Math 110.107, Calculus II (Biological and Social Sciences)

Fall 2010 Course Lecture Synopses

Week 10:  November 1 through November 5

http://www.mathematics.jhu.edu/brown/courses/f10/107.htm

• Monday, November 1:  First, I reiterated the Second Derivative Test for functions of two variables via the use of the Hessian of a function, which I defined as in the book.  This was to facilitate the homework assignment for this week as well as offer a cleaner way to state the theorem.  Then, continuing in Section 10.6, I talked about the vector calculus version of the Extreme Value Theorem (IVT).  Using the Calculus I version, I talked a bit about how and why it works, why one needs a closed interval and a continuous function, and the process one employs to locate the extreme values of a continuous function on a closed interval.  I then discussed the two dimensional notion of a closed domain as an open domain (where each point has some small disk centered at it that is completely inside the domain, together with its boundary (points that are considered in the domain, but do not have any small open disk centered at the point that does not contain both points inside the domain and outside it.)  The EVT is the same for vector calculus also:  The a continuous function defined on a closed domain has both a maximum and a minimum value.  Locating these extrema uses the same process also:  First calculate all the places where the derivative is zero (although it is more common to calculate all of the places where the gradient vector is the 0-vector).  This means calculating all the places of $f(x,y)$.  Throw in all of the places where $\nabla f(x,y)=\left[\begin{array}{c}0\\ 0\end{array}\right]$ is not defined, as well as all of the places where the function restricted to the boundary is also extreme, and pick the greatest and least values.  The caveat in this process is how to evaluate the extrema on the boundary.  In practice, this may not be often easy. But for boundaries of domains that are rectangles, or circles, there are ways.  The trick is to write the function restricted to either a piece of the boundary or the entire boundary as a function of one variable.  Then you can use calculus I techniques to find the extrema there.  I used Example 7 of the book to illustrate.  Next class, I will also go over the circular boundary in Example 8 also, and use Mathematica to show what the graphs of these functions look like.
• Wednesday, November 3: Today, I started with a rehash of example 7 from the text, and showed the graph of the function in Mathematica.  I then set up and did the same for Example 8 in the text, showing the graph and explaining how in this case, finding critical points on the boundary involved a conversion of the function on the boundary to a function of one angular coordinate (via the polar coordinate transformation).  I then spent time on the following problem:  Find three non-negative numbers whose sum is 90 and whose product is as large as possible.  I didn’t solve this problem.  Instead, I set it up as a exercise in restating the problem in a way that allows for calculation.  At first, you will have to maximize a function $P(x,y,z)$ on the domain given by $x\ge 0$, $y\ge 0$, and $z\ge 0$, and subject to the condition that $x+y+z=90$.  Here, the domain seems to be infinite.  But the constraint is a tell.  Remove one of the variables by $z=90-(x+y)$, and $P\left( x,y,90-x-y\right)=xy(90-x-y)$ is now a function of only two variables.  Plus, the domain is now a triangular region in the first quadrant of the $xy$-plane bounded by $x\ge 0$, $y\ge 0$ and $x+y\le 90$. To see this, remember that $z\ge 0$, so that $z=90-(x+y)\ge 0$ means that $x+y\le 90$. This is now a closed domain, and we are back to situation like the other examples.  In this case, it is even easier, since on each of the three edges of this triangle, one of the coordinates is 0.  Hence the product $P(x,y)=0$ there.  But for anything inside the triangle, $P(x,y)>0$.  Hence there will be a maximum and it will occur inside the triangle.  Simply solve the equation $\nabla P(x,y)=\left[\begin{array}{c}0\\ 0\end{array}\right]$. At this point (there will be only one here), the product will be maximized.  I again used Mathematica to show the graph of $P(x,y)$ on the triangle.  I finished with an introduction to Section 11.1, setting up the idea of a system of linear differential equations, in general, both as a system and as a single matrix equation.  I then simplified the system to a homogeneous one (no extra functions of the independent variable $t$), and then to an autonomous system (all of the coefficient functions are constants).  I then further simplified the system so that the coefficient matrix is square, and then further limited the discussion to $2\times 2$-systems (two equations in two unknowns). I talked about a good place to visualize solutions;  for $\dot{\vec{x}}(t)=A\vec{x}(t)$, where $\vec{x}(t)=\left[\begin{array}{c} x_1(t)\\ x_2(t)\end{array}\right]$, a solution curve will look like a parameterized curve in the $x_1x_2$-plane (this is called the phase space of the system).  I talked about whether equilibrium solutions exist and how the properties of the matrix can help to find them, and I finished with some basic notions of how a direction field can help to understand the nature of the solutions to the system.  Next time, we will construct direction fields and start the process of solving $2\times 2$ systems of linear, homogeneous, autonomous, differential equations.