# Tutor profile: Carson H.

## Questions

### Subject: Russian

Translate from English to Russian: "On Wednesdays, I go to the library for two hours, then I go to the cafe, and then to my friend Katya's house, where we watch movies and eat dinner. We have been doing this for 4 weeks, and we have watched 9 movies. But this week, Katya is going to Moscow and she will be there for ten days, so I will go to my sister's and watch a movie with her. She said to come over at 6:00 and to bring popcorn."

First I will translate the sentence and then break down the translation. "По средам, я хожу в библиотеку, потом хожу в кафе, и потом к моей подруге, Кати, где мы смотрим фильмы и обедаем. Мы так делали четыре недели, и посмотрели девять фильмов. Но в этой неделе, Катя поедет в Москву и там будет десять дней, поэтому я пойду к моей сестре и мы с ней посмотрим фильм. Она сказала, чтобы я пришла в шесть и принесла попкорн." По средам--We use this construction (по+dat) to describe a repeated action on a particular day. This is an example of syntax that does not translate directly from English, as you would never say "на средах" in Russian, despite that being the proper syntax in English. я хожу в библиотеку--the verb ходить is used because this is a repeated or habitual action, which requires imperfective aspect. We use the preposition в (as opposed to на) because it is appropriate for describing presence at or motion towards a library, and we decline the noun from библиотека to библиотеку because destinations take the accusative case in Russian. к моей подруге, Кати--We use к+dat here because that is the convention when a specific person's home is the destination. мы смотрим фильмы и обедаем--Again, imperfective aspect is used because this is a habitual action (it happens every Wednesday) and because the actions of eating and watching are concurrent. четыре недели...девять фильмов--The words неделя and фильм must be put into genitive singular and genitive plural, respectively, because in Russian, objects with quantities of 2-4 take genitive singular and quantities 5-20 take genitive plural. (Don't ask me why; I didn't make the rules, :( lol) Катя поедет в Москву--We use the verb поехать (as opposed to пойти) because it is a journey out of town, certainly taking place by vehicle, and we use perfective aspect because it is a one-time occurrence in the future. мы с ней посмотрим фильм.--"Мы с+instr" is a common idiomatic way of expressing mutual activity. The perfective aspect is used here because it is a one-time specific action in the future, and follows the action of going to the sister's house. Она сказала, чтобы--When someone else's instructions are relayed in this way (he/she/they told me to, asked me to, said to, etc), it is common to use the particle чтобы and then put the requested actions in past tense. я пришла в шесть и принесла попкорн.--We use the perfective verbs here because they refer to one-time future actions, and use the prefix при- because it indicates arrival.

### Subject: Pre-Calculus

Apply the following transformations to f(x)=|x|: shift two units to the right; shift one unit down; horizontally compress by a factor of 1/2.

In order to apply these transformations, we need to have an understanding for how different transformations on a graph--things like a shift up or down, or a vertical stretch--affect an equation. To do that, we can look at what I like to call the ABC's of transformations. For any function, we can have four different types of transformations: Vertical stretch/compression, horizontal stretch/compression, vertical shift, and horizontal shift. We can also reflect the function over the x- or y-axis. Those transformations are represented by certain modifications to the way that the function is written. So for some function f(x), we can transform it according to this model: A \cdot f[B(x-C)]+D, where A, B, C, & D all represent one of the transformations I mentioned above. (For our specific function, f(x)=|x|, it looks like A|B(x-C)|+D.) Let's break them down! A--Vertical stretch or compression. This modification stretches the function up or squishes it down, like pulling on or squishing a piece of silly putty. If A is negative (A<0), then our function is reflected over the x-axis. B--Horizontal stretch or compression. This modification stretches the function out or squishes it together, like like stretching and compressing a slinky. If B is negative (B<0), then our function is reflected over the y-axis. (Note: Horizontal stretches and compressions are tricky, because the actual factor of stretch or compression is equal to the reciprocal of B! For example, if B=3, then our function is compressed by a factor of 1/3. On the other hand, if B=1/3, it is stretched by a factor of 3.) C--Horizontal shift. This modification shifts the function left or right, moving it back and forth on the x-axis. If C is positive (C>0) we shift to the right; if C is negative (C<0), we shift to the left. (Note: We have to be careful with C because it is preceded by a negative sign in our model! This means that when we shift to the left and C is negative, it will have a plus sign in front of it in the function. For example, if C=-1, our model would be f(x-(-1)), and -(-1)=+1! So remember that C's sign in the function will be the opposite of what it ACTUALLY is. C is tricky like that...) D--Vertical shift. This modification shifts the function up or down, moving it back and forth on the y-axis. If D is positive (D>0) we shift up; if D is negative (D<0), we shift down. Great! Now that we've defined all of our different transformations and we know how they affect the function, let's apply that information to our problem. :) We're supposed to shift two units right, one unit down, and horizontally compress by a factor of 2. So a shift to the right means that we're shifting horizontally, which means we're using C. And if we're moving two units, that means C=2. Next, a shift down means vertical shift, so we're using D! Since we're moving down, D is negative, and since the shift is one unit, that means D=-1, specifically. Now for our horizontal compression. If our compression factor is 1/2, that means B must be the reciprocal of that--B=2. So last but not least, we plug everything into our function. We know our model is f(x)=A|B(x-C)|+D, so all we have to do is replace the letters with our numbers, and viola! So doing this, we get f(x)=|2(x-2)|-1 And that's it; that's the show. :)

### Subject: Calculus

For the function g(x)=\ln ((x^2-5)^5), find the equation of the line tangent to the curve at the point (x, \ln1024)

The slope of a line tangent to a curve at a point is found by taking the derivative of the function (the curve) and then plugging in the x-value of the point in question. That being said, our first task in this problem will be to find the derivative of g(x). To begin, we must apply the chain rule for natural logarithms, which states that \frac{d}{dx}[\ln(f(x))]=\frac{1}{f(x)}\cdot f'(x) (Or, as my calculus teacher used to say, "The derivative of the natural log of stuff is one over the stuff times the derivative of the stuff," where f(x) is the "stuff".) So we have to identify our "stuff". Well, it's whatever is inside the natural logarithm, so it must be (x^2-5)^5. Awesome! So we take one over that, and we get \frac{1}{(x^2-5)^5} Now, we have to multiply that times the derivative of the "stuff", so the derivative of (x^2-5)^5. For that, we need yet another chain rule. (Yikes!) \frac{d}{dx}[(f(x))^n]=n\cdot (f(x))^{n-1}\cdot f'(x) So now, we're looking ONLY at the (x^2-5)^5 part of our function, and we're trying to take its derivative using the rule on the previous line. So, we apply the power rule to the outside layer of the function, the part that is to the fifth power, and we get 5\cdot (x^2-5)^4 Now, we take that and multiply it times the derivative of what's inside the parentheses. Again, we use the power rule to take that derivative. So we have 5\cdot (x^2-5)^4 \cdot 2x Whew! So all of that was our derivative of the "stuff" from our original expression. So let's get back to that, shall we? :) We had \frac{1}{(x^2-5)^5} as our one over the "stuff", so now we simply multiply by the derivative of the "stuff" to get \frac{1}{(x^2-5)^5} \cdot 5\cdot (x^2-5)^4 \cdot 2x We can simplify this some by dividing out four of our factors of x^2-5 and then multiplying 5\cdot 2x to get our final answer for the derivative, g'(x)= \frac{10x}{x^2-5} Swell! So now that we have the derivative we need to find the slope of our tangent line. Remember, the slope of a line tangent to a curve at a point is equal to the derivative at that point. So, to find our slope, we simply plug our x-coordinate (x=1) into the derivative! So our slope, m, is m= \frac{10 \cdot 1}{1^2-5}= \frac{10}{-4} = -\frac{5}{2} Awesome!! Armed with this information, we can now use the point-slope form of a line to find our final equation. (Recall that the point-slope formula is y-y_1=m(x-x_1), where (x_1,y_1) is a point on the line.) We know our slope, m=-\frac{5}{2}, and our point, (1,\ln1024), so we can plug them into our point-slope form and solve. y-\ln1024=-\frac{5}{2}(x-1) \rightarrow y-\ln1024=-\frac{5}{2}x+\frac{5}{2} \rightarrow y=-\frac{5}{2}x+\frac{5}{2}+\ln1024 And that's it! Our final answer is y=-\frac{5}{2}x+\frac{5}{2}+\ln1024

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