Economic Principles II : Math Memo Prof. Séverine Toussaert - Section 2.002 July 8, 2013 Computing a Rate of Change Suppose we wanted to know by how much a variable increased relative to some reference point. One such variable could be prices in year 2012 compared to prices in the previous year. Assume P 11 = 20 and P 12 = 30. By how much did the price increase? The rate of change is given by : P 12 P 11 30 20 = = 10 P 11 20 20 =0.5 So the price increased by one-half. If you wanted to know in terms of percentage change, you just need to multiply by 100 : Thus the price increased by 50%. P 12 P 11 P 11 100 = 50 If on the other hand P 11 = 30 and P 12 = 20, then we would have a percentage change in price of: P 12 P 11 20 30 100 = 100 = 10 100 = 33 P 11 30 30 Thus the price decreased by 33%. Notice that we obtain a different percentage change depending on the direction of change; the reason is that the reference price is different if we look at an increase from 20 to 30 compared to a decrease from 30 to 20. To obtain a percentage change of 50% in the second case, the price would have to decrease from 30 to 15. Solving linear equations and non-linear equations Alinearequationisanequationoftheform: Y = a + bx 1
where X and Y are variables of interest and a, b are constants. Linear equations you will see many times during the class are : and Q S = a + bp Q D = c where b>0 and d>0. Those equations describe linear supply and demand curves respectively. As you might have expected if you know the basic laws of supply and demand, the supply curve describes a positive relation between prices and quantities while the demand curve describes a negative relation. Suppose we were interested in knowing what are the equilibrium price and quantity (Q,P ) that is, the price and quantity that prevail when supply and demand are equal. How would you do? You would just solve for (Q,P ) by imposing : dp Q S = Q D = Q, a + bp = c dp, bp + dp =, (b + d)p =, P = b + d Finally plugging in the equilibrium value of P in one of the 2 equations, we obtain Q = Q S (P )= a + b. As a sanity check, you should always verify that you obtain the same value for Q if you use the other equation : Q = Q D (P )=c d.sincethisisnotobvious,let sverifyit! a + b c d = a = c + b d = = ab+ad+bb bc+dc dc+ad = ad+bc = ad+bc ( Q = a + bp Whenever you have to solve for a system of linear equations with 2 unknowns like, Q = b dp remember to substitute for one variable in one equation using the other equation (here we did that with Q) and then solve for the remaining unknown (here P ). Think about grouping all the terms where the unknown appears on one side and the constant terms on the side. Once you have solved for this unknown, simply plug your solution in one of the equations to solve for the other unknown. 2
Non-linear equations For linear equations, X and Y are related in an additive way. We will also encounter from time to time equations of the form : c = XY Here X and Y are related in a multiplicative way. For instance, suppose a firm wanted to produce output Y using a production function F (K, L) =Y where F (K, L) = 1 2KL where K and L refer to the capital and labor inputs. If we wanted to know what is the quantity of capital necessary to produce a level of output Y = 10 for every quantity of labor, how would you express the relationship? Well, we have : 10 = 1 2 KL, 20 = KL, K = 20 L The relationship says that when the quantity of labor used increases, less capital is necessary to produce an output of 10. For instance, when L =1, we need K = 20 but when L = 10, only K =2is needed. Straight lines Every straight line has an equation of the following form : where : Y = a + bx Y is called the dependent variable and X the independent variable. a is the vertical intercept and marks the point where the graph of this equation crosses the vertical axis. Notice that when X =0,wehaveY = a. b is the slope of the line and says how much Y changes every time X changes by one unit. For instance when X = 3, we have Y = a + b 3 = a +3b. Now if X increases by one unit (so that X = 4), weobtainy = a + b 4=a +4b. Thus a unit change in X that is 4X =(4 3) = 1, generatesachangeiny of 4Y =(a +4b) (a +3b) =b. Now if X = 5, we obtain Y = a +5b and you can see that a change 4X = (5 3) = 2 implies a change 4Y =(a +5b) (a +3b) =2b. More generally, a change 4X = n induces a change 4Y = nb so the slope of a straight line is just : b = 4Y 4X 3
What happens as we change b? When b>0, thelineisupwardsloping:asx increases by one unit, Y increases by b units. When b<0, the line is downward sloping : as X increases by one unit, Y decreases by b units. When b =0,Xhas no effect on Y ; the equation becomes Y = a so the line is just a horizontal line. Also notice that as b increases, the line gets steeper. In the limit, when b = 1,the line becomes vertical. What happens as we change a? When a =0, the line goes through the origin (if X =0,thenY = 0) When a>0, thegraphofthelinecrossestheverticalaxisabove0(y = a>0). When a<0, the graph of the line crosses the vertical axis below 0 (Y = a<0). As a increases (decreases), the line shifts upwards (downwards) in a parallel way. Now suppose I ask you to represent the equation Q = 50 5P in the plane (Q, P ). NoticethatIaskto consider P as the dependent variable and Q as the independent variable that is, X := Q and Y := P (While it might not seem natural to put quantities on the X axis and prices on the Y axis, every economic textbook follows this convention). So the first thing you should do is rewrite the equation to make P appear on the LHS i.e. Q = 50, 5P = 50 Q, P = 50 1 5 5 Q.Simplifyingweobtain: 5P P = 1 5 Q + 10 Now you are ready to plot the equation! You can do so by taking 2 points (for instance you can take (Q, P )=(0, 10) and (Q, P ) = (50, 0) to draw your line). Similarly, suppose I ask you to represent the equation R = P C C + P F F in the plane (F, C) so that X := F and Y := C. AgainyouhavetoisolateC on the LHS to obtain : P C C = R P F F Notice that the slope here is 4C will appear several times during class. 4F = P F P C, C = R P C P F P C F. Remember this relation and how to obtain it because that 4
Curved lines Consider for instance the curved line with equation : Y = b p X + a Again, a is the vertical intercept. But now what is the slope of the curve at a given point (X, Y )? Take X =1so that Y = b p 1+a = b + a. Now what happens as we increase X by one unit so that X =2and Y = b p 2+a?Observethat4X =(2 1) = 1 implies 4Y =(b p 2+a) (b + a) = b p 2 b = b( p 2 1) ' 0.41b. Youmightfeeltemptedtoconcludethattheslopeat(X, Y )=(1,b+ a) is simply 4Y 4X ' 0.41b 1 = 0.41b. Note however that this is only an approximation of the slope at (X, Y )=(1,b+ a) (the true slope is indeed 0.5b). Now suppose we considered only an increase of X from 1 to 1.5 so that 4X = 0.5 and 4Y = (b p 1.5+a) (b + a) =b( p 1.5 1) ' 0.22b. Then the approximation of the slope of Y = b p X + a at (X, Y )=(1,b+ a) would be 4Y 4X ' 0.22b 0.5 =0.44b which is a better approximation than for 4X =1. As we consider smaller and smaller deviations around (X, Y ) =(1,b + a) so that 4X becomes very small, the approximations of the slope at (X, Y )=(1,b+ a) get closer to 0.5b. In the limit (when 4X! 0), weobtaintheslopeofthecurveat(x, Y )=(1,b+ a). This slope corresponds to the slope of the straight line which is tangent to the curve at (X, Y )=(1,b+ a). Notice that I am talking about the slope of the curve at a specific point. Indeed, since we are talking about a curve, its slope changes as we move along the curve. For instance when X =4, you can verify that the slope of the curve at (X, Y )=(4, 2b + a) is equal to 0.125b. More generally as X increases, the slope gets flatter and flatter (when X = 10000, the slope of the curve at (X, Y ) = (10000, 100b + a) =0.005b). Eventually, the slope of the curve when X!1is 0 which corresponds to the slope of the horizontal line which is tangent to the curve at that point. What you need to remember about the slope of a curved line at a given point (X, Y ) is that : It can be approximated locally by a straight line which is tangent to the curve at that point. The straight line that approximates the curved line changes as we move along the curve. One of the consequences of having a curved line is that you can no longer use only 2 points to draw it. Think about the example : 10 = 1 20 2KL, K = L. Here you will need many points to get an idea of the shape of the curve that is behind this equation. Some useful identities (a + b) 2 = a 2 +2ab + b 2 (a b) 2 = a 2 2ab + b 2 a b a c = a b+c 5
ab a c = a b c (ab) c = a c b c p a = a 1/2 ln(a b) =ln(a)+ln(b) ln a b = ln(a) ln(b) ln(1) = 0 Some basics in geometry area of a rectangle = width height area of a triangle = 1 2 base height 6