"In a compartment model, what doe a determinant of 0 mean?"
On page 612, the book explains that there are three possibilities when the determinant is 0. Matter gets stuck in compartment 1, matter gets stuck in compartment 2, or the matter in the system is a constant.
How do you compute the Hessian matrix of a function $f$?
The formula for the Hessian is on page 551 and the components of the matrix are second derivatives of $f$. If we write $\mathrm{Hess}f(x,y)=\begin{bmatrix} a & b \\ c & d \end{bmatrix} $, then compute $a$ by taking two derivatives in $x$, compute $d$ by taking two derivatives in $y$, and compute $b$ and $c$ by taking one derivative in $x$ and then taking a second derivative in $y$.
"One of the exercises generated a value of 2 for every boundary. Does that mean it is a max and min, or there is none?"
You have to compare the value 2 to the value obtained at the critical point on the interior. If the value of the critical point is greater than 2, then the minimum is attained all along the boundary and the maximum was attained at the critical point. If the value of the critical point is less than 2, then the opposite. In the situation that there are multiple critical points, then you must compare all the values attained.
How do you solve for critical points involving trigonometric functions?
Trigonometric functions aren't much different than any other function. If, for example, we have $\sin(x)=0$, then we know the zeros happen whenever $x$ is a multiple of $\pi$. Similarly $\cos(x)=0$ happens for $x=\pi/2+k\pi$ for any integer $k$. While we may get infinitely many solutions, its possible these correspond to just a finite number of points.
What is the equation of the line perpendicular to the level curve of $f(x,y)=xy+3x-5y$ at $(2,3)$?
First we compute the gradient. $\nabla f(x,y)=(y+3,x-5)$. At $(2,3)$ this is $(6,-3)$. Thus, the line in this direction is $y-3=-2/5 (x-2)$. Note that you have to make the distinction between this question and the question which asks for the line tangent to the graph of the function.
*A student e-mailed me and pointed out a mistake. I accidentally plugged in 3 for x and 2 for y instead of 2 for x and 3 for y. Thanks!
What will your quiz on Thursday cover?
My quiz on Thursday will try to emulate what you're expected to know.
When do you have to normalize a vector?
When computing the directional derivative, the direction which you're concerned with must be normalized. On page 543, you'll see that $u$ must be a unit vector. Otherwise, if the question just asks "What direction does $f(x,y)$ increase most rapidly at $(x_0,y_0)$" like example 7 on page 544 then you don't have to normalize. If the question asks for the unit vector as in example 8, then you have to normalize.
*I might have made an error with this my homework solutions.
**On 20131031, I fixed the errors I detected in my solutions to 10.5. You'll find the updated file in the Dropbox "Solutions to 10.5 [Updated, for reals this time].pdf". I have left the old file for comparison purposes.
Are the maxima and minima of a function restricted to a square always at the corners?
No, not necessarily. The first practice exam I assigned gives a problem to show a max or min can happen at a point other than the corner. When finding maxima and minima inside a square, you find critical points on the interior and must determine candidates along each edge by parametrizing each edge separately.
What will be on the upcoming exam?
The professor has informed you what will be on the exam.
A student asked about #20 in 11.1.
This will be solved much like example 2 on page 596. The eigenvalues are 2 and 1. Then find corresponding eigenvectors $v_1$ and $v_2$. The solution will then be $$x(t)=c_1 e^{2t} v_1 + c_2 e^{t} v_2 $$.
How is completing the square helpful?
After completing the square, you'll have changed your function to $$f(x,y)=A(x-h)^2+B(y-k)^2 +C$$. This function will only have critical point at $(h,k)$. If $A>0$ and $B>0$ then its a minimum at the critical point, because it looks like a bowl facing upwards, possibly a squished bowl though. It'd be nice and circular if $A=B$. If $A < 0$ and $B < 0$ then the bowl points downwards. Finally if you have opposite signs for $A$ and $B$ then its a saddle point.
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| $f(x,y)=x^2+2x+y^2+2y+3$ |
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| $f(x,y)=-2x^2-4x-3y^2+2y-1$ |
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| $f(x,y)=x^2-4x-y^2+2y-1$ |
What is the purpose of a level curve?
The purpose of a level curve is to help picture the function. Thus, given a specific height $c$, we wish to determine all the points $(x,y)$ such that $f(x,y)=c$. Example: $f(x,y)=x^2+y^2$. What is the level curve through the point $(1,1,2)$? Well $f(x,y)=2$ is a circle of radius $\sqrt{2}$! We could also graph this information to illustrate our understanding. See the pictures below, which we can interpret as level curves to the graphs above. If we didn't have the ability to plot in 3D, we could still use the plots below to picture the function. (Using computers to plot in 3D, we can look at level surfaces of functions from $\mathbb{R}^3$ to $\mathbb{R}$!)
What is a level curve? What is the gradient? (This student knows how to compute these mathematically, but was curious about their interpretation)
First picture the graph of the function and then try to imagine just part of that graph which is at a certain height. That is a level curve. In real life, that's like trying to picture a mountain, and then trying to imagine only the part of the mountain which is 1,000 meters high, and then the part which is 500 meters high, etc. Continuing with our mountain analogy, the gradient is the steepest direction you'd have to travel along the mountain.
Below are level curves for the above three functions. Darker represents being less. So you can start to picture which direction you have to go for steepest increase (from dark to light).
If the restriction is on a closed and bounded set, like the boundary of a circle, then you know there must be a maximum and a minimum (and you'll get at least two critical points). If the restriction is on an unbounded set, then you'll want to check the limit at the end of the unbounded region or use other knowledge to determine whether you have a maximum or minimum. On this note, a critical point from the Lagrange multiplier method does not necessarily give a max or min.
What do you do when your function has multiple critical points?
It depends on the question. If you're only looking for absolute maximum and absolute minimum, it suffices to evaluate the function at the critical points and compare. If you're looking to determine where there are local maxima and minima, then you should evaluate the Hessian at each critical point.
I don't understand how to do #3 on the first practice exam.
For the first practice exam, the idea to #3 is that $$\lim _{h \to 0} \frac{\sin h}{h} = 1$$.
Evaluate this function at the given point: $$f(x,y)= \frac{2x}{x^2 + y^2}$$ at $(2,3)$.
You get $$\frac{4}{4+9}=\frac{4}{13}$$
How do you solve multivariable limits?
It depends. If you know the function is continuous, you can evaluate the function at the point. For example when dealing with polynomials or exponential functions. If the function is an indeterminant form like 0 over 0 (remember one variable calculus) then you have to figure out a way around it. Maybe it involves simplification. If you suspect the limit doesn't exist then you try to restrict the limit along a path and show it has different values. A quick and easy way is looking along the $y$ axis, along the $x$ axis, and along any line $y=mx$. Another possibility when trying to look at the limit of $(x,y)$ going to $(0,0)$ is to use the substitution $x=r \cos \theta$ and $y=r \cos \theta$ and look at the limit as $r$ goes to $0$. [More generally if we want $(x,y)$ going to $(a,b)$ then we would have $x=a+r\cos \theta$ and $y=a+r\sin \theta$. This method doesn't always work, but it may come in handy if you understand it.
How do you solve differential equation that begin with the second derivative of the function? Example: #24 on page 618
ReplyDeleteIn this course, you can do it the way it is shown on page 616. Following those steps we create the system:
Delete$\frac{dx}{dt}=v$
$\frac{dv}{dt}=-9x$
After doing all the work, we'll get the solution
$x(t)=\frac{12}{3}\sin(3t)=4\sin(3t)$
(Also if you can send me an e-mail, I can't identify who you are under the alias "student")
How/when do i use a Lagrange multiplier?
ReplyDeleteLagrange multiplier is useful when you want to maximize or minimize a function restricted to a constraint. This contrasts when trying to maximize or minimize a function when restricted to an open set.
DeleteExample: $f:\mathbb{R}^2 \to \mathbb{R}$. Then we could consider using Lagrange multiplier on the boundary of a circle: $x^2+y^2=9$. But we wouldn't use it for the interior of a circle $x^2+y^2 < 9$.
Remark: In the example, the dimension of the domain is 2, the dimension of the boundary of the circle is 1, and the dimension of the interior of the circle is 2.