Presentations
6.1 - 6.8Pacing Tracker
- The quizzes have included questions for 28 objectives. How many have you passed? What are you plans to master those that you haven't mastered yet?
- Once you've passed 26 objectives, you have hit 80% of the total objectives (needed to pass the class with a grade of A/B/C).
- We've finished units 1 through 5. Have you started your self-directed learning project for each unit?
- The 6th project can be over any topic from the entire semester. Feel free to get started on this one as soon as you have an idea.
- Remember you can submit only one SDL project per week. Plan ahead and don't let yourself get behind.
Operators - Discussion
- What is an operator? A function whose domain is a function. When we do something to functions, rather than call it a function of functions, we call it an operator on functions.
- The derivative $\frac{df}{dx}$ operates on a function $f$ to produce a new function $f'$.
- The partial derivative $\frac{\partial f}{\partial x}$ operates on a function $f(x,y)$ to produce a new function $f_x(x,y)$.
- The integral $\int_a^b f(x) dx$ operates on a function $f(x)$ to produce a number.
- The gradient $\vec \nabla f$ operates on a function $f(x,y,z)$ to produce a vector field $(f_x,f_y, f_z)$.
- The del operator $\vec \nabla = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right)$ or $\vec \nabla = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$ can be used in a variety of ways to produce an operator.
- Three new quantities (A great book: "Div, Grad, Curl, and all that" by H. M. Schey - Electrostatics)
- The gradient of a function $f$, namely $\vec \nabla f = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)f = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)$, abuses the notion of scalar multiplication to produce an operator on a function $f$.
- The divergence of a vector field $\vec F$ is $\vec \nabla \cdot \vec F$, read "del dot F." This operator helps understand whether a vector field $\vec F$ is expanding or contracting. This helps understand gas diffusion. Things that are incompressible (a common assumption about liquids) have zero divergence which means $\vec \nabla \cdot \vec F=0$.
- The curl of a vector field $\vec F$ is $\vec \nabla\times \vec F$, read "del cross F.". This quantity tracks the rate and direction of rotation that a vector field $\vec F$ causes. Irrotational vector fields have zero curl. We'll see that every gradient field has zero curl, which is precisely related to why we never saw imaginary eigenvalues in the optimization unit.
We've used all the symbols above before, but now we'll use them in a new way. This will be a common theme in the 6th unit, namely we'll review things we've seen all semester, and use them in a slightly new way.
Brain Gains
Let $$\vec \nabla = \left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right).$$
- For the vector field $\vec F(x,y,z) = (3xy,2x+4y+5z,y+xz^2)$, compute the derivative $D \vec F$ (you'll get a 3 by 3 matrix).
Solution
The derivative is $$D\vec F = \begin{bmatrix} 3y &3x&0 \\ 2 &4& 5\\ z^2 &1& 2xz \\ \end{bmatrix}. $$
- For the vector field $\vec F(x,y,z) = (3xy,2x+4y+5z,y+xz^2)$, compute $\vec \nabla \cdot \vec F$.
Solution
We compute $$\begin{align} \vec \nabla \cdot \vec F &=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\cdot (3xy,2x+4y+5z,y+xz^2)\\ &=\frac{\partial}{\partial x}(3xy)+\frac{\partial}{\partial y}(2x+4y+5z)+\frac{\partial}{\partial z}(y+xz^2)\\ &=(3y)+(4)+(2xz) .\end{align}$$ Note this is just the sum of the diagonal entries of $$D\vec F = \begin{bmatrix} \color{red}{3y} &3x&0 \\ 2 &\color{red}{4}& 5\\ z^2 &1& \color{red}{2xz} \\ \end{bmatrix}. $$
- For the vector field $\vec F(x,y,z) = (3xy,2x+4y+5z,y+xz^2)$, compute $\vec \nabla \times \vec F$.
Solution
We compute $$\begin{align} \vec \nabla \times \vec F &=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\times (3xy,2x+4y+5z,y+xz^2)\\ &=\left( \frac{\partial}{\partial y}(y+xz^2) - \frac{\partial}{\partial z}(2x+4y+5z), \frac{\partial}{\partial z}(3xy) - \frac{\partial}{\partial x}(y+xz^2), \frac{\partial}{\partial x}(2x+4y+5z) - \frac{\partial}{\partial y}(3xy) \right)\\ &=\left( (1) - (5), (0) - (z^2), (2) - (3x) \right) .\end{align}$$ Notice how each component is the difference of two terms in $$D\vec F = \begin{bmatrix} 3y &\color{blue}{3x}&\color{green}{0} \\ \color{blue}{2} &4& \color{red}{5}\\ \color{green}{z^2} &\color{red}{1}& 2xz \\ \end{bmatrix}. $$
- For the vector field $\vec F = (4x+2xy,x^2+2y)$, find a function $f(x,y)$ so that $\vec \nabla f = \vec F$. We call $f$ a potential for $\vec F$. In other words, if we know $f_x = 4x+2xy$ and $f_y = x^2+2y$, then what is $f$?
Solution
Such a function is $f(x,y) = 2x^2+x^2y+y^2$.
- Since $f_x = 4x+2xy$, anti differentiation with respect to $x$ yields $f(x,y) = 2x^2+x^2y +C(y)$, where $C(y)$ is any function of $y$ so that the partial with respect to $x$ is zero.
- Since $f_y = x^2+2y$, anti differentiation with respect to $y$ yields $f(x,y) = x^2y+y^2+D(x)$, where $D(x)$ is any function of $x$ so that the partial with respect to $y$ is zero.
Letting $C(y) = y^2$ and $D(x)=2x^2$ gives the same function $f(x,y) = 2x^2+x^2y+y^2$ in both steps above. A quick check reveals $\vec \nabla f = (4x+2xy,x^2+2y)$, which equals $\vec F$ as needed.
We could have used $f(x,y) = 2x^2+x^2y+y^2+12$, or rather $f(x,y) = 2x^2+x^2y+y^2+C$ for any constant $C$. The problem asked for a potential, not all potentials.
- For the vector field $\vec F = (4x+2xy,x^2+2y)$ (same as above), find the difference in a potential from $(0,1)$ to $(-1,2)$. Then explain why $$ f(-1,2) - f(0,1) = \int_{(0,1)}^{(-1,2)}(4x+2xy)dx+(x^2+2y)dy .$$
Solution
We already found a potential is $f(x,y) = 2x^2+x^2y+y^2$. The difference is $f(-1,2) - f(0,1) = [2+2+4] - [0+0+1] = 7$.
We know that $df = (4x+2xy)dx+(x^2+2y)dy$, which represents a small change in potential. Summing small changes in potential gives a total change in potential, which gives $$f(-1,2) - f(0,1) = \int_C df = \int_{(0,1)}^{(-1,2)}(4x+2xy)dx+(x^2+2y)dy .$$ The last part of the computation above is the formula for the work done by $\vec F = (4x+2xy,x^2+2y)$ from $(0,1)$ to $(-1,2)$. For a vector field $\vec F$ with a potential $f$, the work done along a smooth curve $C$, which starts at $A$ and ends at $B$, is given by $$f(B) - f(A) = \int_C df = \int_A^B \vec F\cdot d\vec r.$$
- Does every vector field have a potential?
Solution
Nope. That is precisely what you'll be examining in groups today. However, when a vector field has a potential, it makes work computations much simpler. This is because the fundamental theorem of line integrals states that for a piecewise smooth curve $C$ that starts at $A$ and ends at $B$, we have $$\int_C \vec \nabla f \cdot d\vec r = f(B) - f(A).$$ We can read the above as "the work done by $\vec F = \vec \nabla f$ along $C$ is the difference in the potential between the start and end of the path.
The proof of the theorem above follows:
- The chain rule tells us that $\frac{df}{dt} = \vec \nabla f\cdot \frac{d\vec r}{dt}$. This means a little change in $f$ is equal to $df = \vec \nabla f\cdot d\vec r$.
- Summing up little changes in $f$ over a curve $C$ that starts at $A$ and ends at $B$ gives the total change in $f$ from $A$ to $B$. This is precisely what we mean by $$\int_C df = f(B) - f(A).$$
- Because $df = \vec \nabla f\cdot d\vec r$, we can rewrite the above as $$\int_C \vec \nabla f \cdot d\vec r = f(B) - f(A).$$
Group problems
- Let $f = xy^2+3x$.
- Compute $\vec F = \vec \nabla f$. (We'll use capital $F$ for vector fields, and lower case $f$ for real-valued functions, i.e. the output is a single real number, rather than a vector.)
- Compute $D^2f$ (so $D\vec F$). It's a square matrix.
- Compute $\int_{ (2,1) }^{ (1,3) } df$ (total change in $f$ from $(2,1)$ to $(1,3)$). [Check: Plugging the points into $f$ yields $12-8 = 4$.]
- Let $\vec F = (2xy+4, x^2+2y)$.
- Compute $D\vec F$. It's a square matrix.
- Find a real-valued function $f$ so that $\vec \nabla f = \vec F $. In particular, this means $df = \vec F \cdot d\vec r$. [Check: $f = x^2y+4x+y^2$.]
- Find the work done by $\vec F$ (so $\int_C \vec F\cdot d\vec r$) to get from $(2,0)$ to $(0,3)$. (Or simpler, just compute $\int_C df$, the total change in $f$ from $(2,0)$ to $(0,3)$, which you can do because $df = \vec F \cdot d\vec r$. ) [Check: $9-8=1$.]
- Let $\vec F = (2x+3y, 4x+5y)$.
- Compute $D\vec F$.
- Why it is impossible to find a function $f$ so that $\vec F = \vec \nabla f$.
- Given a vector field $\vec F$, what condition must be true about $D\vec F$ for there to be a function $f$ such that $\vec\nabla f = \vec F$? We call such function $f$ a potential for $\vec F$. When $\vec F$ has a potential, we say that $\vec F$ is a gradient field.
- Compute the derivative of each vector field $\vec F$ below (obtaining a square matrix). Then find a potential for $\vec F$ or explain why the vector field has no potential.
- $\vec F = (2x,3y)$ [Check: $D\vec F = \begin{bmatrix}2&0\\0&3\end{bmatrix}$ and $f = x^2+\frac{3}{2}y^2$ yields $\vec \nabla f = (2x,3y)$. We can quickly verify that $\vec\nabla f = \vec F$ by a direct computation. ]
- $\vec F = (2y,3x)$
- $\vec F = (3y,3x)$
- $\vec F = (4x,5y,6z)$
- $\vec F = (4x,5z,6y)$
- $\vec F = (4x,5z,5y)$
- $\vec F = (2x-y,-x+4y)$
- $\vec F = (y^2+2x,2xy)$
- $\vec F = (x+yz,xz+4yz,xy+2y^2)$
- $\vec F = (x+yz,4yz,xy+2y^2)$
- $\vec F = (x+yz,xz+4yz,xy)$
- $\vec F = (yz,xz+4yz,xy+2y^2)$
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