z = f(x, y) = c + mx + ny

Transcription

1 Linear Functions In single-variable calculus, we studied linear equations because the allowed us to approimate complicated curves with a fairl simple linear function. Our goal will be to use linear functions in two dimensions to approimate complicated surfaces. A linear function in one variable could be written in point-slope form. That is, we said that = 0 + m( 0 ). Once we knew a point on the linear and the slope, we had all the information necessar to specif the equation. When dealing with two variables, we have a similar point-slope form. This time, however, we have two slopes: one in the -direction and another in the -direction. Equation of a Linear Function If a linear function (a plane) passes through the point ( 0, 0, z 0 ) and has slope m in the -direction and slope n in the -direction, the equation of the linear function (plane) is given b: z = f(, ) = z 0 + m( 0 ) + n( 0 ) Letting c = z 0 m 0 n 0, we can also write f(, ) in the form z = f(, ) = c + m + n Just as two points uniquel determine the equation of a line in 2-space, three points uniquel determine the equation of a plane in 3-space. Eample 1: Find the equation of the plane passing through the points (1, 2, 3), (1, 4, 1), (2, 2, 5). Notice that the first two points have the same -value. So, we can use them to find the slope of the plane in the -direction. As changes from 2 to 4, the z-value changes from 3 to 1. Thus, the slope in the -direction is given b n = Dz/D = (1 3)/(4 2) = -1. The first and third points share the same -value. Thus, we can use them to find the slope of the plane in the -direction. As changes from 1 to 2, the z-value changes from 3 to 5. Thus, the slope in the -direction is given b m = Dz/D = (5 3)/(2 1) = 2. 1

2 Because the plane passes through the point (1, 2, 3), we have that its equation is given b z = 3 + 2( 1) ( 2) = A graph of the plane appears below. Figure 1: Graph of z = f(, ) = We notice that for a fied value of, the graph slopes upwards (in accordance with the fact that it has a slope of +2 in the -direction) and for a fied value of, the graph slopes downward (in accordance with the fact that it has a slope of -1 in the -direction). Recall when we first introduced functions of two variables, we presented them in terms of a table. There are some ke characteristics that we can look for in a table of values that allow us to determine if we are dealing with a linear function. Tables of Values of Linear Functions A table of values of a linear function satisfies the following two properties: 1. All the columns have the same slope 2. All the rows have the same slope (usuall diff. from the slope of the cols) Eample 2: Show that the following table corresponds to a linear function. Find its equation

3 Notice that for a fied -value, the function increases b 3 for ever one unit increase in. That is, we have that n = Dz/D = 3/1 = 3. Also, for a fied -value, the function decreases b 2 for ever two unit increase in. That is, m = Dz/D = -2/2 = -1. Also, we observe that the function passes through the point (0, 0, 4). Using the second form of the equation of a linear function, we have that the equation of the linear function is given b z = Eample 3: The following table contains values from a linear function. Fill in the blank and determine the equation of the linear function _ Notice that for a fied -value, as increases from 1 to 2 (D = 1), the z-value increases b 2. This tells us that m = Dz/D = 2/1 = 2. From this, we can deduce that the lower-left entr of the table is 16. Notice, also, that for a fied -value, as increases from 2 to 4 (D = 2), the z-value increases b 10. This tells us that the n = Dz/D = 10/2 = 5. From this, we see that the middle entr of the top row will be = 19 and the middle entr of the bottom row will be = 21. Thus, the completed table is as follows: We have found that m = 2 and n = 5. Using the fact that the function passes through the point (1, 2, 14), we can use the first form of the equation of a linear function to see that the function is given b z = f(, ) = ( 1) + 5( 2) =

4 In the previous section, we gave an eample where the contour diagram of a linear function was specified and the reader was asked to graph the equation. All linear functions have contour diagrams that share some ke properties. We list them below. Contour Diagrams of Linear Functions Contour diagrams of linear functions satisf the following two properties: 1. The level curves are parallel straight lines 2. Each of the level curves are evenl spaced Eample 4: Below are three contour diagrams. Which one(s) correspond to a linear function? For those that do not, what propert is violated? I II III Graph I cannot correspond to a linear function since the level curves are not parallel straight lines. The contours, however, are evenl spaced. This actuall corresponds to the 2 2 graph of the function f (, ). Graph II does correspond to a linear function. The level curves are parallel straight lines and the contours are evenl spaced. Graph III cannot correspond to a linear function since the level curves are not evenl spaced. The contours are parallel straight lines, though. 4

5 It is also possible to determine the equation of the linear function b eamining the contour diagram. Eample 5: Find the equation of the linear function whose contour diagram is below. The equation of the linear function is given b z = z 0 + m( 0 ) + n( 0 ). Notice that when = 0 and = 0, we have that z = 0. So, we have the point ( 0, 0, z 0 ) = (0, 0, 0) on the equation. As we move out one unit in the positive -direction, the function value decreases b 2. This implies that n = -2. As we move out two units in the -direction, the function value increases b 2. This implies that m = 2/2 = 1. So, the equation of our linear function is given b z = 0 + 1( 0) 2( 0) = 2. 5