = \iint_S (\vec \nabla \times \vec F)\cdot \vec n d\sigma$ - Stokes's Theorem
- $\int_C \vec F\cdot \vec n ds = \iint_R (\vec \nabla \cdot \vec F)dA$ - Green's theorem
- $\iint_R \vec F\cdot \vec n d\sigma = \iiint_D (\vec \nabla \cdot \vec F)dV$ - Divergence Theorem
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