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Jessica B.

Recent College Graduate and Tutor for 4 years

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Calculus

TutorMe

Question:

If $$\lim_{h\to0}\frac{arctan(a+h)-arctan(a)}{h}=\frac{1}{17}$$, what is the value of $$a$$ ?

Jessica B.

Answer:

Don't hurt yourself. This problem is easier than it looks. You just have to recall a few formulas. First off, if you've gone through a bit of calculus, this limit may look familiar to you; in fact, it is. It's the formula for the definition of derivative! $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=f'(x)$$ So looking at this you can determine that the formula in the question is also known as the derivative of $$arctan(a)$ $. Now that you've gotten here, you might be thinking "oh, the derivative is here, so let's just take the integral and use the chain rule and--" Nope, don't do that. Just recall the derivative rules for the "arc" functions. You might want to write them all down from "sin" to "cot" but for now, we'll only need the one for arctan: $$\frac{d}{dx}arctan(x)=\frac{1}{1+x^{2}}$$ So now, applying this to what we've already reasoned: $$\frac{d}{dx}arctan(a)=\frac{1}{1+a^{2}}$$=$$\frac{1}{17}$$ So if $$\frac{1}{1+a^{2}}$$=$$\frac{1}{17}$$ , it also stands to reason that $$1+a^{2}=17$$ (this can be proven by multiplying both sides of the equation by $$17$$ and $$a^{2}$$) So now you just solve for x by subtracting 1 from both sides and taking the square root! $$a^{2}=16$$ $$a=4, -4$$ :)

Economics

TutorMe

Question:

Suppose there are 2 firms in an industry. Their cost functions are as follows: $$ C_{1}(y_{1}) = (y_{1})^{2}$$ for $$y_{1}≥0 $$ $$C_{2}(y_{2}) = 12y_{2}$$ for $$y_{2}≥0$$ The inverse demand function for the industry is $$p = 100-y_{t}$$ If the firms choose their prices simultaneously (and not sequentially) what will be the equilibrium output for each firm, and the market price?

Jessica B.

Answer:

Let's first identify certain phrases in the problem that are going to indicate what type of problem you’re going to want to set up. First off you’re seeing “2 firms in an industry”. That should tell you “duopoly.” And you can familiarize yourself with that kind of procedure in microeconomics if you haven’t yet. The next thing that you can immediately apply is “equilibrium output” and “market price” – this tells you we are going to be dealing with either a Cournot or Stackelberg model of working out quantities, and not the Bertrand model of the price. The final deciding term is “simultaneously” which tells us we are going to be using the Cournot model of equilibrium. To start this model, you’ll need to find the reaction function of each firm. To do this, you start with their profit functions then take their partial derivative with respect to the y of each firm (y1 or y2 in this case): $$Profit = revenue – cost = price*(quantity – cost)$$ $$Π_{1} = P*(y_{1}) – C_{1}(y_{1})$$ $$= (100-y_{t})(y_{1}) – y_{1}^{2} = (100-(y_{1}+y_{2}))y_{1} – y_{1}^{2}$$ $$= 100y_{1} – y_{1}^{2} – y_{1}y_{2} – y_{1}^{2}$$ $$\frac{∂Π}{∂y1} = 100 – 2y_{1} – y_{2} – 2y_{1} $$ $$100 = y_{2} + 4y_{1}$$ $$Π_{2} = P*(y_{2}) – C_{1}(y_{2})$$ $$ = (100-(y_{1}+y_{2}))y_{2} – 12y_{2}$$ $$= 88y_{2} – y_{2}^{2} – y_{1}y_{2}$$ $$\frac{∂Π}{∂y2} = 88– y_{1} – 2y_{2} $$ $$88= 2y_{2} + y_{1}$$ So now we have these two reaction functions: $$100 = y_{2} + 4y_{1}$$ $$88= 2y_{2} + y_{1}$$ To find the Cournot equilibrium, use either substitution or elimination like you would do in any other system of equations (In this case we'll use substitution) $$88= 2(100-4y_{1} )+ y_{1}$$ $$88= 200-8y_{1}+ y_{1}$$ $$88= 200-7y_{1}$$ $$-112= -7y_{1}$$ $$16= y_{1}$$ Plug this value into the equation for $$y_{2}$$: $$100 = y_{2} + 4(16)$$ $$100 = y_{2} + 64$$ $$36= y_{2}$$ So the equilibrium outputs are $$y_{1}=16$$, $$y_{2}=36$$ $$y_{t}=y_{1}+y_{2}=16+36=52$$ Now we plug that into the original inverse demand equation: $$P=100-y_{t}=100-(52)=48$$

Algebra

TutorMe

Question:

Solve the system of equations: $$x^{2} + 2y = 31$$ $$x + y = 16$$

Jessica B.

Answer:

There are two ways to solve such a system; substitution and elimination. The easiest way to do this particular one is the substitution method since we have the simple equation $$x + y = 16$$ that can be easily solved for x or y. We'll go for solving for y: $$y = 16 - x$$ . Now we can plug this into the other equation: $$x^{2} + 2(16-x) = 31$$ Now expand and get all the terms on one side (we can anticipate having to do some factoring with this quadratic equation) : $$x^{2} + 32 - 2x - 31 = 0$$ --> $$x^{2} - 2x + 1 = 0$$ This is a classic $$ax^{2} + bx + c = 0$$ quadratic equation. Now the best way to factor this out is using the box method. Make a 2x2 box. Place $$x^{2}$$ in the top left corner and 2 in the bottom right: __________ | | | | $$x^{2}$$ | | __________ | | | | | $$2$$ | __________ Now to fill in the other two squares, figure out two numbers that add up to equal the coefficient b and multiply to equal the coefficients a*c. then place an "x" next to them. In this example, $$b = -2$$ and $$a*c = 1$$ . now another systems of equations could solve this but....let's just save time by reasoning through it. What two numbers multiply to equal 1? 1 and 1. But they can't add up to a negative number! Oh, wait... -1 and -1 do, and they multiply to make positive 1! So the two terms are $$-1x$$ and $$-1x$$. Plug them in: __________ | | | | $$ x^{2}$$ | $$-x$$ | Now that we've got the box filled, we are going to fill in the outside on the __________ top and left. This is easier to show than explain: | | | | $$-x$$ | $$2$$ | __________ $$x$$ $$-1$$ __ ^____ ^ __ | ^ | ^ | $$x$$ < | $$x^{2}$$< | $$-x$$ | __^________ | ^ | ^ | -$$1$$ < |$$-x$$< < < $$2$$ | __________ if you can gather from the ^ and < arrows, I have pulled the greatest common factor from each row and column. on the top it is $$(x - 1)$$ and on the left is $$(x - 1)$$. So now we have factored the equation $$x^{2} - 2x + 1 = 0$$ into $$(x - 1)(x - 1) = 0$$. We will set each factor equal to 0 (or just one of the factors, since they're the same: $$x - 1 = 0$$ ; $$x = 1$$ This is the only value of x. So now we can plug it into the equation x + y = 16 and we get $$x = 1$$ ; $$y = 15$$ . Congrats! you've solved it. Plug the values into the other equation to double check $$(1)^{2} + 2(15) = 1 + 30 = 31$$ . :)

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