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Tutor profile: Fares E.

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Fares E.
Comp. Engineering Junior at NC State University with Extensive Calculus and Mechanics Background
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Questions

Subject: Pre-Calculus

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Question:

Determine the equation of the vertical asymptote of the following rational function: y = (x-3)(x+2)/(x+3)(x-3). If there is a hole, identify it.

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Fares E.
Answer:

Asymptotes can be very challenging for new students to understand and conceptualize. We learn early on in mathematics that you cannot divide anything by 0, because all that does is yield infinity, and infinity is not a number on any number-line. By definition, a rational function always has an equation with an independent variable (x) in the denominator, meaning there is a value we can plug into x that produces a denominator equal to zero, which produces an output of the function (y) equal to infinity. Since there is no way for us to reach that high (numbers are unending and infinity doesn't exist anywhere on the y-axis), all we are left with are all the x-values that come right before producing a zero-denominator and right after. The closer we come from both sides (negative + positive) to an x-value that produces a zero-denominator, the closer y gets to negative and positive infinity on either side, but since x never actually reaches that x-value, we say that it just comes infinitely close. Thus, by definition, it produces a vertical asymptote: vertical because it is y that is approaching infinity in the upward and downward directions. Now, onto solving the question. You may notice that (x-3) is in both the numerator and denominator, so they can cancel out, and we are now left with y = (x+2)/(x+3). We see that the denominator holds (x+3), so what value can we assign to x that would make that equal to zero? Well, x+3=0 makes x=-3, and that is your answer! The closer x gets to -3 from the left side (e.g. x=-3.000...1), the closer y gets to positive infinity (upward). Why positive? Because both the numerator and denominator are negative at this x-value yielding a positive result. Likewise, the closer x gets to -3 from the right side (e.g. x=-2.999...9), the closer y gets to negative infinity (downward). Why negative? Because now the denominator is positive (-2.999...9 + 3 is tiny, but positive), while the numerator remains negative, yielding a negative result. Go ahead and graph the function on Desmos and see for yourself! Now for the "hole" part. Well, technically, there are no actual holes in graphs; it more or less has to do specifically with how a function is expressed in a question prior to putting a rational function into its lowest terms (cancelling like-terms out), as we did earlier. We see from the original question that (x-3) exists both in the numerator and denominator, meaning if we were to plug in [x-3=0 -> x=3] into the function, it would produce y = 0/0. So, what are we to do with this? A zero in any numerator yields a zero output, and a zero in any denominator yields an infinitely-large output; it makes no mathematical sense. So, we say that this is a "hole" in the graph. It is a point that simply doesn't exist, and it isn't an asymptote because y doesn't approach infinity as x approaches that value (both the numerator and denominator approach zero so the "effect" is cancelled out and things look normal). So there is your answer: x = 3.

Subject: Discrete Math

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Question:

An irrational number is one with an infinite number of decimal places that do not repeat. Prove that the square root of 2 is irrational.

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Fares E.
Answer:

One of my favorite topics in Discrete Math is Proofs, because it allows us to understand how impossible-to-grasp mathematical concepts are proven with certainty. Take this question for example - it would seem like the only possible way to prove that the square root of 2 is irrational would be to write out an infinite sequence of decimal places; how else would you prove that its unending?! Well, this is where Proof by Contradiction can help out. This is a very useful method in Discrete Math that first assumes the opposite of the proposed hypothesis to be true - in this case that the square root of 2 is rational. Then, the method suggests that we should progress by listing out the exact conclusions we would be making based on the mathematical definition of what we are assuming - in this case, that the square root of two can be expressed as a/b, where both a and b are integers with no common factors (simplest form of a fraction). "a" cannot be zero since the square root of any positive value cannot be 0, and b must also be non-zero, since we cannot divide by 0. The final stage of this method is to find some kind of conclusion that is contradictory to our previous assumptions, proving that what we assumed was false, and thus that the opposite - what the question was asking for - is true. If the square root of 2 can be expressed as a/b, then squaring both sides says that 2 can be expressed as (a^2)/(b^2). If this is the case, that means that we can say a^2 = 2(b^2), and we know that anything multiplied by 2 gives an even number. So, regardless of what integer b^2 is, a^2 must be even, and if a^2 is even, then a must be even because the square root of any even number (that we know is an integer) must also be even [conclusion 1: a is even]. If a is even, we can express it as a = 2c, where c is some arbitrary integer (since anything multiplied by 2 is even). Using our previous equation, it means 2(b^2) = (2c)^2 => b^2 = 2c^2. Since c^2 in this case is multiplied by 2, it must give us an even number, meaning b^2 is even; and if b^2 is even, then b must be even for the same reason given above [conclusion 2: b is also even]. Although all may seem good and consistent, these two conclusions coexisting gives rise to a major contradiction. As mentioned earlier in our assumption, a and b have to be non-zero integers with no common factors, but if they are both even, then they should logically be able to be divided by 2 each, thus sharing a common factor. For the square root of 2 to truly be rational, it should be able to be represented by a/b where at least one of them is odd. Thus, by contradiction, we have proven that the square root of 2 cannot be a rational number.

Subject: Physics

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Question:

If the Moon is pulled-on by Earth's gravity, why doesn't it fall?

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Fares E.
Answer:

This is because the Moon is in orbit, but what does that really mean? Being in orbit implies that an object is in circular motion, but we know in classical physics that objects cannot experience circular motion without an external centripetal force, a force that pulls an object toward the middle of a circular path. An object also cannot experience circular motion without having a constant "straight-path" velocity. So, with those two concepts in mind, imagine the Earth as a point in the middle of a large circle: the Moon's circular path. What keeps the Moon from falling directly through the radius? Well, the Moon has a velocity that is always in tangent with this circle at any given moment, and gravity in this case is always perpendicular to that velocity that always causes the Moon to accelerate toward the center. Thus, the net motion of the Moon in that moment is somewhat diagonal in-between those two vectors. If we could split this never-ending event into "time-frames" and take a look at the following frame, we would see that the Moon along with it's straight-path velocity moved and tilted diagonally, still experiencing a centripetal force perpendicular to that. The cycle continues forever, and if you could split these events into infinitely-short time-frames, you end up with a perfect circle with no edges, as opposed to a polygon. By the way, the same concept goes for any spinning object, even if the centripetal force is not gravitational. So why doesn't the Moon "lose energy" where it's velocity becomes so small that it is overpowered by force of Earth's gravity, causing it to spiral rapidly toward the Earth? Well, according to Newton's first law of motion, an object in motion stays in motion unless acted upon by an external force. Here on Earth, we are familiar with things slowing down due to friction or air-resistance, but in the case of the Moon, it exists well beyond Earth's atmosphere and experiences no resistance that counteracts it's velocity (a force in the opposite direction). Earth's gravity is always perpendicular to its velocity and doesn't counteract it. So, in theory, it can go on forever.

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