# Tutor profile: Caitlin D.

## Questions

### Subject: Calculus

How are differentials and integrals related to each other and what is so important about them?

Calculus is a major foundation in mathematics, taking up at least 3 full classes before one may be considered a master at it. Calculus I deals mostly with derivatives, Calculus II deals with integrals, and Calculus III combines these in such a way that we may apply our knowledge of them in 2 dimensions to 3 dimensions. This sounds extremely complicated if one has never heard of these or doesn't understand how they are related, but like anything else, breaking down each subject within each course into its basic components will lead to understanding. Derivatives and integrals can be best compared to one another by analyzing graphs. A graph of a basic exponential curve looks like a line that starts at x=0, is mostly flat (horizontal), and then increases rapidly around a curve to be nearly vertical as x approaches a certain value. Choosing to look only at positive values of x and y (without the z-axis as a factor), one can analyze exactly how rapidly the graph grows (rate of change) at each tiny, infinite point of the line. The derivatives all combined together from the various points may represent a new graph, the derivative of the original function. Each function also has an algebraic way to solve for its derivative of the original equation. Integrals analyze the exact area under a curve on a graph, for instance, the area under the exponential from x=0 to x's final value (asymptote where the graph is near vertical). This area can be calculated by looking at the graphs and following a pattern of drawing tiny squares of definite areas under the curve and then adding all of the areas together. This can also be done algebraically, similar to derivatives. Derivatives and integrals are related algebraically. The derivative of a function is a new function, and the integral of that new function is the original. Ex) 1. The original function: $$f(x)=e^{2x}$$ Its derivative: $$f'(x)=2e^{2x}$$ 2. The original function (note that this was chosen to be the derivative from problem 1): $$f(x)=2e^{2x}$$ Its integral: $$\int{f(x)}=e^{2x}$$ (note that this is the same as the original function of problem 1) Derivatives are important as they lead to topics such as differential equations, in which multiple derivatives are considered within one equation that needs a solution. Integrals are important as they lead to computing areas not only under curves in 2-D, but also of volumes of objects in 3-D. Combinations of integrals lead to these solutions which are very complicated but extremely useful to applied mathematics, as in physics and engineering.

### Subject: Physics

How should one begin to solve physics problems including those that deal with kinematics, projectile motion, energy conservation, momentum, and forces (Physics I concepts)?

Physics seeks to answer questions about the motions and nature of the universe. These problems which are given in classrooms are based on real-life situations and understanding how to do the most basic ones allows you to improve your skills to then tackle more difficult ones. Most problems seen in Physics I involve Newton's Laws of Motion which include concepts about the tendencies of objects to move or remain still, to exert a push or pull on objects around them, and the general equation which explains what exactly a force is, $$F=ma$$. To approach general physics problems, it is important to move through the question being asked and underline key information about the situation. This includes values with units, any given equations, and what exactly the question is asking. These values are what allow you to compute an answer (often times a new value with a unit). The next step is to create a drawing of the situation, whether it be of a ball rolling down a hill or a lamp hanging from the ceiling. The underlined values should be given in the picture as well, so that you can quickly visualize exactly what the situation looks like and relatively what the answer should come out to be. For example, when examining forces, you may be asked to analyze the normal force of a ball which sits still on a table. After underlining and drawing a picture which includes the direction of forces on the ball, identify your knowledge of the subject. However much the table is pushing up onto the ball, the ball must be pushing the same amount back on the table (Newton's 3rd Law), and if the ball is not moving, the normal force of the table onto the ball must be equal to the force of gravity on the ball (they cancel out). This is what keeps the ball from dropping straight through the table. Combining your understanding of Newton's Laws with basic math, you can conclude that the normal force of the ball onto the table must be equal to the force of gravity on the ball, so long as those are the only two forces acting on it in the vertical direction. Similar methods may be applied to other situations, such as analyzing a football being thrown across a field (where will it land exactly and when?) and figuring out how much kinetic energy it will have if it collides with another football in the air. Each physics problem is different, but solving them is not unique. Breaking the problem down, figuring out what it is you're looking for (as seen through drawings), and using a specific concept or question to get there is the key to solving these.

### Subject: Differential Equations

What is a "second-order, linear, homogeneous differential equation" and how can I solve it? Ex) $$2y''+3y'+y=0$$

Break down each part of the title of that type of equation. A differential equation refers to an equation that involves both regular variables such as y, and derivatives of that variable such as y' or y''. Linear refers to the fact that there is no multiplicity of the variable y in the equation other than 1 (in other words- y is not squared or cubed, etc.). Second-order refers to the fact that the highest amount of differentials of the variable y used is 2. This is indicated by the notation '' used with y in the first term. The term homogeneous refers to the fact that the equation is set equal to 0 rather than anything else. To solve these kinds of equations, we can first notice that each coefficient of the equation is constant (there is no x in front of any term with a y in it). This means that we may use the method of constant coefficients to solve for the equation that represents the answer. The equation $$y=e^{mx}$$ is always a solution to these types of problems and all we need to do now is figure out what the m represents. To start, we consider y equal to $$e^{mx}$$. Therefore, $$y'=me^{mx}$$ and $$y''=m^2e^{mx}$$. We then algebraically plug these into the original equation to yield $$2m^2e^{mx}+3me^{mx}+e^{mx}=0$$. We can next factor out the term $$e^{mx}$$ from the left hand side to result in $$e^{mx}(2m^2+3m+1)=0$$. Divide both sides by $$e^{mx}$$ and we are left with $$2m^2+3m+1=0$$. To solve for m, we use methods of factoring if possible, and quadratic equation when not, to find two solutions to the equation. In this case, $$m=-1, -\frac{1}{2}$$. Put this all together and the final solution will be a combination of each term $$e^{mx}$$ with its own m. There also will be unknown constants in front of each term, as these may be filled in when the problem will provide initial conditions for the solution (such as $$y(0)=0$$ or $$y'(0)=0$$). The general solution for the differential equation $$2y''+3y'+y=0$$ is: $$y=Ae^{-x}+Be^{-\frac{1}{2}x}$$ where A and B are unknown constants.

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