Section14_6

# Section14_6 - Section 14.6 Tangent planes and differentials...

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Unformatted text preview: (3/23/08) Section 14.6 Tangent planes and differentials Overview: In this section we study linear functions of two variables and equations of tangent planes to the graphs of functions of two variables. Then we discuss differentials with two variables and their use in estimating errors. Topics: • Linear functions of two variables • Level curves of linear functions • Zooming in on level curves of a nonlinear z = f ( x , y ) • Equations of tangent planes • Normal vectors • Differentials and error estimates Linear functions of two variables Recall that a function y = f ( x ) of one variable is linear if its graph in an xy-plane is a line, and that in this case its derivative is constant and equals the slope of the line. In studying such functions, we frequently use either the slope-intercept equation y = mx + b for the line, where m is the slope and b the y-intercept of the line (Figure 1), or the point-slope equation y = y + m ( x- x ) where m is the slope and ( x ,y ) a point on the line (Figure 2). x y b Slope = m x y ( x ,y ) Slope = m The point-slope equation The slope-intercept equation y = mx + b y = y + m ( x- x ) FIGURE 1 FIGURE 2 A function z = f ( x,y ) of two variables is linear if its graph in xyz-space is a plane. We found equations of planes in Section 13.5 by using their normal vectors. Here we will need the next theorem, which gives equations for planes in terms of the slopes of their cross sections in the x- and y-directions. 342 Section 14.6, Tangent planes and differentials p. 343 (3/23/08) Theorem 1(a) (The slope-intercept equation of a plane) Suppose that the z-intercept of a plane is b , that the slope of its vertical cross sections in the positive x-direction is m 1 , and that the slope of its vertical cross sections in the positive y-direction is m 2 (Figure 3). Then the plane has the equation, z = m 1 x + m 2 y + b. (1) (b) (The point-slope equation of a plane) Suppose that a plane contains the point ( x ,y ,z ) , that the slope of its vertical cross sections in the positive x-direction is m 1 , and that the slope of its vertical cross sections in the positive y-direction is m 2 (Figure 4). Then the plane has the equation, z = z + m 1 ( x- x ) + m 2 ( y- y ) . (2) The slope-intercept equation The point-slope equation FIGURE 3 FIGURE 4 Proof of part (a): We consider the case where the z-intercept b , the slopes m 1 and m 2 , and x and y are all positive, as in Figure 3. To obtain an equation for the plane we need to find the z-coordinate of the point R with xy-coordinates ( x,y ). The z-coordinate of the point P is b . The horizontal run from P to Q is x and the line segment PQ , which is on a vertical cross section of the plane in the x-direction, has slope m 1 . Consequently the z-coordinate of Q is z = b + m 1 x . The horizontal run from Q to R is y and the line segment QR , which is on a vertical cross section in the y-direction, has slope m 2 , so that the z-coordinate of R is z = b + m 1 x + m 2 y , as stated in...
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Section14_6 - Section 14.6 Tangent planes and differentials...

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