# Tutor profile: Travis H.

## Questions

### Subject: Basic Math

Suppose we had two medium pizzas, one pepperoni and the other Hawaiian. The pepperoni pizza was divided evenly into 6 slices, while the Hawaiian pizza into 4 slices. John ate two slices of pepperoni and one slice of Hawaiian. How much pizza did John end up eating?

The first step to solving a problem is to give numerical values to certain parts of the problem. We know John ate two slices of a six-slice pizza, or $$\frac{2}{6}$$ the pizza. Similarly, we know John ate one slice of a four-slice pizza, or $$\frac{1}{4}$$ the pizza. How much pizza John ate is synonymous to asking what is the sum of $$\frac{2}{6}$$ and $$\frac{1}{4}$$. To add fractions, we need to have common denominators. Upon exploration, we find that our lowest common denominator is $$12$$. So we have $$ \begin{align*} \frac{2}{6} + \frac{1}{4} &= \left (\frac{2}{6} \cdot 1 \right ) + \left (\frac{1}{4} \cdot 1 \right )\\ &= \left (\frac{2}{6}\cdot \frac{2}{2} \right )+ \left ( \frac{1}{4} \cdot \frac{3}{3} \right )\\ &= \frac{4}{12} + \frac{3}{12}\\ &= \frac{4 + 3}{12}\\ &= \frac{7}{12} \end{align*} $$ We can safely and confidently conclude that John eating $$\frac{2}{6}$$ of one pizza and $$\frac{1}{4}$$ of another that he ate a total of $$\frac{7}{12}$$ of a pizza.

### Subject: Calculus

Change-of-variables are often used in mathematics but certainly can feel daunting and confusing. Let's run through a classic example and see how it should feel. Suppose we are asked to evaluate the integral $$\int \sqrt{3x + 1} \, dx$$. How might we use a change of variable to solve?

In this classic problem, a strong candidate for our $$u$$-substitution appears to be the $$3x + 1$$ underneath the radical. Let's see what happens when we use that for our substitution. First, we let $$u = 3x + 1$$, which leads to its derivative being $$du = 3 \, dx$$, which leads to $$\frac{1}{3} du = dx$$ Following through, our integral now is $$\begin{align*} \int \sqrt{3x + 1} \, dx &= \int \sqrt{u} \cdot \frac{1}{3} \, du \\ &= \frac{1}{3} \int u^{\frac{1}{2}} \, du \\ &= \frac{1}{3} \cdot \left (\frac{2}{3} u^\frac{3}{2} \right )\\ &= \frac{2}{9} u^{\frac{3}{2}}\\ &= \frac{2}{9} (3x +1)^{\frac{3}{2}} + C\\ \end{align*} $$ It is always worth checking your answer by taking the derivative of your just-found solution. We hope to get back to where we started! If the derivative of our solution is the original integrand, then we've done everything correctly. To begin, we have $$ \begin{align*} \frac{d}{dx} \bigg [\frac{2}{9} (3x +1)^{\frac{3}{2}} + C \bigg ] &= \frac{d}{dx} \bigg [\frac{2}{9} (3x +1)^{\frac{3}{2}} \bigg ] + \frac{d}{dx} \bigg [C \bigg ]\\ &= \frac{d}{dx} \bigg [\frac{2}{9} (3x +1)^{\frac{3}{2}} \bigg ] + 0\\ &= \frac{d}{dx} \bigg [\frac{2}{9} (3x +1)^{\frac{3}{2}} \bigg ]\\ &= \bigg [\frac{3}{2} \cdot \frac{2}{9} (3x +1)^{\frac{1}{2}} \cdot 3 \bigg ]\\ &= \bigg [\frac{18}{18} \cdot (3x +1)^{\frac{1}{2}} \bigg ]\\ & = (3x +1)^{\frac{1}{2}} \end{align*}$$ Just as we wanted!

### Subject: Algebra

Let's explore the common algebraic mistake: $$(x + y)^2 = x^2 + y^2$$. This is not true, but let's confirm why.

Let's recall our properties of exponents. While it is true that $$(xy)^2 = x^2y^2$$ via the distributive property for exponents, it does not apply when there is a sum inside the parenthetical. We must recall that something raised to a power is a shorthand for repeated multiplication. That is, $$\begin{align*} (x+y)^2 &= (x+y)\cdot(x+y)\\ &= (x\cdot x) + (x\cdot y) + (y\cdot x) + (y\cdot y) \, \, \, \text{ (FOIL) }\\ &= x^2 + 2xy + y^2 \end{align*}$$ So remember! While it may seem natural to bring those exponents down to each term, we must be delicate with our exponents and the properties that come with them.

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