# Tutor profile: John J.

## Questions

### Subject: Applied Mathematics

How would I solve $$\dfrac{dy}{dx}=\dfrac{1}{y-yx^2}$$?

This is a first-order differential equation. But it's even nicer than that - it's of a particular form that's called "separable". That just means we can write the right-hand side of the equation in the form $$\dfrac{f(x)}{g(y)}$$, where we decide to divide by the function $$g(y)$$ instead of multiply because it makes the resulting formula a little nicer - namely, we get $$g(y)dy=f(x)dx$$ and we can integrate both sides of the equation to get our solution. Notice that each term in the denominator of the right-hand side above has a single factor of $$y$$ - that means we can factor it out to obtain $(\dfrac{dy}{dx}=\dfrac{1}{y(1-x^2)}$) If we let $$f(x)=\dfrac{1}{1-x^2}$$ and $$g(y)=y$$ we see this equation is indeed separable - note the weirdness with the dividing by $$g(y)$$, which is why $$g(y)\neq \dfrac{1}{y}$$. Then we want to integrate both sides of $(y\ dy=\dfrac{1}{1-x^2}dx$) The left side of this equation is simple - we get $$\dfrac{1}{2}y^2+C_1$$. The right-hand side is a bit more complex, but you just have to use partial fraction decomposition - everyone's favorite - to see that $$\dfrac{1}{1-x^2}=\dfrac{1}{2(1-x)}-\frac{1}{2(1+x)}$$. This integrates to $(\frac{1}{2}(\ln(1-x)-\ln(1+x)+C_2=\ln(\dfrac{1-x}{1+x})+C_2$) where that last simplification comes as a consequence of the properties of logarithms. Getting everything together, we can combine those two arbitrary constants $$C_1$$ and $$C_2$$ into one constat $$C$$ (the sum or difference of two arbitrary constants is still an arbitrary constant!) on the right-hand side to get $(y^2=\ln(\dfrac{1-x}{1+x})+C$) or $(y=\pm \sqrt{\ln(\dfrac{1-x}{1+x})+C}$)

### Subject: Calculus

Why are integrals sometimes negative? If they are areas, shouldn't they always be positive?

The answer rests in the definition of the integral as a limit of Riemann sums, or approximations. The first lecture on integrals typically goes through an area computation by approximating the area under a curve with a bunch of rectangles, and then adding those areas together. In the end, you let the number of rectangles go to infinity while each individual rectangle's area goes to 0, and you arrive at the actual area under a curve. But how were those approximating rectangles constructed? You partitioned the domain of the function into several sub-intervals of a known width (typically denoted $$\Delta x$$), and the height of any given rectangle was defined to be the value of the function at some x-value its corresponding sub-interval. Well, that definition is exactly what is used for the overall definition of the definite integral, except nowhere in the definition do we require the function always have positive values. By allowing the function to take negative values, the "heights" of some of these rectangles are now negative so there are regions where a definite integral will be negative - in this case, it's typically referred to as "Signed area" instead of just area, to remind you that it's a more abstract notion than a simple rectangular area. I typically remember this, and work with integrals theoretically, by thinking of them more as displacements than areas. The same definition, when applied to a velocity function, will give you the area under the graph of the velocity function, but this can be interpreted as the displacement the particle or other entity underwent in a given timeframe. Since a negative velocity means you are travelling backwards, a negative integral just means you travelled a certain distance backwards! This has always made it easier for me to wrap my head around some of the abstract properties of the definite integral.

### Subject: Algebra

How do I solve $$3x^2+2x-1=5$$ using the quadratic formula?

First of all, the quadratic formula requires that the equation be of the form (...)=0, not =5, so the first step would be to move the 5 over to the left-hand side by subtraction to get $$3x^2+2x-6=0$$. Then you have to identify the required quantities for the quadratic formula - $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$. By convention, we assume that a quadratic equation is written in the form $$ax^2+bx+c=0$$, so the $$a,b,c$$ in the above formula reference this standard form. That is, $$a$$ is the leading coefficient, $$b$$ the middle coefficient, and $$c$$ is the constant term. So, by looking at our equation, we see $$a=3, b=2, c=-6$$. Thus, by substituting into our formula, $$x=\dfrac{-2\pm\sqrt{2^2-4(3)(-6)}}{2(3)}=\dfrac{-2\pm\sqrt{4+72}}{6}=\dfrac{-2\pm\sqrt{76}}{6}$$