Units and Dimensions (Solutions)

Units and Dimensions We already know that a graph lacking units is not very useful. A time of t = 2 is not very useful, as we do not know if that is 2 years, hours, or nanoseconds. Thus, when dealing with physical quantities it is very important to include units in our equations, as the units give meaning to our measurements. Unfortunately, the system of units used in the United States is based on the archaic imperial system of units developed in the United Kingdom (which was later abandoned in favor of the SI system), we are forced to deal excessively with conversion factors between units. Example 1 Convert 1 mile into centimeters. Solution In order to complete this conversion, we need to remember and utilize a number of conversion factors between units. Recall that 2.54 cm ≈ 1 in, 12 in = 1 ft, and 5280 ft = 1 mile. If we divide each of these equations by the unit on the right-hand side, we will have three equations that all equal 1. Since 1 is the multiplicative identity, we can use these equations as conversion factors. We can begin with 1 mile, and multiply it by all three factors to obtain a result in centimeters. in cm ft · 12 · 2.54 mile ft in Notice that all of the units cancel out leaving only centimeters, so multiplying the coefficients we get the result 1 mile ≈ 1 mile · 5280 1 mile ≈ 160, 934 cm Example 2 Convert 1 C to ml given that 2 C = 1 pint and 1 pint = .473176 liters. Solution We need to work out this example as above, and multiply 1 C by a number of conversion factors to express it in milliliters. 1C≈1C· 1 pint ml liter · .473176 · 103 2 C pint liter 1 C = 236.588 ml Dimensions are fundamental physical quantities: length, time, mass, force, etc. Units are used to measure these dimensions. For instance, miles and centimeters both measure length, whereas Newtons and pounds both measure force. We can use conversion factors to convert between different units that both measure the same quantity, but we cannot devise general factors to convert between units measuring different dimensions. The reason for this is that the relationships between dimensions are situational. If we consider a triangle and a square, the dimension of area has a different relationship with the length of each figure’s sides. Thus, to convert between different dimensions, we must find a conversion that is specific to the situation we are dealing with. Example 3 Suppose the circumference of a circle is a cm. Find the area of the circle. Solution First, we must recall the relationship that the circumference of a circle a = 2πr, where r is the radius of the circle. Next, recall that the area A = πr2 . Thus a cm 2 a2 ) = cm2 2π 4π Example 4 Suppose a cylinder has a radius of 2 cm, and a height of 3 cm. Find the volume of the cylinder. A = πr2 = π · ( Solution The volume of a cylinder V = A · h where A is the cross-sectional area of the cylinder, and h is the height of the cylinder. From before, we know that A = πr2 . Thus, V = A · h = πr2 · h = π(2 cm)2 · (3 cm) = 12π cm3 Dimensional analysis is an extremely useful tool for scientists to quickly check the validity of their calculations. Dimensional analysis is the process of examining the dimensions in an equation to make sure they are sensible. In example 4, in order to calculate a volume, we needed three lengths. This is dimensionally consistent, as volume = length3 . On the other hand, if we were trying to use an equation to calculate volume where the right-hand side had units of Newton · meter, it would be immediately obvious using dimensional analysis that the equation was incorrect. It is important to note that dimensional consistency does not guarantee a correct relationship, only that inconsistency guarantees a nonsensical one. If a function accepts an input with a specific type of dimension, we can perform a similar sanity check on the values we are inputting in the function. For instance, if we have a function f (t) that calculates population given an input of time, it is nonsensical to choose an input of force to the equation. Nevertheless, if one does blindly chose such an input to the function, the resulting equation will be dimensionally inconsistent. Similarly, it is important to be concious of the order in which compositions are formed, as it is likely one of the two possible compositions will be nonsense. Unlike dimensions, we can easily alter the units of the inputs that a function accepts, simply by using a conversion factor as we have before. The result will be a scaled function, which can be a simple way to make the results of an experiment easier to understand. Example 5 Suppose the function b(t) = 2 · t where b is measured in millions of bacteria, and t is measured in hours. Rewrite b as a function of thousands of bacteria, and rewrite both of those functions as functions of minutes rather than hours. Solution Since we are rewriting the function in terms of thousands of bacteria rather than millions, we should choose a new letter B to represent the new function. To find B, we multiply b by an appropriate conversion factor. B(t) = 2 · t million bacteria 1000 thousand bacteria · hour million bacteria thousand bacteria hour Now if we want to modify these two functions, we should define two new functions, in terms of m rather than t, to represent that we are modifying the independent variable, rather than the dependent variable. Once again, we need only to multiply by an appropriate conversion factor. B(t) = 2000 · t b(m) = 2 · t million bacteria 1 hour · hour 60 min b(m) = 1 million bacteria 30 min Similarly, we find B(m) = 100 thousand bacteria 3 min In all of these graphs, the function looks the same, except for that the labels on the graph have changed. Since we are multiplying either the dependent or independent variable by a constant, we are scaling the values on the graph. Now if we make a different type of conversion, such as adding a value to either the dependent or independent variable, it will have a different effect on the graph. Example 6 Suppose you have a very special type of heater, which takes a celcius input temperature from one room, and outputs a celcius temperature that is double the input to the second room TO (TI ) = 2 · TI . Rewrite the function in terms of a Kelvin input, output, and both input and output. Graph these functions. Solution Recall that a Kelvin temperature is related to a celsius temperature by the equation TK = 273.15 + TC . If we rewrite the function in terms of a Kelvin input (TO (TIK )) we find TO (TIK ) = 2 · (TIK − 273.15) Similarly, if we write the output in terms of Kelvin with a Celcius input (TOK (TI )) we find TOK (TI ) = 273.15 + 2 · TI Finally, if we write both the input and output in terms of Kelvin (TOK (TIK )) we find TOK (TIK ) = 273.15 + 2 · (TIK − 273.15) If we look at the graphs of all of these functions, we see that they look identical except for that the values on the horizontal and vertical axes are shifted either higher or lower. Finally, it is important to note that doubling a Celcius temperature is very different from doubling a Kelvin temperature! 2 · TK = 2 · (273.15 + TC ) = 546.30 + 2 · TC ⇒ 2 · TK 6= 2 · TC As we can see above, the reason that doubling Kelvin is not the same as doubling Celcius is the presence of the additive term to convert between the two. If we were converting between units where the conversion was multiplicative (such as inches to feet), doubling in terms of one unit would be identical to doubling in terms of the other. However, as we have just seen, if the conversion is additive or subtractive, this is most certainly not true. The previous examples illustrate that we can scale a function by multiplying either the dependent or independent variable by a constant, or shift it by adding a constant to the independent or dependent variable. The following tables summarize these processes for a function f (x) and a constant a. Scaling Vertical: a · f (x) Horizontal: f (a · x) Shifting Vertical: f (x) + a Horizontal: f (x + a) |a| > 1 amplification compression in x a>0 shifted up by a shifted left by a |a| < 1 attenuation expansion in x a<0 shifted down by a shifted right by a It is important to be careful when dealing with scaling and shifting of functions, because the way in which the function is changed may not always be intuitive. For instance, if a function f (t) is shifted horizontally by an a > 0, say a = 1, our intuition might say that f (t + 1) should equal the original function shifted to the right. However, if we think about things happening in time, it makes sense that the function should actually be shifted left, because by adding 1 to the current time, we are making things occur sooner than in the original function. In addition to applying this type of reasoning, by inputting a few values into both the original and modified function, one can usually deduce the effects of scaling and shifting on the function.