10.1 Setting up a differential equation & 10.2 Solutions of differential equations

Main Points:
Differential equations can be used to answer questions or find new information about a function when only bits and pieces of the vital information is given about the function. The solution for a differential equation is an equation that satisfies the differential equation which expresses the slope. To solve numerically a differential equation, one must be given a value for a variable in the function. Once you have this you can compute various data points and form a function that reflects that data. Finding a function for the data isn't always possible, but to check if your equation works, substitute it into both sides of the differential equation and see if they equal each other. To find the arbitrary constant of the equation you derived simply substitute in a value given to you and solve for the constant.

Challenges:
I still don't know how to find an equation that satisfies a differential equation. From what I read, it seems that we need to memorize function forms for different differential equation forms. That doesn't seem very mathematical to me. Can't we solve from scratch for the function. How did they come up with those solutions or forms for functions in the first place?

Reflections:
It seems like a great way to solve a problem when only part of the info is given, but I don't know how this can be useful when there isn't a set way to solve for an equation that works. However, I do appreciate the versatility of this method and how it can be used with different info given and in different situations.

9.6 Constrained Optimization

Main Points:
Optimization problem solving is often constrained by certain external things, this is why we need constrained optimization. If the function can be maximized or minimized with the function of a constraint, then that max or min occurs at the point where the constraint function is tangent to the contour of the function being maximized, or at an endpoint of the constraint function. You can also the method of Lagrange Multipliers. The lagrange multiplier represents the change in the optimum value of the function in question when the constraint is increased by one. You can use this fact to construct other equations that relate this fact in relation to each variable (increasing just x or just y) and solve for all three variables with the three equations. The Lagrangian function can also be used, but unfortunately I have no idea how. The best way for me to describe this process is basically finding where the partial derivatives of x, y and Lambda equal 0 (when the function can't increase anymore) . New symbols frighten me.

Challenges:
I don't fully understand the process of solving with Lagrange multipliers. The part I'm having trouble with is how they derive the equations for the Lagrange, and how that relates to x and y of both the original function and the constraint function.

Reflections:
These methods will be very usefull in real world applications, I'm sure. What I need is a person talking through these explanations rather than a book. When in this world of our's are there no constraints?

4.3 Global Maxima and Minima

Main Points:
Global Maxima occure at the greatest value for a function not constrained by a certain interval (except by the endpoints of the function if it has endpoints). Global Minima occure at the lowest value for a function not constrained by a certain interval. To find the global maxima or minima, compare all critical points of the function as well as the endpoints. If the function continues to infinite or -infinite, then there is no global maxima or minima respectively.

Challenges:
When I looked at the gas consumption example in the book, and I saw the graph, my first reaction was, "hey 20 mph must be most efficient since they're talking about global maxima and minima, but I quickley realised the problem is slightly more complicated. They were looking for most efficient mpg's not gallons per hour, the book made a good point that maximizing or minimizing a function definately goes beyond finding the highest point on a graph.

Reflections:
I think this will help more so than just a local max or min in solving real world problems that require maximizing or minimizing something. I could be wrong, but this concept seems to be more applicable in more realistic problem solving.

1.3 Rates of Change 2.4 The Second Derivative 4.1 Local Maxima and Minima & 4.2 Inflection Points

Main Points:
The average rate of change is defined as the change in y divided by the change in x. The average rate of change for a linear function is its slope. A function is linear if the rate of change is constant on all intervals. A function is increasing if the values of y increases with x, the opposite is true if y decreases as x increases. A function is concave up if it bends upward from left to right, the opposite is true if the function bends downward from left to right. If the second derivative of a function is greater than 0 on an interval, then f ' is increasing, so the graph is concave up. If the second derivative is less than 0, then the opposite is true. The second derivative can be thought as the rate of change of the slope. A critical point is one where f '(x) = 0. This means that the tangent line has a slope of 0 at this point (indicating a possible minimum or maximum). You can test for a local maxima or minima by using the first derivative test. If f changes from increasing to decreasing, then the critical point is a maxima, if the function changes from decreasing to increasing, then the critical point is a minima. If the second derivative is positive at the critical point, then it is a minima. If the second derivative is negative, then the critical point is a maxima. The point at which a function changes concavity is called an inflection point. A point of inflection is where the second derivative equals zero. The sign of the second derivative changes from one side of an inflection point to the other.

Challenges:
The only thing that was challenging was trying to wrap my head around the idea of a rate of change for a rate of change. I understand what it means, but it's a little hard to visiualize at first. One can think of an exponential function when trying to understand this.

Reflections:
This material seems like it could have many practical applications in the real world, especially finding the maximum or minimum of a function. I'm wondering how finding an inflection point of a function could be usefull. Second derivatives could be usefull for finding information about exponential functions that represent real world problems.

Supplementary notes

Main Points:
The gradient of a function is a vector that consists of the partial derivative of the function in respect to x and the partial derivative in respect to y. The partial derivative in the x direction makes up the x component of the vector and the partial derivative in the y direction makes up the y component of the vector. Directional derivatives are a representation of the change in x and y in any direction given by a unit vector. To find the directional derivative one must find the gradient of the function and then multiply the x and y components of the gradient at a given point by the x and y components of the unit vector being used. If you look at a graph of gradient vectors of a function you'll notice that they are perpendicular to the function and they point in the direction of greatest increase and the direction opposite would represent the greatest decrease. As the slope of a function increases, so does the gradient vector's magnitude.

Challenges:
The statements made about the graphical representation of gradient vectors were a little confusing. The direction that they point is stated as the direction of greatest increase for the function, but when I tried to visualize this rule with the given graph, I couldn't figure out how this was true.

Reflections:
It seems to me that there are real world applications for math that can solve for the rate of change of a function in any direction. I hope we can see some of the applications of this math, so it will be more than just a bunch of numbers and coordinates.

Linear Algebra for math 135 - Scalar Multiplication, Vector Addition, and The Dot Product

Main Points:
Vectors are comprised of a point with a magnitude from the origin to that point. The notation can be simply written as the point would such as (2,3). Vectors can be multiplied too. If multiplied by a scalar, you simply multiply the x and y by the number and the new coordinates represent the new vector. If you add two or more vectors together you simply add all the x values together and all the y values together to get the resultant vector. To find a dot product, you multiply the x values together and add the product of the y, z etc. values (x*x + y*y + z*z etc.). The result is a scalar. If two lines are perpindicular, the dot product will be 0. The length of a vector (a) is equal to the square root of (a^2). To find the angle between two vectors (u and v), use the equation u*v=(length of u)*(length of v)*cos(the angle between the two vectors).

Challenges:
I think it may be a typo that is screwing me up right now. The reading says that to find the angle between u and u you use the equation u * v = (length of u)*(length of u)*cos(angle). Where did that v come in? Is it a typo?, because I can't make any sense of it.

Reflections:
I thought the dot product section of the reading was the most interesting. I thought it was interesting how we could calculate the angle between two vectors with this simple equation they gave involving simply the length and magnitude of the vectors. I just thought that was neat.

3.5 Derivatives of Periodic Functions, 9.3 Partial Derivatives & 9.4 Computing Partial Derivatives Algebraically

Main Points:
The derivative of the sin function is the cos function. This is true because the slopes of these functions are periodic. The partial derivative is the slope of a function dependent on one variable (x or y). The partial derivative can be written as (f (a+h, b) - f (a, b))/h this can be used to find the partial derivative from a data table. This method can also be used on a contour diagram. All one must do is find the interval between contours and the difference between them in the x or y direction. To find the partial derivative of a function algebraically, one must plug in the value for x or y that is static, and then find the derivative of the function after that value has been substituted in. Since the result of a partial derivative is a function itself, you can find the 2nd order partial derivative of that function too. Also, f ' (x,y) (a,b) = f ' (y,x) (a,b).

Challenges:
The second order partial derivative is a little bit confusing. I understand the concept of a derivative of a derivative or how fast the slope is increasing. It's kind of like an acceleration. The part that I was having slight troubles with was just how that concept applies to a second derivative. I just don't understand if this is saying how fast the x increases in terms of y or what. I just need some clarifying on what it basically means.

Reflections:
When thinking about 2nd derivatives while I was writing the Challenges section, I thought that a really good way to think of a 2nd derivative is that it's like an acceleration (yes I'm taking physics). Acceleration is exactly what a 2nd derivative is because it's how fast your position changing is changing.